pre-calculus senior high school teaching guide

Table of Contents

About This Teaching Guide 1

DepEd Curriculum Guide for Precalculus 2

Unit 1: Analytic Geometry (19 one-hour sessions) 6

Lesson 1.1: Introduction to Conic Sections and Circles . . . . . . . . 7

1.1.1: An Overview of Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.1.2: Definition and Equation of a Circle . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.3: More Properties of Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.1.4: Situational Problems Involving Circles. . . . . . . . . . . . . . . . . . . . 12

Lesson 1.2: Parabolas. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.2.1: Definition and Equation of a Parabola . . . . . . . . . . . . . . . . . . . . 19

1.2.2: More Properties of Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.2.3: Situational Problems Involving Parabolas . . . . . . . . . . . . . . . . 27

Lesson 1.3: Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1.3.1: Definition and Equation of an Ellipse . . . . . . . . . . . . . . . . . . . . . 32

1.3.2: More Properties of Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.3.3: Situational Problems Involving Ellipses . . . . . . . . . . . . . . . . . . . 40

Lesson 1.4: Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.4.1: Definition and Equation of a Hyperbola . . . . . . . . . . . . . . . . . . 45

1.4.2: More Properties of Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

1.4.3: Situational Problems Involving Hyperbolas . . . . . . . . . . . . . . . 54

Lesson 1.5: More Problems on Conic Sections . . . . . . . . . . . . . . . . 59

1.5.1: Identifying the Conic Section by Inspection. . . . . . . . . . . . . . . 59

1.5.2: Problems Involving Di↵erent Conic Sections . . . . . . . . . . . . . . 61

iii

Lesson 1.6: Systems of Nonlinear Equations . . . . . . . . . . . . . . . . . . 66

1.6.1: Review of Techniques in Solving Systems of LinearEquations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

1.6.2: Solving Systems of Equations Using Substitution . . . . . . . . . 68

1.6.3: Solving Systems of Equations Using Elimination . . . . . . . . . . 71

1.6.4: Applications of Systems of Nonlinear Equations . . . . . . . . . . 75

Unit 2: Mathematical Induction (10 one-hour sessions) 80

Lesson 2.1: Review of Sequences and Series . . . . . . . . . . . . . . . . . . . 81

Lesson 2.2: Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

2.2.1: Writing and Evaluating Sums in Sigma Notation . . . . . . . . . 86

2.2.2: Properties of Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Lesson 2.3: Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . 95

2.3.1: Proving Summation Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

2.3.2: Proving Divisibility Statements. . . . . . . . . . . . . . . . . . . . . . . . . . . 100?2.3.3: Proving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

Lesson 2.4: The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

2.4.1: Pascal’s Triangle and the Concept of Combination. . . . . . . . 108

2.4.2: The Binomial Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

2.4.3: Terms of a Binomial Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . 114?2.4.4: Approximation and Combination Identities . . . . . . . . . . . . . . . 116

Unit 3: Trigonometry (29 one-hour sessions) 121

Lesson 3.1: Angles in a Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

3.1.1: Angle Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

3.1.2: Coterminal Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

3.1.3: Arc Length and Area of a Sector . . . . . . . . . . . . . . . . . . . . . . . . . 128

Lesson 3.2: Circular Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

3.2.1: Circular Functions on Real Numbers . . . . . . . . . . . . . . . . . . . . . 135

3.2.2: Reference Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

Lesson 3.3: Graphs of Circular Functions and SituationalProblems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

3.3.1: Graphs of y = sin x and y = cosx . . . . . . . . . . . . . . . . . . . . . . . . 144

3.3.2: Graphs of y = a sin bx and y = a cos bx . . . . . . . . . . . . . . . . . . . 146

3.3.3: Graphs of y = a sin b(x� c) + d andy = a cos b(x� c) + d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

3.3.4: Graphs of Cosecant and Secant Functions . . . . . . . . . . . . . . . . 154

3.3.5: Graphs of Tangent and Cotangent Functions . . . . . . . . . . . . . 159

3.3.6: Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

Lesson 3.4: Fundamental Trigonometric Identities. . . . . . . . . . . . .176

3.4.1: Domain of an Expression or Equation . . . . . . . . . . . . . . . . . . . . 176

3.4.2: Identity and Conditional Equation . . . . . . . . . . . . . . . . . . . . . . . 178

3.4.3: The Fundamental Trigonometric Identities . . . . . . . . . . . . . . . 180

3.4.4: Proving Trigonometric Identities . . . . . . . . . . . . . . . . . . . . . . . . . 183

Lesson 3.5: Sum and Di↵erence Identities . . . . . . . . . . . . . . . . . . . . .189

3.5.1: The Cosine Di↵erence and Sum Identities . . . . . . . . . . . . . . . . 189

3.5.2: The Cofunction Identities and the Sine Sum andDi↵erence Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

3.5.3: The Tangent Sum and Di↵erence Identities . . . . . . . . . . . . . . . 195

Lesson 3.6: Double-Angle and Half-Angle Identities . . . . . . . . . . . 199

3.6.1: Double-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

3.6.2: Half-Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

Lesson 3.7: Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . 208

3.7.1: Inverse Sine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

3.7.2: Inverse Cosine Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

3.7.3: Inverse Tangent Function and the Remaining InverseTrigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

Lesson 3.8: Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 233

3.8.1: Solutions of a Trigonometric Equation . . . . . . . . . . . . . . . . . . . . 234

3.8.2: Equations with One Term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

3.8.3: Equations with Two or More Terms . . . . . . . . . . . . . . . . . . . . . . 241

Lesson 3.9: Polar Coordinate System . . . . . . . . . . . . . . . . . . . . . . . . . 251

3.9.1: Polar Coordinates of a Point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252

3.9.2: From Polar to Rectangular, and Vice Versa . . . . . . . . . . . . . . . 257

3.9.3: Basic Polar Graphs and Applications . . . . . . . . . . . . . . . . . . . . . 260

References 273

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Unit 1

Analytic Geometry

https://commons.wikimedia.org/wiki/File%3ASan Juanico Bridge 2.JPG

By Morten Nærbøe (Own work)[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)

or GFDL (http://www.gnu.org/copyleft/fdl.html)],via Wikimedia Commons

Stretching from Samar to Leyte with a total length of 2.16 kilometers, the SanJuanico Bridge has served as one of the main thoroughfares of economic and socialdevelopment in the country since its completion in 1973. Adding picturesquee↵ect on the whole architecture, geometric structures are subtly built to serveother purposes. The arch-shaped support on the main span of the bridge helpsmaximize its strength to withstand mechanical resonance and aeroelastic flutterbrought about by heavy vehicles and passing winds.

https://commons.wikimedia.org/wiki/File%3ASan_Juanico_Bridge_2.JPG

Lesson 1.1. Introduction to Conic Sections and Circles

Time Frame: 4 one-hour sessions

Learning Outcomes of the Lesson

At the end of the lesson, the student is able to:

(1) illustrate the di↵erent types of conic sections: parabola, ellipse, circle, hyper-bola, and degenerate cases;

(2) define a circle;

(3) determine the standard form of equation of a circle;

(4) graph a circle in a rectangular coordinate system; and

(5) solve situational problems involving conic sections (circles).

Lesson Outline

(1) Introduction of the four conic sections, along with the degenerate conics

(2) Definition of a circle

(3) Derivation of the standard equation of a circle

(4) Graphing circles

(5) Solving situational problems involving circles

Introduction

We introduce the conic sections, a particular class of curves which sometimesappear in nature and which have applications in other fields. In this lesson, wediscuss the first of their kind, circles. The other conic sections will be covered inthe next lessons.

1.1.1. An Overview of Conic Sections

We introduce the conic sections (or conics), a particular class of curves whichoftentimes appear in nature and which have applications in other fields. Oneof the first shapes we learned, a circle, is a conic. When you throw a ball, thetrajectory it takes is a parabola. The orbit taken by each planet around the sunis an ellipse. Properties of hyperbolas have been used in the design of certaintelescopes and navigation systems. We will discuss circles in this lesson, leavingparabolas, ellipses, and hyperbolas for subsequent lessons.

• Circle (Figure 1.1) - when the plane is horizontal

• Ellipse (Figure 1.1) - when the (tilted) plane intersects only one cone to forma bounded curve

7

• Parabola (Figure 1.2) - when the plane intersects only one cone to form anunbounded curve

• Hyperbola (Figure 1.3) - when the plane (not necessarily vertical) intersectsboth cones to form two unbounded curves (each called a branch of the hyper-bola)

Figure 1.1 Figure 1.2 Figure 1.3

We can draw these conic sections (also called conics) on a rectangular co-ordinate plane and find their equations. To be able to do this, we will presentequivalent definitions of these conic sections in subsequent sections, and use theseto find the equations.

There are other ways for a plane and the cones to intersect, to form what arereferred to as degenerate conics : a point, one line, and two lines. See Figures 1.4,1.5 and 1.6.


1.1.2. Definition and Equation of a Circle

A circle may also be considered a special kind of ellipse (for the special case whenthe tilted plane is horizontal). For our purposes, we will distinguish betweenthese two conics.

See Figure 1.7, with the point C(3, 1) shown. From the figure, the distanceof A(�2, 1) from C is AC = 5. By the distance formula, the distance of B(6, 5)from C is BC =

p(6� 3)2 + (5� 1)2 = 5. There are other points P such that

PC = 5. The collection of all such points which are 5 units away from C, formsa circle.

8

Figure 1.7 Figure 1.8

Let C be a given point. The set of all points P having the samedistance from C is called a circle. The point C is called the center ofthe circle, and the common distance its radius.

The term radius is both used to refer to a segment from the center C to apoint P on the circle, and the length of this segment.

See Figure 1.8, where a circle is drawn. It has center C(h, k) and radius r > 0.A point P (x, y) is on the circle if and only if PC = r. For any such point then,its coordinates should satisfy the following.

PC = rp

(x� h)2 + (y � k)2 = r

(x� h)2 + (y � k)2 = r2

This is the standard equation of the circle with center C(h, k) and radius r. Ifthe center is the origin, then h = 0 and k = 0. The standard equation is thenx2 + y2 = r2.

Example 1.1.1. In each item, give thestandard equation of the circle satisfy-ing the given conditions.(1) center at the origin, radius 4

(2) center (�4, 3), radiusp7

(3) circle in Figure 1.7

(4) circle A in Figure 1.9

(5) circle B in Figure 1.9

(6) center (5,�6), tangent to the y-axis Figure 1.9

9

(7) center (5,�6), tangent to the x-axis

(8) It has a diameter with endpoints A(�1, 4) and B(4, 2).

Solution. (1) x2 + y2 = 16

(2) (x+ 4)2 + (y � 3)2 = 7

(3) The center is (3, 1) and the radius is 5, so the equation is (x�3)2+(y�1)2 =25.

(4) By inspection, the center is (�2,�1) and the radius is 4. The equation is(x+ 2)2 + (y + 1)2 = 16.

(5) Similarly by inspection, we have (x� 3)2 + (y � 2)2 = 9.

(6) The center is 5 units away from the y-axis, so the radius is r = 5 (you canmake a sketch to see why). The equation is (x� 5)2 + (y + 6)2 = 25.

(7) Similarly, since the center is 6 units away from the x-axis, the equation is(x� 5)2 + (y + 6)2 = 36.

(8) The center C is the midpoint of A and B: C =��1+4

2 , 4+22

�=�32 , 3

�. The

radius is then r = AC =q�

�1� 32

�2+ (4� 3)2 =

q294 . The circle has

equation�x� 3

2

�2+ (y � 3)2 = 29

4 . 2

Seatwork/Homework 1.1.2

Find the standard equation of the circle being described in each item.

(1) Center at the origin, radiusp11 Answer: x2 + y2 = 11

(2) Center (�6, 7), tangent to the y-axis Answer: (x+ 6)2 + (y � 7)2 = 36

(3) It has a diameter with endpoints A(�3, 2) and B(7, 4).

Answer: (x� 2)2 + (y � 3)2 = 26

1.1.3. More Properties of Circles

After expanding, the standard equation✓x� 3

2

◆2

+ (y � 3)2 =29

4

can be rewritten asx2 + y2 � 3x� 6y � 5 = 0,

an equation of the circle in general form.

If the equation of a circle is given in the general form

Ax2 + Ay2 + Cx+Dy + E = 0, A 6= 0,

10

orx2 + y2 + Cx+Dy + E = 0,

we can determine the standard form by completing the square in both variables.

Completing the squareTeaching Notes

Recall thetechnique ofcompleting thesquare. This wasintroduced inGrade 9.

in an expression like x2 + 14x means determiningthe term to be added that will produce a perfect polynomial square. Since thecoe�cient of x2 is already 1, we take half the coe�cient of x and square it, andwe get 49. Indeed, x2 + 14x + 49 = (x + 7)2 is a perfect square. To completethe square in, say, 3x2 + 18x, we factor the coe�cient of x2 from the expression:3(x2 + 6x), then add 9 inside. When completing a square in an equation, anyextra term introduced on one side should also be added to the other side.

Example 1.1.2. Identify the center and radius of the circle with the given equa-tion in each item. Sketch its graph, and indicate the center.

(1) x2 + y2 � 6x = 7

(2) x2 + y2 � 14x+ 2y = �14

(3) 16x2 + 16y2 + 96x� 40y = 315

Solution. The first step is to rewrite each equation in standard form by complet-ing the square in x and in y. From the standard equation, we can determine thecenter and radius.

(1)

x2 � 6x+ y2 = 7

x2 � 6x+ 9 + y2 = 7 + 9

(x� 3)2 + y2 = 16

Center (3, 0), r = 4, Figure 1.10

(2)

x2 � 14x+ y2 + 2y = �14

x2 � 14x+ 49 + y2 + 2y + 1 = �14 + 49 + 1

(x� 7)2 + (y + 1)2 = 36

Center (7,�1), r = 6, Figure 1.11

(3)

16x2 + 96x+ 16y2 � 40y = 315

16(x2 + 6x) + 16

✓y2 � 5

2y

◆= 315

16(x2 + 6x+ 9) + 16

✓y2 � 5

2y +

25

16

◆= 315 + 16(9) + 16

✓25

16

◆

11

16(x+ 3)2 + 16

✓y � 5

4

◆2

= 484

Teaching Notes

A common mistakecommitted by

students is to add 9

and 2516 only. They

often forget the

multiplier outside

the parenthesis.

(x+ 3)2 +

✓y � 5

4

◆2

=484

16=

121

4=

✓11

2

◆2

Center��3, 54

�, r = 5.5, Figure 1.12. 2


In the standard equation (x � h)2 + (y � k)2 = r2, both the two squaredterms on the left side have coe�cient 1. This is the reason why in the precedingexample, we divided by 16 at the last equation.


Identify the center and radius of the circle with the given equation in each item.Sketch its graph, and indicate the center.

(1) x2 + y2 � 5x+ 4y = 46

Answer: center�52 ,�2

�, radius 15

2 = 7.5, Figure 1.13

(2) 4x2 + 4y2 + 40x� 32y = 5

Answer: center (�5, 4), radius 132 = 6.5, Figure 1.14


1.1.4. Situational Problems Involving Circles

We now consider some situational problems involving circles.

12

?Example 1.1.3. A street with two lanes, each 10 ft wide, goes through asemicircular tunnel with radius 12 ft. How high is the tunnel at the edge of eachlane? Round o↵ to 2 decimal places.

Figure 1.15

Solution. We draw a coordinate system with origin at the middle of the highway,as shown in Figure 1.15. Because of the given radius, the tunnel’s boundary ison the circle x2 + y2 = 122. Point P is the point on the arc just above the edgeof a lane, so its x-coordinate is 10. We need its y-coordinate. We then solve102 + y2 = 122 for y > 0, giving us y = 2

p11 ⇡ 6.63 ft. 2

Example 1.1.4. A piece of a broken plate was dug up in an archaeological site.It was put on top of a grid, as shown in Figure 1.16, with the arc of the platepassing through A(�7, 0), B(1, 4) and C(7, 2). Find its center, and the standardequation of the circle describing the boundary of the plate.

Teaching Notes

A perpendicularbisector of asegment is the linethat passesthrough themidpoint of thesegment and isperpendicular tothe segment.


Solution. We first determine the center. It is the intersection of the perpendicularbisectors of AB and BC (see Figure 1.17). Recall that, in a circle, the perpen-dicular bisector of any chord passes through the center. Since the midpoint M

13

of AB is��7+1

2 , 0+42

�= (�3, 2), and m

AB

= 4�01+7 = 1

2 , the perpendicular bisectorof AB has equation y � 2 = �2(x+ 3), or equivalently, y = �2x� 4.

Since the midpoint N of BC is�1+72 , 4+2

2

�= (4, 3), and m

BC

= 2�47�1 = �1

3 ,the perpendicular bisector of BC has equation y � 3 = 3(x� 4), or equivalently,y = 3x� 9.

The intersection of the two lines y = 2x � 4 and y = 3x � 9 is (1,�6) (bysolving a system of linear equations). We can take the radius as the distance ofthis point from any of A, B or C (it’s most convenient to use B in this case). Wethen get r = 10. The standard equation is thus (x� 1)2 + (y + 6)2 = 100. 2


?1. A single-lane street 10 ft wide goes through a semicircular tunnel with radius9 ft. How high is the tunnel at the edge of each lane? Round o↵ to 2 decimalplaces. Answer: 7.48 ft

2. An archeologist found the remains of an ancient wheel, which she then placedon a grid. If an arc of the wheel passes through A(�7, 0), B(�3, 4) and C(7, 0),locate the center of the wheel, and the standard equation of the circle definingits boundary. Answer: (0,�3), x2 + (y + 3)2 = 58

Exercises 1.1

1. Identify the center and radius of the circle with the given equation in eachitem. Sketch its graph, and indicate the center.

(a) x2 + y2 = 49 Answer: center (0, 0), r = 7

(b) 4x2 + 4y2 = 25 Answer: center (0, 0), r = 52

(c)�x� 7

4

�2+�y + 3

4

�2= 169

16 Answer: center�74 ,�

34

�, r = 13

4

(d) x2 + y2 � 12x� 10y = �12 Answer: center (6, 5), r = 7

(e) x2 + y2 + 8x� 9y = 6 Answer: center (�4, 4.5), r = 132

(f) x2 + y2 + 10x+ 12y = �12 Answer: center (�5,�6), r = 7

(g) 2x2 + 2y2 � 14x+ 18y = 7 Answer: center (3.5,�4.5), r = 6

(h) 4x2 + 4y2 � 20x+ 40y = �5 Answer: center (2.5,�5), r =p30

(i) 9x2 + 9y2 + 42x+ 84y + 65 = 0 Answer: center��7

3 ,�143

�, r = 2

p5

14

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

2. FindTeaching Notes

To determine theequation of acircle, we just needto determine thecenter and theradius.

the standard equation of the circle which satisfies the given conditions.

(a) center at the origin, radius 2p2 Answer: x2 + y2 = 8

(b) center at (15,�20), radius 9 Answer: (x� 15)2 + (y + 20)2 = 81

(c) center at (5, 6), through (9, 4) Answer: (x� 5)2 + (y � 6)2 = 20

Solution. The radius is the distance from the center to (9, 4):p

(5� 9)2 + (6� 4)2 =p20.

(d) center at (�2, 3), tangent to the x-axis

Answer: (x+ 2)2 + (y � 3)2 = 9

(e) center at (�2, 3), tangent to the y-axis

Answer: (x+ 2)2 + (y � 3)2 = 4

(f) center at (�2, 3), tangent to the line y = 8

Answer: (x+ 2)2 + (y � 3)2 = 25

15

Solution. We need to determine the radius. This is best done bysketching the center and line, to see that the center (�2, 3) is 5 unitsaway from the nearest point on the line, (�2, 8) (which is the point oftangency).

(g) center at (�2, 3), tangent to the line x = �10

Answer: (x+ 2)2 + (y � 3)2 = 64

(h) center in the third quadrant, tangent to both the x-axis and y-axis,radius 7 Answer: (x+ 7)2 + (y + 7)2 = 49

(i) a diameter with endpoints (�9, 2) and (15, 12)

Answer: (x� 3)2 + (y � 7)2 = 169

(j) concentric with x2 + y2 + 2x� 4y = 5, radius is 7

Answer: (x+ 1)2 + (y � 2)2 = 49

Solution. Two circles are said to be concentric if they have the samecenter. The standard equation of the given circle is(x+ 1)2 + (y � 2)2 = 10. Thus, the circle we’re looking for has center(�1, 2) and radius 7.

(k) concentric with x2 + y2 � 8x� 10y = �16 and 4 times the area

Answer: (x� 4)2 + (y � 5)2 = 100

Solution. The given circle has standard equation

(x� 4)2 + (y � 5)2 = 52.

Its radius is 5, so its area is 25⇡ sq. units. The circle we are lookingfor should have area 100⇡ sq. units, so its radius is 10.

(l) concentric with x2 + y2 � 10x � 6y = �2, same radius asx2 + y2 � 14x+ 6y = �33 Answer: (x� 5)2 + (y � 3)2 = 25

(m) center at C(3, 4), tangent to the line y = 13x� 1

3

Answer: (x� 3)2 + (y � 4)2 = 10

Solution. (A sketch will greatly help in understanding the argument.)If P is the point of tangency,

Teaching Notes

The radius drawnto a point on the

circle isperpendicular to

the line tangent tothe circle at that

point.

then line CP is perpendicular to the giventangent line. Since the tangent line has slope 1

3 , line CP has slope �3.Because it passes through C, line CP has equation y� 4 = �3(x� 3),or y = �3x + 13. Solving the system {y = 1

3x � 13 , y = �3x + 13}

yields x = 4 and y = 1, the coordinates of P . The radius is thenCP =

p(3� 4)2 + (4� 1)2 =

p10.

(n) center at (�4, 3), tangent to the line y = �4x� 30

Answer: (x+ 4)2 + (y � 3)2 = 17

16

Solution. (Similar to the previous problem) Let P be the point oftangency, so line CP is perpendicular to the tangent line. The tangentline has slope �4, so line CP has slope 1

4 . Line CP passes through C,so it has equation y� 3 = 1

4(x+4), or y = 14x+4. Solving the system

{y = �4x� 30, y = 14x+4} yields x = �8 and y = 2, the coordinates

of P . The radius is then CP =p(�4 + 8)2 + (3� 2)2 =

p17.

?3. A seismological station is located at (0,�3), 3 km away from a straightshoreline where the x-axis runs through. The epicenter of an earthquakewas determined to be 6 km away from the station.

(a) Find the equation of the curve that contains the possible location ofthe epicenter. Answer: x2 + (y + 3)2 = 62

(b) If furthermore, the epicenter was determined to be 2 km away fromthe shore, find its possible coordinates (rounded o↵ to two decimalplaces). Answer: (±3.32, 2)

Solution. Since the epicenter is 6 units away from (0,�3), it could be anyof the points of a circle with center (0,�3) and radius 6. The equation isthen x2 + (y+3)2 = 62. Next, we solve this equation for x if y = 2, and weget x2 = 62 � (2 + 3)2 = 11, and so x = ±

p11 ⇡ ±3.32.

4. A ferris wheel is elevated 1 m above ground. When a car reaches the highestpoint on the ferris wheel, its altitude from ground level is 31 m. How faraway from the center, horizontally, is the car when it is at an altitude of25 m? Answer: 12 m

Solution. The ferris wheel, as shownbelow, is drawn 1 unit above the x-axis (ground level), center on the y-axis, and highest point at y = 31.The diameter is thus 30, and the ra-dius 15. We locate the center at(0, 16), and write the equation of thecircle as x2 + (y � 16)2 = 152.

If y = 25, we have x2 + (25� 16)2 =152, so x2 = 152 � 92 = 144, andx = ±12. (Clearly, there are twopoints on the ferris wheel at an altitude of 25 m.) Thus, the car is 12 maway horizontally from the center.

?5. A window is to be constructed as shown, with its upper boundary the arcof a circle having radius 4 ft and center at the midpoint of base AD. If the

17

vertical side is to be 34 as long as the base, find the dimensions (vertical side

and base) of this window. Round o↵ your final answer to 1 decimal place.

Answer: base 4.44 ft, side 3.33 ft

Solution. We put two lines corresponding to the x-axis and y-axis, as shown,with the origin coinciding with the midpoint of the window’s base. Thisorigin is the center of the circle containing the arc. The equation of the circleis then x2 + y2 = 16. Let n be length of the base AD, so the side AD haslength 3

4n. Point B then has coordinates�n

2 ,3n4

�. Therefore,

�n

2

�2+�3n4

�2=

16. Solving this for n > 0 yields n = 16p13. The base is then n ⇡ 4.44 ft and

the side 34n ⇡ 3.33 ft.

4

Lesson 1.2. Parabolas




(1) define a parabola;

(2) determine the standard form of equation of a parabola;

(3) graph a parabola in a rectangular coordinate system; and

(4) solve situational problems involving conic sections (parabolas).

Lesson Outline

(1) Definition of a parabola

(2) Derivation of the standard equation of a parabola

18

(3) Graphing parabolas

(4) Solving situational problems involving parabolas

Introduction

A parabola is one of the conic sections. We have already seen parabolas whichopen upward or downward, as graphs of quadratic functions. Here, we will seeparabolas opening to the left or right. Applications of parabolas are presentedat the end.

1.2.1. Definition and Equation of a Parabola

Consider the point F (0, 2) and the line ` having equation y = �2, as shown inFigure 1.18. What are the distances of A(4, 2) from F and from `? (The latteris taken as the distance of A from A

`

, the point on ` closest to A). How aboutthe distances of B(�8, 8) from F and from ` (from B

`

)?

AF = 4 and AA`

= 4

BF =p

(�8� 0)2 + (8� 2)2 = 10 and BB`

= 10

There are other points P such that PF = PP`

(where P`

is the closest point online `). The collection of all such points forms a shape called a parabola.


Let F be a given point, and ` a given line not containing F . The set ofall points P such that its distances from F and from ` are the same, iscalled a parabola. The point F is its focus and the line ` its directrix.

Consider a parabola with focus F (0, c) and directrix ` having equation x = �c.See Figure 1.19. The focus and directrix are c units above and below, respectively,the origin. Let P (x, y) be a point on the parabola so PF = PP

`

, where P`

is the

19

point on ` closest to P . The point P has to be on the same side of the directrixas the focus (if P was below, it would be closer to ` than it is from F ).

PF = PP`p

x2 + (y � c)2 = y � (�c) = y + c

x2 + y2 � 2cy + c2 = y2 + 2cy + c2

x2 = 4cy

The vertex V is the point midway between the focus and the directrix. Thisequation, x2 = 4cy, is then the standard equation of a parabola opening upwardwith vertex V (0, 0).

Suppose the focus is F (0,�c) and the directrix is y = c. In this case, apoint P on the resulting parabola would be below the directrix (just like thefocus). Instead of opening upward, it will open downward. Consequently, PF =p

x2 + (y + c)2 and PP`

= c � y (you may draw a version of Figure 1.19 forthis case). Computations similar to the one done above will lead to the equationx2 = �4cy.

We collect here the features of the graph of a parabola with standard equationx2 = 4cy or x2 = �4cy, where c > 0.

(1) vertex : origin V (0, 0)

• If the parabola opens upward, the vertex is the lowest point. If theparabola opens downward, the vertex is the highest point.

(2) directrix : the line y = �c or y = c

• The directrix is c units below or above the vertex.

(3) focus : F (0, c) or F (0,�c)

• The focus is c units above or below the vertex.

• Any point on the parabola has the same distance from the focus as ithas from the directrix.

(4) axis of symmetry : x = 0 (the y-axis)

20

• This line divides the parabola into two parts which are mirror imagesof each other.

Example 1.2.1. Determine the focus and directrix of the parabola with thegiven equation. Sketch the graph, and indicate the focus, directrix, vertex, andaxis of symmetry.(1) x2 = 12y (2) x2 = �6y

Solution. (1) The vertex is V (0, 0) and the parabola opens upward. From 4c =12, c = 3. The focus, c = 3 units above the vertex, is F (0, 3). The directrix,3 units below the vertex, is y = �3. The axis of symmetry is x = 0.

(2) The vertex is V (0, 0) and the parabola opens downward. From 4c = 6, c = 32 .

The focus, c = 32 units below the vertex, is F

�0,�3

2

�. The directrix, 3

2 unitsabove the vertex, is y = 3

2 . The axis of symmetry is x = 0.

Example 1.2.2. What is the standard equation of the parabola in Figure 1.18?

Solution. From the figure, we deduce that c = 2. The equation is thus x2 =8y. 2

21


1. Give the focus and directrix of the parabola with equation x2 = 10y. Sketchthe graph, and indicate the focus, directrix, vertex, and axis of symmetry.Answer: focus

�0, 52

�, directrix y = �5

2

2. Find the standard equation of the parabola with focus F (0,�3.5) and directrixy = 3.5. Answer: x2 = �14y

1.2.2. More Properties of Parabolas

The parabolas we considered so far are “vertical” and have their vertices at theorigin. Some parabolas open instead horizontally (to the left or right), and somehave vertices not at the origin. Their standard equations and properties are givenin the box. The corresponding computations are more involved, but are similarto the one above, and so are not shown anymore.

In all four cases below, we assume that c > 0. The vertex is V (h, k), and itlies between the focus F and the directrix `. The focus F is c units away fromthe vertex V , and the directrix is c units away from the vertex. Recall that, forany point on the parabola, its distance from the focus is the same as its distancefrom the directrix.

22

(x� h)2 = 4c(y � k) (y � k)2 = 4c(x� h)

(x� h)2 = �4c(y � k) (y � k)2 = �4c(x� h)

directrix `: horizontal directrix `: vertical

axis of symmetry: x=h, vertical axis of symmetry: y=k, horizontal

The following observations are worth noting.

• The equations are in terms of x � h and y � k: the vertex coordinates aresubtracted from the corresponding variable. Thus, replacing both h and kwith 0 would yield the case where the vertex is the origin. For instance, thisreplacement applied to (x�h)2 = 4c(y�k) (parabola opening upward) wouldyield x2 = 4cy, the first standard equation we encountered (parabola openingupward, vertex at the origin).

• If the x-part is squared, the parabola is “vertical”; if the y-part is squared,the parabola is “horizontal.” In a horizontal parabola, the focus is on the leftor right of the vertex, and the directrix is vertical.

• If the coe�cient of the linear (non-squared) part is positive, the parabolaopens upward or to the right; if negative, downward or to the left.

23

Example 1.2.3. Figure 1.20 shows the graph of parabola, with only its focusand vertex indicated. Find its standard equation.

Teaching Notes

In finding theequation of a

parabola, we justneed to determinethe vertex and the

value of c.

What is its directrix and itsaxis of symmetry?

Solution. The vertex is V (5,�4) and the focus is F (3,�4). From these, wededuce the following: h = 5, k = �4, c = 2 (the distance of the focus from thevertex). Since the parabola opens to the left, we use the template (y � k)2 =�4c(x� h). Our equation is

(y + 4)2 = �8(x� 5).

Its directrix is c = 2 units to the right of V , which is x = 7. Its axis is thehorizontal line through V : y = �4.

Figure 1.20

The standard equation (y+ 4)2 = �8(x� 5) from the preceding example canbe rewritten as y2 + 8x + 8y � 24 = 0, an equation of the parabola in generalform.

If the equation is given in the general form Ax2+Cx+Dy+E = 0 (A and Care nonzero) or By2+Cx+Dy+E = 0 (B and C are nonzero), we can determinethe standard form by completing the square in both variables.

Example 1.2.4. Determine the vertex, focus, directrix, and axis of symmetryof the parabola with the given equation. Sketch the parabola, and include thesepoints and lines.

24

(1) y2 � 5x+ 12y = �16

(2) 5x2 + 30x+ 24y = 51

Solution. (1) We complete the square on y, and move x to the other side.

y2 + 12y = 5x� 16

y2 + 12y + 36 = 5x� 16 + 36 = 5x+ 20

(y + 6)2 = 5(x+ 4)

The parabola opens to the right. It has vertex V (�4,�6). From 4c = 5, weget c = 5

4 = 1.25. The focus is c = 1.25 units to the right of V : F (�2.75,�6).The (vertical) directrix is c = 1.25 units to the left of V : x = �5.25. The(horizontal) axis is through V : y = �6.

(2) We complete the square on x, and move y to the other side.

5x2 + 30x = �24y + 51

5(x2 + 6x+ 9) = �24y + 51 + 5(9)

5(x+ 3)2 = �24y + 96 = �24(y � 4)

(x+ 3)2 = �24

5(y � 4)

In the last line, we divided by 5 for the squared part not to have any coe�-cient. The parabola opens downward. It has vertex V (�3, 4).

25

From 4c = 245 , we get c = 6

5 = 1.2. The focus is c = 1.2 units below V :F (�3, 2.8). The (horizontal) directrix is c = 1.2 units above V : y = 5.2. The(vertical) axis is through V : x = �3.

Example 1.2.5. A parabola has focus F (7, 9) and directrix y = 3. Find itsstandard equation.

Solution. The directrix is horizontal, and the focus is above it. The parabolathen opens upward and its standard equation has the form (x� h)2 = 4c(y� k).Since the distance from the focus to the directrix is 2c = 9 � 3 = 6, then c = 3.Thus, the vertex is V (7, 6), the point 3 units below F . The standard equation isthen (x� 7)2 = 12(y � 6). 2


1. Determine the vertex, focus, directrix, and axis of symmetry of the parabolawith equation x2� 6x+5y = �34. Sketch the graph, and include these pointsand lines.

Answer: vertex (3,�5), focus (3,�6.25), directrix y = �3.75, axis x = 3

26

2. A parabola has focus F (�2,�5) and directrix x = 6. Find the standardequation of the parabola. Answer: (y + 5)2 = �16(x� 2)

1.2.3. Situational Problems Involving Parabolas

We now solve some situational problems involving parabolas.

Example 1.2.6. A satellite dish has a shape called a paraboloid, where eachcross-section is a parabola. Since radio signals (parallel to the axis) will bounceo↵ the surface of the dish to the focus, the receiver should be placed at the focus.How far should the receiver be from the vertex, if the dish is 12 ft across, and 4.5ft deep at the vertex?

27

Solution. The second figure above shows a cross-section of the satellite dish drawnon a rectangular coordinate system, with the vertex at the origin. From theproblem, we deduce that (6, 4.5) is a point on the parabola. We need the distanceof the focus from the vertex, i.e., the value of c in x2 = 4cy.

x2 = 4cy

62 = 4c(4.5)

c =62

4 · 4.5 = 2

Thus, the receiver should be 2 ft away from the vertex. 2

Example 1.2.7. The cable of a suspension bridge hangs in the shape of aparabola. The towers supporting the cable are 400 ft apart and 150 ft high.If the cable, at its lowest, is 30 ft above the bridge at its midpoint, how high isthe cable 50 ft away (horizontally) from either tower?

Solution. Refer to the figure above, where the parabolic cable is drawn withits vertex on the y-axis 30 ft above the origin. We may write its equation as(x � 0)2 = a(y � 30); since we don’t need the focal distance, we use the simplervariable a in place of 4c. Since the towers are 150 ft high and 400 ft apart, wededuce from the figure that (200, 150) is a point on the parabola.

x2 = a(y � 30)

2002 = a(150� 30)

a =2002

120=

1000

3

The parabola has equation x2 = 10003 (y � 30), or equivalently,

y = 0.003x2 + 30. For the two points on the parabola 50 ft away from thetowers, x = 150 or x = �150. If x = 150, then

y = 0.003(1502) + 30 = 97.5.

28

Thus, the cable is 97.5 ft high 50 ft away from either tower. (As expected, weget the same answer from x = �150.) 2


?1. A satellite dish in the shape of a paraboloid is 10 ft across, and 4 ft deep atits vertex. How far is the receiver from the vertex, if it is placed at the focus?Round o↵ your answer to 2 decimal places. (Refer to Example 1.2.6.)

Answer: 1.56 ft

Exercises 1.2

1. Determine the vertex, focus, directrix, and axis of symmetry of the parabolawith the given equation. Sketch the graph, and include these points and lines.

(a) x2 = �4y

(b) 3y2 = 24x

(c)�y + 5

2

�2= �5

�x� 9

2

�

(d) x2 + 6x+ 8y = 7

(e) y2 � 12x+ 8y = �40

(f) 16x2 + 72x� 112y = �221

Answer:

Item Vertex Focus Directrix Axis of Symmetry

(a) (0, 0) (0,�1) y = 1 x = 0

(b) (0, 0) (2, 0) x = �2 y = 0

(c) (4.5,�2.5) (3.25,�2.5) x = 5.75 y = �2.5

(d) (�3, 2) (�3, 0) y = 4 x = �3

(e) (2,�4) (5,�4) x = �1 y = �4

(f) (�2.25, 1.25) (�2.25, 3) y = �0.5 x = �2.25

29

(a) (b) (c)

(d) (e) (f)

2. Find the standard equation of the parabola which satisfies the given conditions.Teaching Notes

It is helpful todraw a diagram for

each item.(a) vertex (1,�9), focus (�3,�9) Answer: (y + 9)2 = �16(x� 1)

(b) vertex (�8, 3), directrix x = �10.5 Answer: (y � 3)2 = 10(x+ 8)

(c) vertex (�4, 2), focus (�4,�1) Answer: (x+ 4)2 = �12(y � 2)

(d) focus (7, 11), directrix x = 1 Answer: (y � 11)2 = 12(x� 4)

(e) focus (7, 11), directrix y = 4 Answer: (x� 7)2 = 14(y � 7.5)

(f) vertex (�5,�7), vertical axis of symmetry, through the point P (7, 11)

Answer: (x+ 5)2 = 8(y + 7)

Solution. Since the axis is vertical and P is above the vertex, then theparabola opens upward and has equation of the form (x+5)2 = 4c(y+7).We plug the coordinates of P : (7 + 5)2 = 4c(11 + 7). We then get c = 2.Thus, we have (x+ 5)2 = 8(y + 7).

(g) vertex (�5,�7), horizontal axis of symmetry, through the point P (7, 11)

Answer: (y + 7)2 = 27(x+ 5)

Solution. Since the axis is horizontal and P is to the right of the vertex,then the parabola opens to the right and has equation of the form (y +

30

7)2 = 4c(x + 5). We plug the coordinates of P : (11 + 7)2 = 4c(7 + 5).We then get c = 6.75. Thus, we have (y + 7)2 = 27(x+ 5).

3. A satellite dish shaped like a paraboloid, has diameter 2.4 ft and depth 0.9 ft.If the receiver is placed at the focus, how far should the receiver be from thevertex? Answer: 0.4 ft

4. If the diameter of the satellite dish from the previous problem is doubled, withthe depth kept the same, how far should the receiver be from the vertex?

Answer: 1.6 ft?5. A satellite dish is shaped like a paraboloid, with the receiver placed at the

focus. It is to have a depth of 0.44 m at the vertex, with the receiver placed0.11 m away from the vertex. What should the diameter of the satellite dishbe? Answer: 0.88 m

?6. A flashlight is shaped like a paraboloid, so that if its light bulb is placed atthe focus, the light rays from the bulb will then bounce o↵ the surface in afocused direction that is parallel to the axis. If the paraboloid has a depth of1.8 in and the diameter on its surface is 6 in, how far should the light sourcebe placed from the vertex? Answer: 1.25 in

7. The towers supporting the cable of a suspension bridge are 1200 m apart and170 m above the bridge it supports. Suppose the cable hangs, following theshape of a parabola, with its lowest point 20 m above the bridge. How high isthe cable 120 m away from a tower? Answer: 116 m

4

Lesson 1.3. Ellipses




(1) define an ellipse;

(2) determine the standard form of equation of an ellipse;

(3) graph an ellipse in a rectangular coordinate system; and

(4) solve situational problems involving conic sections (ellipses).

Lesson Outline

(1) Definition of an ellipse

(2) Derivation of the standard equation of an ellipse

31

(3) Graphing ellipses

(4) Solving situational problems involving ellipses

Introduction

An ellipse is one of the conic sections that most students have not encounteredformally before, unlike circles and parabolas. Its shape is a bounded curve whichlooks like a flattened circle. The orbits of the planets in our solar system aroundthe sun happen to be elliptical in shape. Also, just like parabolas, ellipses havereflective properties that have been used in the construction of certain structures(shown in some of the practice problems). We will see some properties andapplications of ellipses in this section.

1.3.1. Definition and Equation of an Ellipse

Consider the points F1(�3, 0) and F2(3, 0), as shown in Figure 1.21. What is thesum of the distances of A(4, 2.4) from F1 and from F2? How about the sum ofthe distances of B (and C(0,�4)) from F1 and from F2?Teaching Notes

You may reviewthe distance

formula.AF1 + AF2 = 7.4 + 2.6 = 10

BF1 +BF2 = 3.8 + 6.2 = 10

CF1 + CF2 = 5 + 5 = 10

There are other points P such that PF1 + PF2 = 10. The collection of all suchpoints forms a shape called an ellipse.

Figure 1.21Figure 1.22

Let F1 and F2 be two distinct points. The set of all points P , whosedistances from F1 and from F2 add up to a certain constant, is calledan ellipse. The points F1 and F2 are called the foci of the ellipse.

32

Given are two points on the x-axis, F1(�c, 0) and F2(c, 0), the foci, both cunits away from their center (0, 0). See Figure 1.22. Let P (x, y) be a point onthe ellipse. Let the common sum of the distances be 2a (the coe�cient 2 willmake computations simpler). Thus, we have PF1 + PF2 = 2a.

PF1 = 2a� PF2p(x+ c)2 + y2 = 2a�

p(x� c)2 + y2

x2 + 2cx+ c2 + y2 = 4a2 � 4ap(x� c)2 + y2 + x2 � 2cx+ c2 + y2

ap

(x� c)2 + y2 = a2 � cx

a2⇥x2 � 2cx+ c2 + y2

⇤= a4 � 2a2cx+ c2x2

(a2 � c2)x2 + a2y2 = a4 � a2c2 = a2(a2 � c2)

b2x2 + a2y2 = a2b2 by letting b =pa2 � c2, so a > b

x2

a2+

y2

b2= 1

When we let b =pa2 � c2, we assumed a > c. To see why this is true, look at

4PF1F2 in Figure 1.22. By the Triangle Inequality, PF1 + PF2 > F1F2, whichimplies 2a > 2c, so a > c.

We collect here the features of the graph of an ellipse with standard equationx2

a2+

y2

b2= 1, where a > b. Let c =

pa2 � b2.

(1) center : origin (0, 0)

(2) foci : F1(�c, 0) and F2(c, 0)

• Each focus is c units away from the center.

• For any point on the ellipse, the sum of its distances from the foci is 2a.

(3) vertices : V1(�a, 0) and V2(a, 0)

• The vertices are points on the ellipse, collinear with the center and foci.

33

• If y = 0, then x = ±a. Each vertex is a units away from the center.

• The segment V1V2 is called the major axis. Its length is 2a. It dividesthe ellipse into two congruent parts.

(4) covertices : W1(0,�b) and W2(0, b)

• The segment through the center, perpendicular to the major axis, is theminor axis. It meets the ellipse at the covertices. It divides the ellipseinto two congruent parts.

• If x = 0, then y = ±b. Each covertex is b units away from the center.

• The minor axis W1W2 is 2b units long. Since a > b, the major axis islonger than the minor axis.

Example 1.3.1. Give the coordinates of the foci, vertices, and covertices of theellipse with equation

x2

25+

y2

9= 1.

Sketch the graph, and include these points.

Solution. With a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =pa2 � b2 = 4.

foci: F1(�4, 0), F2(4, 0) vertices: V1(�5, 0), V2(5, 0)

covertices: W1(0,�3), W2(0, 3)

Example 1.3.2. Find the (standard) equation of the ellipse whose foci areF1(�3, 0) and F2(3, 0), such that for any point on it, the sum of its distancesfrom the foci is 10. See Figure 1.21.

Solution. We have 2a = 10 and c = 3, so a = 5 and b =pa2 � c2 = 4. The

equation isx2

25+

y2

16= 1. 2

34


1. Give the coordinates of the foci, vertices, and covertices of the ellipse with

equationx2

169+

y2

25= 1. Sketch the graph, and include these points.

Answer: foci: F1(�12, 0) and F2(12, 0), vertices: V1(�13, 0) and V2(13, 0),covertices: W1(0,�5) and W2(0, 5)

2. Find the equation in standard form of the ellipse whose foci are F1(�8, 0) andF2(8, 0), such that for any point on it, the sum of its distances from the foci

is 20. Answer:x2

100+

y2

36= 1

1.3.2. More Properties of Ellipses

The ellipses we have considered so far are “horizontal” and have the origin as theircenters. Some ellipses have their foci aligned vertically, and some have centersnot at the origin. Their standard equations and properties are given in the box.The derivations are more involved, but are similar to the one above, and so arenot shown anymore.

In all four cases below, a > b and c =pa2 � b2. The foci F1 and F2 are c

units away from the center. The vertices V1 and V2 are a units away from thecenter, the major axis has length 2a, the covertices W1 and W2 are b units awayfrom the center, and the minor axis has length 2b. Recall that, for any point onthe ellipse, the sum of its distances from the foci is 2a.

35

Center Corresponding Graphs

(0, 0)

x2

a2+

y2

b2= 1

x2

b2+

y2

a2= 1

(h, k)

(x� h)2

a2+

(y � k)2

b2= 1

(x� h)2

b2+

(y � k)2

a2= 1

major axis: horizontal major axis: vertical

minor axis: vertical minor axis: horizontal

In the standard equation, if the x-part has the bigger denominator, the ellipseis horizontal. If the y-part has the bigger denominator, the ellipse is vertical.

Example 1.3.3. Give the coordinates of the center, foci, vertices, and coverticesof the ellipse with the given equation. Sketch the graph, and include these points.

36

(1)(x+ 3)2

24+

(y � 5)2

49= 1

(2) 9x2 + 16y2 � 126x+ 64y = 71

Solution. (1) From a2 = 49 and b2 = 24, we have a = 7, b = 2p6 ⇡ 4.9, and

c =pa2 � b2 = 5. The ellipse is vertical.

center: (�3, 5)

foci: F1(�3, 0), F2(�3, 10)

vertices: V1(�3,�2), V2(�3, 12)

covertices: W1(�3� 2p6, 5) ⇡ (�7.9, 5)

W2(�3 + 2p6, 5) ⇡ (1.9, 5)

(2) We first change the given equation to standard form.

9(x2 � 14x) + 16(y2 + 4y) = 71

9(x2 � 14x+ 49) + 16(y2 + 4y + 4) = 71 + 9(49) + 16(4)

9(x� 7)2 + 16(y + 2)2 = 576

(x� 7)2

64+

(y + 2)2

36= 1

37

We have a = 8 and b = 6. Thus, c =pa2 � b2 = 2

p7 ⇡ 5.3. The ellipse is

horizontal.

center: (7,�2)

foci: F1(7� 2p7,�2) ⇡ (1.7,�2)

F2(7 + 2p7,�2) ⇡ (12.3,�2)

vertices: V1(�1,�2), V2(15,�2)

covertices: W1(7,�8), W2(7, 4)

Example 1.3.4. The foci of an ellipse are (�3,�6) and (�3, 2). For any pointon the ellipse, the sum of its distances from the foci is 14. Find the standardequation of the ellipse.

Solution. The midpoint (�3,�2) of the foci is the center of the ellipse. Theellipse is vertical (because the foci are vertically aligned) and c = 4. From thegiven sum, 2a = 14 so a = 7. Also, b =

pa2 � c2 =

p33. The equation is

(x+ 3)2

33+

(y + 2)2

49= 1. 2

Example 1.3.5. An ellipse has vertices (2 �p61,�5) and (2 +

p61,�5), and

its minor axis is 12 units long. Find its standard equation and its foci.

38

Solution. The midpoint (2,�5) of the vertices is the center of the ellipse, which ishorizontal. Each vertex is a =

p61 units away from the center. From the length of

the minor axis, 2b = 12 so b = 6. The standard equation is(x� 2)2

61+(y + 5)2

36=

1. Each focus is c =pa2 � b2 = 5 units away from (2,�5), so their coordinates

are (�3,�5) and (7,�5). 2


1. Give the coordinates of the center, foci, vertices, and covertices of the ellipsewith equation 41x2 + 16y2 + 246x � 192y + 289 = 0. Sketch the graph, andinclude these points.

Answer: center C(�3, 6), foci F1(�3, 1) and F2(�3, 11), vertices V1(�3, 6 �p41) and V2(�3, 6 +

p41), covertices W1(�7, 6) and W2(1, 6)

2. An ellipse has vertices (�10,�4) and (6,�4), and covertices (�2,�9) and(�2, 1). Find its standard equation and its foci.

Answer:(x+ 2)2

64+

(y + 4)2

25= 1, foci (�2�

p39,�4) and (�2 +

p39,�4)

39

1.3.3. Situational Problems Involving Ellipses

We now apply the concept of ellipse to some situational problems.

?Example 1.3.6. A tunnel has the shape of a semiellipse that is 15 ft high atthe center, and 36 ft across at the base. At most how high should a passing truckbe, if it is 12 ft wide, for it to be able to fit through the tunnel? Round o↵ youranswer to two decimal places.

Solution. Refer to the figure above. If we draw the semiellipse on a rectangularcoordinate system, with its center at the origin, an equation of the ellipse whichcontains it, is

x2

182+

y2

152= 1.

To maximize its height, the corners of the truck, as shown in the figure, wouldhave to just touch the ellipse. Since the truck is 12 ft wide, let the point (6, n)be the corner of the truck in the first quadrant, where n > 0, is the (maximum)height of the truck. Since this point is on the ellipse, it should fit the equation.Thus, we have

62

182+

n2

152= 1

n2 = 152✓1� 62

182

◆

n = 10p2 ⇡ 14.14 ft 2

Example 1.3.7. The orbit of a planet has the shape of an ellipse, and on oneof the foci is the star around which it revolves. The planet is closest to the starwhen it is at one vertex. It is farthest from the star when it is at the other vertex.Suppose the closest and farthest distances of the planet from this star, are 420million kilometers and 580 million kilometers, respectively. Find the equation ofthe ellipse, in standard form, with center at the origin and the star at the x-axis.Assume all units are in millions of kilometers.

40

Solution. In the figure above, the orbit is drawn as a horizontal ellipse withcenter at the origin. From the planet’s distances from the star, at its closestand farthest points, it follows that the major axis is 2a = 420 + 580 = 1000(million kilometers), so a = 500. If we place the star at the positive x-axis,then it is c = 500 � 420 = 80 units away from the center. Therefore, we getb2 = a2 � c2 = 5002 � 802 = 243600. The equation then is

x2

250000+

y2

243600= 1.

The star could have been placed on the negative x-axis, and the answer wouldstill be the same. 2


?1. The arch of a bridge is in the shape of a semiellipse, with its major axis at thewater level. Suppose the arch is 20 ft high in the middle, and 120 ft across itsmajor axis. How high above the water level is the arch, at a point 20 ft fromthe center (horizontally)? Round o↵ to 2 decimal places. Refer to Example1.3.6. Answer: 18.86 ft

Exercises 1.3

1. Give the coordinates of the center, vertices, covertices, and foci of the ellipsewith the given equation. Sketch the graph, and include these points.

(a)x2

169+

y2

25= 1

41

(b)x2

144+

y2

169= 1

(c) 4x2 + 13y2 = 52

(d)(x+ 7)2

16+

(y � 4)2

25= 1

(e) 9x2 + 16y2 + 72x� 96y + 144 = 0

(f) 36x2 + 20y2 � 144x+ 120y � 396 = 0

Answer:

Item Center Vertices Covertices Foci

(a) (0, 0) (±13, 0) (0,±5) (±12, 0)

(b) (0, 0) (0,±13) (±12, 0) (0,±5)

(c) (0, 0) (±p13, 0) (0,±2) (±3, 0)

(d) (�7, 4) (�7,�1) (�11, 4) (�7, 1)

(�7, 9) (�3, 4) (�7, 7)

(e) (�4, 3) (�8, 3) (�4, 0) (�4±p7, 3)

(0, 3) (�4, 6)

(f) (2,�3) (2,�9) (2± 2p5,�3) (2,�7)

(2,�3) (2, 3) (2, 1)

(a) (b) (c)

(d) (e) (f)

42

2. Find the standard equation of the ellipse which satisfies the given conditions.

(a) foci (�7, 6) and (�1, 6), the sum of the distances of any point from the

foci is 14 Answer:(x+ 4)2

49+

(y � 6)2

40= 1

(b) center (5, 3), horizontal major axis of length 20, minor axis of length 16

Answer:(x� 5)2

100+

(y � 3)2

64= 1

(c) major axis of length 22, foci 9 units above and below the center (2, 4)

Answer:(x� 2)2

40+

(y � 4)2

121= 1

(d) covertices (�4, 8) and (10, 8), a focus at (3, 12)

Answer:(x� 3)2

49+

(y � 8)2

65= 1

Solution. The midpoint of the covertices is the center, (3, 8). From thispoint, the given focus is c = 4 units away. Since b = 7 (the distance fromthe center to a covertex), then a2 = b2 + c2 = 65. The ellipse then has

equation(x� 3)2

49+

(y � 8)2

65= 1.

(e) focus (�6,�2), covertex (�1, 5), horizontal major axis

Answer:(x+ 1)2

74+

(y + 2)2

49= 1

Solution. Make a rough sketch of the points to see that the center is tothe right of the given focus, and below the given covertex. The center isthus (�1,�2). It follows that c = 5, b = 7, so a2 = b2 + c2 = 74. The

ellipse then has equation(x+ 1)2

74+

(y + 2)2

49= 1.

3. A semielliptical tunnel has height 9 ft and a width of 30 ft. A truck that isabout to pass through is 12 ft wide and 8.3 ft high. Will this truck be able topass through the tunnel? Answer: No

4. A truck that is about to pass through the tunnel from the previous item is 10ft wide and 8.3 ft high. Will this truck be able to pass through the tunnel?

Answer: Yes

5. An orbit of a satellite around a planet is an ellipse, with the planet at onefocus of this ellipse. The distance of the satellite from this star varies from300, 000 km to 500, 000 km, attained when the satellite is at each of the twovertices. Find the equation of this ellipse, if its center is at the origin, and thevertices are on the x-axis. Assume all units are in 100, 000 km.

Answer: x

2

16 +y

2

15 = 1

43

6. The orbit of a planet around a star is described by the equation x

2

640,000 +y

2

630,000 = 1, where the star is at one focus, and all units are in millions ofkilometers. The planet is closest and farthest from the star, when it is at thevertices. How far is the planet when it is closest to the sun? How far is theplanet when it is farthest from the sun?

Answer: 700 million km, 900 million km

Solution. The ellipse has center at the origin, and major axis on the x-axis.Since a2 = 640, 000, then a = 800, so the vertices are V1(�800, 0) andV2(800, 00). Since b2 = 630, 000, then c =

pa2 � b2 =

p10, 000 = 100. Sup-

pose the star is at the focus at the right of the origin (this choice is arbitrary,since we could have chosen instead the focus on the left). Its location is thenF (100, 0). The closest distance is then V2F = 700 (million kilometers) andthe farthest distance is V1F = 900 (million kilometers).

7. A big room is constructed so that the ceiling is a dome that is semiellipticalin shape. If a person stands at one focus and speaks, the sound that is madebounces o↵ the ceiling and gets reflected to the other focus. Thus, if twopeople stand at the foci (ignoring their heights), they will be able to hear eachother. If the room is 34 m long and 8 m high, how far from the center shouldeach of two people stand if they would like to whisper back and forth and heareach other? Answer: 15 m

Solution. We could put a coordinate system with the floor of the room onthe x-axis, and the center of the room at the origin, as shown in the figures.The major axis has length 34, and the height of the room is half of the minoraxis. The ellipse that contains the ceiling then has equation x

2

172 +y

2

82 = 1. The

distance of a focus from the center is c =pa2 � b2 =

p172 � 82 = 15. Thus,

the two people should stand 15 m away from the center.

8. A whispering gallery has a semielliptical ceiling that is 9 m high and 30 mlong. How high is the ceiling above the two foci? Answer: 5.4 m

44

Solution. As in the previous problem, put a coordinate system with the floorof the room on the x-axis, and the center of the room at the origin. The majoraxis has length 30, and half the minor axis is 9. The ellipse that contains theceiling then has equation x

2

152 +y

2

92 = 1. The distance of a focus from the center

is c =pa2 � b2 =

p152 � 92 = 12. If we put x = 12 in the equation of the

ellipse, we get 122

152 +y

2

92 = 1. Solving for y > 0 yields y = 275 = 5.4. The height

of the ceiling above each focus is 5.4 m.

4

Lesson 1.4. Hyperbolas




(1) define a hyperbola;

(2) determine the standard form of equation of a hyperbola;

(3) graph a hyperbola in a rectangular coordinate system; and

(4) solve situational problems involving conic sections (hyperbolas).

Lesson Outline

(1) Definition of a hyperbola

(2) Derivation of the standard equation of a hyperbola

(3) Graphing hyperbolas

(4) Solving situational problems involving hyperbolas

Introduction

A hyperbola is one of the conic sections that most students have not encoun-tered formally before, unlike circles and parabolas. Its graph consists of twounbounded branches which extend in opposite directions. It is a misconceptionthat each branch is a parabola. This is not true, as parabolas and hyperbolashave very di↵erent features. An application of hyperbolas in basic location andnavigation schemes are presented in an example and some exercises.

1.4.1. Definition and Equation of a Hyperbola

Consider the points F1(�5, 0) and F2(5, 0) as shown in Figure 1.23. What is theabsolute value of the di↵erence of the distances of A(3.75,�3) from F1 and from

45

F2? How about the absolute value of the di↵erence of the distances of B��5, 163

�

from F1 and from F2?

|AF1 � AF2| = |9.25� 3.25| = 6

|BF1 � BF2| =��16

3� 34

3

�� = 6

There are other points P such that |PF1 � PF2| = 6. The collection of all suchpoints forms a shape called a hyperbola, which consists of two disjoint branches.For points P on the left branch, PF2 � PF1 = 6; for those on the right branch,PF1 � PF2 = 6.


Let F1 and F2 be two distinct points. The set of all points P , whosedistances from F1 and from F2 di↵er by a certain constant, is called ahyperbola. The points F1 and F2 are called the foci of the hyperbola.

In Figure 1.24, given are two points on the x-axis, F1(�c, 0) and F2(c, 0), thefoci, both c units away from their midpoint (0, 0). This midpoint is the centerof the hyperbola. Let P (x, y) be a point on the hyperbola, and let the absolutevalue of the di↵erence of the distances of P from F1 and F2, be 2a (the coe�cient2 will make computations simpler). Thus, |PF1 � PF2| = 2a, and so

��p

(x+ c)2 + y2 �p

(x� c)2 + y2�� = 2a.

Algebraic manipulations allow us to rewrite this into the much simpler

x2

a2� y2

b2= 1, where b =

pc2 � a2.

When we let b =pc2 � a2, we assumed c > a. To see why this is true, suppose

that P is closer to F2, so PF1 � PF2 = 2a. Refer to Figure 1.24. Suppose also

46

that P is not on the x-axis, so 4PF1F2 is formed. From the triangle inequality,F1F2 + PF2 > PF1. Thus, 2c > PF1 � PF2 = 2a, so c > a.

Now we present a derivation. For now, assume P is closer to F2 so PF1 > PF2,and PF1 � PF2 = 2a.

Teaching Notes

If it is assumedthat P is closer toF1, then the sameequation will beobtained becauseof symmetry.

PF1 = 2a+ PF2p(x+ c)2 + y2 = 2a+

p(x� c)2 + y2

⇣p(x+ c)2 + y2

⌘2

=⇣2a+

p(x� c)2 + y2

⌘2

cx� a2 = ap

(x� c)2 + y2

(cx� a2)2 =⇣ap(x� c)2 + y2

⌘2

(c2 � a2)x2 � a2y2 = a2(c2 � a2)

b2x2 � a2y2 = a2b2 by letting b =pc2 � a2 > 0

x2

a2� y2

b2= 1

We collect here the features of the graph of a hyperbola with standard equa-tion

x2

a2� y2

b2= 1.

Let c =pa2 + b2.


(1) center : origin (0, 0)

(2) foci : F1(�c, 0) and F2(c, 0)

• Each focus is c units away from the center.

47

• For any point on the hyperbola, the absolute value of the di↵erence ofits distances from the foci is 2a.

(3) vertices : V1(�a, 0) and V2(a, 0)

• The vertices are points on the hyperbola, collinear with the center andfoci.

• If y = 0, then x = ±a. Each vertex is a units away from the center.

• The segment V1V2 is called the transverse axis. Its length is 2a.

(4) asymptotes : y = b

a

x and y = � b

a

x, the lines `1 and `2 in Figure 1.26

• The asymptotes of the hyperbola are two lines passing through the cen-ter which serve as a guide in graphing the hyperbola: each branch ofthe hyperbola gets closer and closer to the asymptotes, in the directiontowards which the branch extends. (We need the concept of limits fromcalculus to explain this.)

• An aid in determining the equations of the asymptotes: in the standardequation, replace 1 by 0, and in the resulting equation x

2

a

2 � y

2

b

2 = 0, solvefor y.

• To help us sketch the asymptotes, we point out that the asymptotes`1 and `2 are the extended diagonals of the auxiliary rectangle drawnin Figure 1.26. This rectangle has sides 2a and 2b with its diagonalsintersecting at the center C. Two sides are congruent and parallel tothe transverse axis V1V2. The other two sides are congruent and parallelto the conjugate axis, the segment shown which is perpendicular to thetransverse axis at the center, and has length 2b.

Example 1.4.1. Determine the foci, vertices, and asymptotes of the hyperbolawith equation

x2

9� y2

7= 1.

Sketch the graph, and include these points and lines, the transverse and conjugateaxes, and the auxiliary rectangle.

Solution. With a2 = 9 and b2 = 7, we havea = 3, b =

p7, and c =

pa2 + b2 = 4.

foci: F1(�4, 0) and F2(4, 0)

vertices: V1(�3, 0) and V2(3, 0)

asymptotes: y =p73 x and y = �

p73 x

The graph is shown at the right. The conju-gate axis drawn has its endpoints b =

p7 ⇡

2.7 units above and below the center. 2

48

Example 1.4.2. Find the (standard) equation of the hyperbola whose foci areF1(�5, 0) and F2(5, 0), such that for any point on it, the absolute value of thedi↵erence of its distances from the foci is 6. See Figure 1.23.

Solution. We have 2a = 6 and c = 5, so a = 3 and b =pc2 � a2 = 4. The

hyperbola then has equationx2

9� y2

16= 1. 2


1. Determine foci, vertices, and asymptotes of the hyperbola with equation

x2

16� y2

20= 1.

Sketch the graph, and include these points and lines, along with the auxiliaryrectangle.

Answer: foci F1(�6, 0) and F2(6, 0), vertices V1(�4, 0) and V2(4, 0), asymp-

totes y =p52 x and y = �

p52 x

2. Find the equation in standard form of the hyperbola whose foci are F1(�4p2, 0)

and F2(4p2, 0), such that for any point on it, the absolute value of the di↵er-

ence of its distances from the foci is 8. Answer:x2

16� y2

16= 1

1.4.2. More Properties of Hyperbolas

The hyperbolas we considered so far are “horizontal” and have the origin as theircenters. Some hyperbolas have their foci aligned vertically, and some have centers

49

not at the origin. Their standard equations and properties are given in the box.The derivations are more involved, but are similar to the one above, and so arenot shown anymore.

Center Corresponding Hyperbola

(0, 0)

x2

a2� y2

b2= 1

y2

a2� x2

b2= 1

(h, k)

(x� h)2

a2� (y � k)2

b2= 1

(y � k)2

a2� (x� h)2

b2= 1

transverse axis: horizontal transverse axis: vertical

conjugate axis: vertical conjugate axis: horizontal

In all four cases above, we let c =pa2 + b2. The foci F1 and F2 are c units

away from the center C. The vertices V1 and V2 are a units away from the center.The transverse axis V1V2 has length 2a. The conjugate axis has length 2b and is

50

perpendicular to the transverse axis. The transverse and conjugate axes bisecteach other at their intersection point, C. Each branch of a hyperbola gets closerand closer to the asymptotes, in the direction towards which the branch extends.The equations of the asymptotes can be determined by replacing 1 in the standardequation by 0. The asymptotes can be drawn as the extended diagonals of theauxiliary rectangle determined by the transverse and conjugate axes. Recall that,for any point on the hyperbola, the absolute value of the di↵erence of its distancesfrom the foci is 2a.

In the standard equation, aside from being positive, there are no other re-strictions on a and b. In fact, a and b can even be equal. The orientation of thehyperbola is determined by the variable appearing in the first term (the positiveterm): the corresponding axis is where the two branches will open. For example,if the variable in the first term is x, the hyperbola is “horizontal”: the transverseaxis is horizontal, and the branches open to the left and right in the direction ofthe x-axis.

Example 1.4.3. Give the coordinates of the center, foci, vertices, and asymp-totes of the hyperbola with the given equation. Sketch the graph, and includethese points and lines, the transverse and conjugate axes, and the auxiliary rect-angle.

(1)(y + 2)2

25� (x� 7)2

9= 1

(2) 4x2 � 5y2 + 32x+ 30y = 1

Solution. (1) From a2 = 25 and b2 = 9, we have a = 5, b = 3, and c =pa2 + b2 =

p34 ⇡ 5.8. The hyperbola is vertical. To determine the asymp-

totes, we write (y+2)2

25 � (x�7)2

9 = 0, which is equivalent to y+2 = ±53(x� 7).

We can then solve this for y.

center: C(7,�2)

foci: F1(7,�2�p34) ⇡ (7,�7.8) and F2(7,�2 +

p34) ⇡ (7, 3.8)

vertices: V1(7,�7) and V2(7, 3)

asymptotes: y = 53x� 41

3 and y = �53x+ 29

3

The conjugate axis drawn has its endpoints b = 3 units to the left and rightof the center.

51

(2) We first change the given equation to standard form.

4(x2 + 8x)� 5(y2 � 6y) = 1

4(x2 + 8x+ 16)� 5(y2 � 6y + 9) = 1 + 4(16)� 5(9)

4(x+ 4)2 � 5(y � 3)2 = 20

(x+ 4)2

5� (y � 3)2

4= 1

We have a =p5 ⇡ 2.2 and b = 2. Thus, c =

pa2 + b2 = 3. The hyperbola

is horizontal. To determine the asymptotes, we write (x+4)2

5 � (y�3)2

4 = 0which is equivalent to y � 3 = ± 2p

5(x+ 4), and solve for y.

center: C(�4, 3)

foci: F1(�7, 3) and F2(�1, 3)

vertices: V1(�4�p5, 3) ⇡ (�6.2, 3) and V2(�4 +

p5, 3) ⇡ (�1.8, 3)

asymptotes: y = 2p5x+ 8p

5+ 3 and y = � 2p

5x� 8p

5+ 3

The conjugate axis drawn has its endpoints b = 2 units above and belowthe center.

52

Example 1.4.4. The foci of a hyperbola are (�5,�3) and (9,�3). For any pointon the hyperbola, the absolute value of the di↵erence of its of its distances fromthe foci is 10. Find the standard equation of the hyperbola.

Solution. The midpoint (2,�3) of the foci is the center of the hyperbola. Eachfocus is c = 7 units away from the center. From the given di↵erence, 2a = 10 soa = 5. Also, b2 = c2 � a2 = 24. The hyperbola is horizontal (because the foci arehorizontally aligned), so the equation is

(x� 2)2

25� (y + 3)2

24= 1. 2

Example 1.4.5. A hyperbola has vertices (�4,�5) and (�4, 9), and one of itsfoci is (�4, 2�

p65). Find its standard equation.

Solution. The midpoint (�4, 2) of the vertices is the center of the hyperbola,which is vertical (because the vertices are vertically aligned). Each vertex isa = 7 units away from the center. The given focus is c =

p65 units away from

the center. Thus, b2 = c2 � a2 = 16, and the standard equation is

(y � 2)2

49� (x+ 4)2

16= 1. 2

53


1. Give the coordinates of the center, foci, vertices, and asymptotes of the hy-perbola with equation 9x2 � 4y2 � 90x� 32y = �305. Sketch the graph, andinclude these points and lines, along with the auxiliary rectangle.

Answer: center C(5,�4), foci F1(5,�4�2p13) and F2(5,�4+2

p13), vertices

V1(5,�10) and V2(5, 2), asymptotes y = �32x+ 7

2 and y = 32x� 23

2

2. A hyperbola has vertices (1, 9) and (13, 9), and one of its foci is (�2, 9). Find

its standard equation. Answer:(x� 7)2

36� (y � 9)2

45= 1

1.4.3. Situational Problems Involving Hyperbolas

We now give an example on an application of hyperbolas.

Example 1.4.6. An explosion is heard by two stations 1200 m apart, located atF1(�600, 0) and F2(600, 0). If the explosion was heard in F1 two seconds beforeit was heard in F2, identify the possible locations of the explosion. Use 340 m/sas the speed of sound.

Solution. Using the given speed of sound, we deduce that the sound traveled340(2) = 680 m farther in reaching F2 than in reaching F1. This is then thedi↵erence of the distances of the explosion from the two stations. Thus, theexplosion is on a hyperbola with foci are F1 and F2, on the branch closer to F1.

54

We have c = 600 and 2a = 680, so a = 340 and b2 = c2 � a2 = 244400.The explosion could therefore be anywhere on the left branch of the hyperbola

x

2

115600 �y

2

244400 = 1. 2


?1. Two stations, located at M(�1.5, 0) and N(1.5, 0) (units are in km), simulta-neously send sound signals to a ship, with the signal traveling at the speed of0.33 km/s. If the signal from N was received by the ship four seconds beforethe signal it received from M , find the equation of the curve containing thepossible location of the ship. Answer: x

2

0.4356 �y

2

1.8144 = 1 (right branch)

Exercises 1.4

1. Give the coordinates of the center, foci, vertices, and the asymptotes of thehyperbola with the given equation. Sketch the graph, and include these pointsand lines.

(a)x2

36� y2

64= 1

(b)y2

25� x2

16= 1

(c) (x� 1)2 � y2 = 4

(d)(y + 2)2

15� (x+ 3)2

10= 1

(e) 3x2 � 2y2 � 42x� 16y = �67

(f) 25x2 � 39y2 + 150x+ 390y = �225

55

Answer:

Item Center Vertices Foci

(a) (0, 0) (±6, 0) (±10, 0)

(b) (0, 0) (0,±5) (0,±p41)

(c) (1, 0) (�1, 0), (3, 0) (1± 2p2, 0)

(d) (�3,�2) (�3,�2±p15) (�3,�7), (�3, 3)

(e) (7,�4) (3,�4), (11,�4) (7± 2p10,�4)

(f) (�3, 5) (�3, 0), (�3, 10) (�3,�3), (�3, 13)

Item Asymptotes

(a) y = ±43x

(b) y = ±54x

(c) y = x� 1, y = �x+ 1

(d) y = ±q

32x± 3

q32 � 2

(e) y = ±q

32x⌥ 7

q32 � 4

(f) y = ± 5p39x± 15p

39+ 5

(a) (b) (c)

56

(d) (e) (f)

2. Find the standard equation of the hyperbola which satisfies the given condi-tions.

(a) foci (�4,�3) and (�4, 13), the absolute value of the di↵erence of thedistances of any point from the foci is 14

Answer:(y � 5)2

49� (x+ 4)2

15= 1

(b) vertices (�2, 8) and (8, 8), a focus (12, 8)

Answer:(x� 3)2

25� (y � 8)2

56= 1

(c) center (�6, 9), a vertex (�6, 15), conjugate axis of length 12

Answer:(y � 9)2

25� (x+ 6)2

36= 1

(d) asymptotes y = 43x+ 1

3 and y = �43x+ 41

3 , a vertex (�1, 7)

Answer:(x� 5)2

36� (y � 7)2

64= 1

Solution. The asymptotes intersect at (5, 7). This is the center. Thedistance of the given vertex from the center is a = 6. This vertex andcenter are aligned horizontally, so the hyperbola has equation of the form(x�h)2

a

2 � (y�k)2

b

2 = 1. The asymptotes consequently have the form y� k =± b

a

(x�h), and thus, have slopes ± b

a

. From the given asymptotes, b

a

= 43 .

Since a = 6, then b = 8. The standard equation is then

(x� 5)2

36� (y � 7)2

64= 1.

(e) asymptotes y = 13x+ 5

3 and y = �13x+ 7

3 , a focus (1, 12)

Answer:(y � 2)2

10� (x� 1)2

90= 1

Solution. The asymptotes intersect at (1, 2). This is the center. Thedistance of the given focus from the center is c = 10. This focus and

57

center are aligned vertically, so the hyperbola has equation of the form(y�k)2

a

2 � (x�h)2

b

2 = 1. The asymptotes consequently have the form y� k =±a

b

(x�h), and thus, have slopes ±a

b

. From the given asymptotes, a

b

= 13 ,

so b = 3a.c2 = 100 = a2 + b2 = a2 + (3a)2 = 10a2

Thus, a2 = 10, and b2 = 9a2 = 90. The standard equation is

(y � 2)2

10� (x� 1)2

90= 1.

3. Two control towers are located at points Q(�500, 0) and R(500, 0), on astraight shore where the x-axis runs through (all distances are in meters).At the same moment, both towers sent a radio signal to a ship out at sea, eachtraveling at 300 m/µs. The ship received the signal from Q 3 µs (microseconds)before the message from R.

(a) Find the equation of the curve containing the possible location of the

ship. Answer:x2

202500� y2

47500= 1 (left branch)

(b) Find the coordinates (rounded o↵ to two decimal places) of the ship if itis 200 m from the shore (y = 200). Answer: (�610.76, 200)

Solution. Since the time delay between the two signals is 3 µs, then the di↵er-ence between the distances traveled by the two signals is 300 · 3 = 900 m. Theship is then on a hyperbola, consisting of points whose distances from Q and R(the foci) di↵er by 2a = 900. With a = 450 and c = 500 (the distance of eachfocus from the center, the origin), we have b2 = c2�a2 = 5002�4502 = 47500.

Since a2 = 202500, the hyperbola then has equation x

2

202500 �y

2

47500 = 1. Sincethe signal from Q was received first, the ship is closer to Q than R, so theship is on the left branch of this hyperbola. Using y = 200, we then solve

x

2

202500 �2002

47500 = 1 for x < 0 (left branch), and we get x ⇡ �610.76.

4

58

Lesson 1.5. More Problems on Conic Sections




(1) recognize the equation and important characteristics of the di↵erent types ofconic sections; and

(2) solve situational problems involving conic sections.

Lesson Outline

(1) Conic sections with associated equations in general form

(2) Problems involving characteristics of various conic sections

(3) Solving situational problems involving conic sections

Introduction

Inspecting the equation can lead us to the right conic section for its graph,and set us on the right step towards analyzing it. We will also look at problemsthat use the properties of the di↵erent conic sections, allowing us to synthesizewhat has been covered so far.

1.5.1. Identifying the Conic Section by Inspection

The equation of a circle may be written in standard form

Ax2 + Ay2 + Cx+Dy + E = 0,

that is, the coe�cients of x2 and y2 are the same. However, it does not followthat if the coe�cients of x2 and y2 are the same, the graph is a circle.

General Equation Standard Equation graph

(A) 2x2 + 2y2 � 2x+ 6y + 5 = 0�x� 1

2

�2+�y + 3

2

�2= 0 point

(B) x2 + y2 � 6x� 8y + 50 = 0 (x� 3)2 + (y � 4)2 = �25 empty set

For a circle with equation (x � h)2 + (y � k)2 = r2, we have r2 > 0. This isnot the case for the standard equations of (A) and (B).

In (A), because the sum of two squares can only be 0 if and only if each squareis 0, it follows that x� 1

2 = 0 and y + 32 = 0. The graph is thus the single point�

12 ,�

32

�.

In (B), no real values of x and y can make the nonnegative left side equal tothe negative right side. The graph is then the empty set.

59

Let us recall the general form of the equations of the other conic sections. Wemay write the equations of conic sections we discussed in the general form

Ax2 +By2 + Cx+Dy + E = 0.

Some terms may vanish, depending on the kind of conic section.

(1) Circle: both x2 and y2 appear, and their coe�cients are the same

Ax2 + Ay2 + Cx+Dy + E = 0

Example: 18x2 + 18y2 � 24x+ 48y � 5 = 0

Degenerate cases: a point, and the empty set

(2) Parabola: exactly one of x2 or y2 appears

Ax2 + Cx+Dy + E = 0 (D 6= 0, opens upward or downward)

By2 + Cx+Dy + E = 0 (C 6= 0, opens to the right or left)

Examples: 3x2 � 12x+ 2y + 26 = 0 (opens downward)

� 2y2 + 3x+ 12y � 15 = 0 (opens to the right)

(3) Ellipse: both x2 and y2 appear, and their coe�cients A and B have the samesign and are unequal

Examples: 2x2 + 5y2 + 8x� 10y � 7 = 0 (horizontal major axis)

4x2 + y2 � 16x� 6y + 21 = 0 (vertical major axis)

If A = B, we will classify the conic as a circle, instead of an ellipse.

Degenerate cases: a point, and the empty set

(4) Hyperbola: both x2 and y2 appear, and their coe�cients A and B have dif-ferent signs

Examples: 5x2 � 3y2 � 20x� 18y � 22 = 0 (horizontal transverse axis)

� 4x2 + y2 + 24x+ 4y � 36 = 0 (vertical transverse axis)

Degenerate case: two intersecting lines

The following examples will show the possible degenerate conic (a point, twointersecting lines, or the empty set) as the graph of an equation following a similarpattern as the non-degenerate cases.

(1) 4x2 + 9y2 � 16x+ 18y + 25 = 0 =) (x� 2)2

32+

(y + 1)2

22= 0

=) one point: (2,�1)

(2) 4x2 + 9y2 � 16x+ 18y + 61 = 0 =) (x� 2)2

32+

(y + 1)2

22= �1

=) empty set

60

(3) 4x2 � 9y2 � 16x� 18y + 7 = 0 =) (x� 2)2

32� (y + 1)2

22= 0

=) two lines: y + 1 = ±2

3(x� 2)

A Note on Identifying a Conic Sectionby Its General Equation

It is only after transforming a given general equation to standardform that we can identify its graph either as one of the degenerateconic sections (a point, two intersecting lines, or the empty set) or asone of the non-degenerate conic sections (circle, parabola, ellipse, orhyperbola).


The graphs of the following equations are (nondegenerate) conic sections. Identifythe conic section.

(1) 5x2 � 3y2 + 10x� 12y = 22 Answer: hyperbola

(2) 2y2 � 5x� 12y = 17 Answer: parabola

(3) 3x2 + 3y2 + 42x� 12y = �154 Answer: circle

(4) 3x2 + 6x+ 4y = 18 Answer: parabola

(5) 7x2 + 3y2 � 14x+ 12y = �14 Answer: ellipse

(6) �4x2 + 3y2 + 24x� 12y = 36 Answer: hyperbola

1.5.2. Problems Involving Di↵erent Conic Sections

The following examples require us to use the properties of di↵erent conic sectionsat the same time.

Example 1.5.1. A circle has center at the focus of the parabola y2+16x+4y =44, and is tangent to the directrix of this parabola. Find its standard equation.

Solution. The standard equation of the parabola is (y + 2)2 = �16(x � 3). Itsvertex is V (3,�2). Since 4c = 16 or c = 4, its focus is F (�1,�2) and its directrixis x = 7. The circle has center at (�1,�2) and radius 8, which is the distancefrom F to the directrix. Thus, the equation of the circle is

(x+ 1)2 + (y + 2)2 = 64. 2

Example 1.5.2. The vertices and foci of 5x2 � 4y2 + 50x + 16y + 29 = 0 are,respectively, the foci and vertices of an ellipse. Find the standard equation ofthis ellipse.

61

Solution. We first write the equation of the hyperbola in standard form:

(x+ 5)2

16� (y � 2)2

20= 1.

For this hyperbola, using the notations ah

, bh

, and ch

to refer to a, b, and c ofthe standard equation of the hyperbola, respectively, we have a

h

= 4, bh

= 2p5,

ch

=p

a2h

+ b2h

= 6, so we have the following points:

center: (�5, 2)

vertices: (�9, 2) and (�1, 2)

foci: (�11, 2) and (1, 2).

It means that, for the ellipse, we have these points:

center: (�5, 2)

vertices: (�11, 2) and (1, 2)

foci: (�9, 2) and (�1, 2).

In this case, ce

= 4 and ae

= 6, so that be

=pa2e

� c2e

=p20. The standard

equation of the ellipse is

(x+ 5)2

36+

(y � 2)2

20= 1. 2


1. Find the standard equation of all circles having center at a focus of 21x2 �4y2 + 84x� 24y = 36 and passing through the farther vertex.

Answer: (x+ 7)2 + (y + 3)2 = 49, (x� 3)2 + (y + 3)2 = 49

2. Find the standard equation of the hyperbola one branch of which has focus andvertex that are the same as those of x2 � 6x+ 8y = 23, and whose conjugateaxis is on the directrix of the same parabola.

Answer:(y � 6)2

4� (x� 3)2

12= 1

Exercises 1.5

1. The graphs of the following equations are non-degenerate conic sections. Iden-tify the conic section.

(a) 5x2 + 7y2 � 40x� 28y = �73 Answer: ellipse

(b) 5y2 + 2x� 30y = �49 Answer: parabola

(c) 3x2 � 3y2 + 12x� 12y = 5 Answer: hyperbola

(d) 3x2 + 3y2 + 12x+ 12y = 4 Answer: circle

62

(e) 2x2 + 24x� 5y = �57 Answer: parabola

2. The graphs of the following equations are degenerate conic sections. What arethe specific graphs?

(a) x2 + 3y2 � 4x+ 24y = �52 Answer: point: (2,�4)

(b) 9x2 � 4y2 + 18x� 16y = 7 Answer: lines: y + 2 = ±32(x+ 1)

(c) 3x2 + 5y2 � 6x� 20y = �25 Answer: empty set

3. An ellipse has equation 25x2 + 16y2 + 150x � 32y = 159. Find the standardequations of all parabolas whose vertex is a focus of this ellipse and whosefocus is a vertex of this ellipse.

Answer: (x + 3)2 = �8(y + 2), (x + 3)2 = 32(y + 2), (x + 3)2 = �32(y � 4),and (x+ 3)2 = 8(y � 4)

Solution. The standard equation of the ellipse is

(x+ 3)2

16+

(y � 1)2

25= 1.

Its center is (�3, 1). Since a = 5 and b = 4, we get c = 3, so the vertices areP (�3,�4) and S(�3, 6), while its foci are Q(�3,�2) and R(�3, 4). We thenget four parabolas satisfying the conditions of the problem. The focal distanceindicated below is the distance from the vertex to the focus.

vertex focus focal distance standard equation

Q(�3,�2) P (�3,�4) 2 (x+ 3)2 = �8(y + 2)

Q(�3,�2) S(�3, 6) 8 (x+ 3)2 = 32(y + 2)

R(�3, 4) P (�3,�4) 8 (x+ 3)2 = �32(y � 4)

R(�3, 4) S(�3, 6) 2 (x+ 3)2 = 8(y � 4)

4. Find the standard equation of the hyperbola whose conjugate axis is on thedirectrix of the parabola y2 +12x+6y = 39, having the focus of the parabolaas one of its foci, and the vertex of the parabola as one of its vertices.

Answer:(x� 7)2

9� (y + 3)2

27= 1

Solution. The standard equation of the parabola is (y + 3)2 = �12(x� 4), soits vertex is V (4,�3), and it opens to the left. With 4c = 12, or c = 3, itsfocus is F (1,�3), and its directrix is x = 7. The hyperbola has its center on

63

x = 7, its conjugate axis, and a vertex at (4,�3). Its center is then C(7,�3).The conjugate axis is vertical so the hyperbola is horizontal, with constantsah

= CV = 3 and ch

= CF = 6, so b2h

= c2h

� a2h

= 27. The standard equationof the required hyperbola is

(x� 7)2

9� (y + 3)2

27= 1.

5. Find the standard equation of the parabola opening to the left whose axiscontains the major axis of the ellipse x2 + 4y2 � 10x � 24y + 45 = 0, whosefocus is the center of the ellipse, and which passes through the covertices ofthis ellipse. Answer: (y � 3)2 = �4(x� 6)

Solution. The standard form of the ellipse is

(x� 5)2

16+

(y � 3)2

4= 1.

Its center (5, 3) is the focus of the parabola. Since b = 2, its covertices areW1(5, 1) and W2(5, 5). The vertex of the parabola, c units to the right of (5, 3),is (5 + c, 3). Its equation can be written as (y� 3)2 = �4c(x� (5 + c)). Since(5, 5) is a point on this parabola, we have (5� 3)2 = �4c(5� (5+ c)). Solvingthis equation for c > 0 yields c = 1. Therefore, the standard equation of therequired parabola is (y � 3)2 = �4(x� 6).

6. Find the standard equation of the ellipse whose major and minor axes are thetransverse and conjugate axes (not necessarily in that order) of the hyperbola

4x2 � 9y2 � 16x� 54y = 29. Answer:(x� 2)2

9+

(y + 3)2

4= 1

Solution. The standard equation of the hyperbola is

(y + 3)2

4� (x� 2)2

9= 1,

with center (2,�3), and constants ah

= 2 and bh

= 3. Since its conjugate axis(which is horizontal and has length 2b

h

= 6) is longer than its transverse axis(length 2a

h

= 4), the ellipse is horizontal. Its major axis has length 2ae

= 6and its minor axis has length 2b

e

= 4, so ae

= 3 and be

= 2. The ellipse sharesthe same center as the hyperbola. Thus, the standard equation of the requiredellipse is

(x� 2)2

9+

(y + 3)2

4= 1.

7. If m 6= �3, 2, find the value(s) of m so that the graph of

(2m� 4)x2 + (m+ 3)y2 = (m+ 3)(2m� 4)

is

64

(a) a circle,

(b) a horizontal ellipse,

(c) a vertical ellipse,

(d) a hyperbola (is it horizontal or vertical?), or

(e) the empty set.

Answer: (a) m = 7, (b) 2 < m < 7, (c) m > 7, (d) �3 < m < 2 (horizontal),(e) m < �3

Solution. It might be helpful to observe that the equation is equivalent to

x2

m+ 3+

y2

2m� 4= 1.

(a) The graph is a circle if m+3 = 2m� 4 > 0 (positive, so the graph is nota point or the empty set). This happens if m = 7.

(b) We require 0 < 2m� 4 < m+ 3. Thus, 2 < m < 7.

(c) We require 0 < m+ 3 < 2m� 4. Thus, m > 7.

(d) We need m+3 and 2m�4 to have di↵erent signs. We consider two cases.

i. If m + 3 < 0 < 2m � 4, then m < �3 AND m > 2, which cannothappen.

ii. If 2m� 4 < 0 < m+ 3, then �3 < m < 2. In this case, the equation

can be written, with positive denominators, asx2

m+ 3� y2

4� 2m= 1.

The hyperbola is horizontal.

(e) The remaining case is when m < �3. In this case, m + 3 < 0 and2m� 4 < 0. This makes x2/(m+ 3) + y2/(2m� 4) negative, never equalto 1. The graph is then the empty set.

4

65

Lesson 1.6. Systems of Nonlinear Equations




(1) illustrate systems of nonlinear equations;

(2) determine the solutions of systems of nonlinear equations using techniquessuch as substitution, elimination, and graphing; and

(3) solve situational problems involving systems of nonlinear equations.

Lesson Outline

(1) Review systems of linear equations

(2) Solving a system involving one linear and one quadratic equation

(3) Solving a system involving two quadratic equations

(4) Applications of systems of nonlinear equations

Introduction

After recalling the techniques used in solving systems of linear equations inGrade 8, we extend these methods to solving a system of equations to systemsin which the equations are not necessarily linear. In this lesson, the equationsare restricted to linear and quadratic types, although it is possible to adapt themethodology so systems with other types of equations. We focus on quadraticequations for two reasons: to include a graphical representation of the solutionand to ensure that either a solution is obtained or it is determined that there isno solution. The latter is possible because of the quadratic formula.

Teaching Notes

Recall that thetask of solving a

system ofequations is

equivalent tofinding points of

intersection.

1.6.1. Review of Techniques in Solving Systems of LinearEquations

Recall the methods we used to solve systems of linear equations.Teaching Notes

Systems of linearequations and

solving them wereintroduced and

studied in Grade 8at the last part of

Quarter I.

There werethree methods used: substitution, elimination, and graphical.

Example 1.6.1. Use the substitution method to solve the system, and sketchthe graphs in one Cartesian plane showing the point of intersection.

8<

:4x+ y = 6

5x+ 3y = 4

Solution. Isolate the variable y in the first equation, and then substitute into thesecond equation.

66

4x+ y = 6

=) y = 6� 4x

5x+ 3y = 4

5x+ 3(6� 4x) = 4

�7x+ 18 = 4

x = 2

y = 6� 4(2) = �2

Example 1.6.2. Use the elimination method to solve the system, and sketch thegraphs in one Cartesian plane showing the point of intersection.

8<

:2x+ 7 = 3y

4x+ 7y = 12

Solution. We eliminate first the variable x. Rewrite the first equation whereinonly the constant term is on the right-hand side of the equation, then multiplyit by �2, and then add the resulting equation to the second equation.

2x� 3y = �7

(�2)(2x� 3y) = (�2)(�7)

�4x+ 6y = 14

�4x+ 6y = 14

4x+ 7y = 12

13y = 26

y = 2

x = �1

2


Use either substitution or elimination method to solve the system, and sketch thegraphs in one Cartesian plane showing the point of intersection.

1.

8<

:x� 3y = 5

2x+ 5y = �1

67

Answer: (2,�1)

2.

8<

:5x+ 3y = 4

3x+ 5y = 9

Answer:�12 ,

32

�

1.6.2. Solving Systems of Equations Using Substitution

We begin our extension with a system involving one linear equation and onequadratic equation. In this case, it is always possible to use substitution bysolving the linear equation for one of the variables.

Example 1.6.3. Solve the following system, and sketch the graphs in one Carte-sian plane. 8

<

:x� y + 2 = 0

y � 1 = x2

68

Solution. We solve for y in terms of x in the first equation, and substitute thisexpression to the second equation.

x� y + 2 = 0 =) y = x+ 2

y � 1 = x2

(x+ 2)� 1 = x2

x2 � x� 1 = 0

x =1±

p5

2

x =1 +

p5

2=) y =

1 +p5

2+ 2 =

5 +p5

2

x =1�

p5

2=) y =

1�p5

2+ 2 =

5�p5

2

Solutions:

1 +

p5

2,5 +

p5

2

!and

1�

p5

2,5�

p5

2

!

The first equation represents a line with x-intercept �2 and y-intercept 2,while the second equation represents a parabola with vertex at (0, 1) and whichopens upward.


Solve each system, and sketch the graphs in one Cartesian plane showing thepoint(s) of intersection.

1.

8<

:x2 + y2 = 16

x� y = 4

Answer: (4, 0) and (0,�4)

Solution. Solving for x in the second equation, we get x = y+4. Substitute

69

this expression into the first equation.

x2 + y2 = 16 =) (y + 4)2 + y2 = 16

y2 + 8y + 16 + y2 = 16

2y2 + 8y = 0

y = 0 or y = �4

Teaching Notes

We substitute eachvalue of y (0 and�4) to the secondequation x� y = 4

(or x = y + 4).

y = 0 =) x = 4 and y = �4 =) x = 0

Solutions: (4, 0) and (0,�4)

2.

8<

:y = x2

x = y2

Answer: (0, 0) and (1, 1)

Solution. Since the equations represent parabolas, we can use either of themto isolate one variable. This is in fact the form in which both equations aregiven. Substituting y = x2 into x = y2, we get

x = y2 =) x = (x2)2

x4 � x = 0

x(x3 � 1) = 0

x = 0 or x = 1

70

x = 0 =) y = 0 and x = 1 =) y = 1

Solutions: (0, 0) and (1, 1)

1.6.3. Solving Systems of Equations Using Elimination

Elimination method is also useful in systems of nonlinear equations. Sometimes,some systems need both techniques (substitution and elimination) to solve them.

Example 1.6.4. Solve the following system:8<

:y2 � 4x� 6y = 11

4(3� x) = (y � 3)2.

Solution 1. We expand the second equation, and eliminate the variable x byadding the equations. Teaching Notes

The variable y

could also beeliminated first bysubtracting thesecond equationfrom the first.

4(3� x) = (y � 3)2 =) 12� 4x = y2 � 6y + 9 =) y2 + 4x� 6y = 38<

:y2 � 4x� 6y = 11

y2 + 4x� 6y = 3

Adding these equations, we get

2y2�12y = 14 =) y2�6y�7 = 0 =) (y�7)(y+1) = 0 =) y = 7 or y = �1.

Teaching Notes

We may actuallysubstitute y = 7and y = �1 (one ata time) into any ofthe two givenequations, andthen solve for x.

Solving for x in the second equation, we have

x = 3� (y � 3)2

4.

y = 7 =) x = �1 and y = �1 =) x = �1

Solutions: (�1, 7) and (�1,�1) 2

71

The graphs of the equations in the preceding example with the points ofintersection are shown below.

Sometimes the solution can be simplified by writing the equations in standardform, although it is usually the general form which is more convenient to use insolving systems of equations. Moreover, the standard form is best for graphing.

We solve again the previous example in a di↵erent way.

Solution 2. By completing the square, we can change the first equation into stan-dard form:

y2 � 4x� 6y = 11 =) 4(x+ 5) = (y � 3)2.8<

:4(x+ 5) = (y � 3)2

4(3� x) = (y � 3)2

Using substitution or the transitive property of equality, we get

4(x+ 5) = 4(3� x) =) x = �1.

Substituting this value of x into the second equation, we have

4[3� (�1)] = (y � 3)2 =) 16 = (y � 3)2 =) y = 7 or y = �1.

The solutions are (�1, 7) and (�1,�1), same as Solution 1. 2

Example 1.6.5. Solve the system and graph the curves:8<

:(x� 3)2 + (y � 5)2 = 10

x2 + (y + 1)2 = 25.

72

Solution. Expanding both equations, we obtain8<

:x2 + y2 � 6x� 10y + 24 = 0

x2 + y2 + 2y � 24 = 0.

Subtracting these two equations, we get Teaching Notes

Because theequationx+ 2y � 8 = 0 isobtained bycombining the twoequations (throughsubstraction), thisequation alsocontains thesolutions of theoriginal system. Infact, this is the linepassing throughthe common pointsof the two circles.

�6x� 12y + 48 = 0 =) x+ 2y � 8 = 0

x = 8� 2y.

We can substitute x = 8� 2y to either the first equation or the second equation.For convenience, we choose the second equation.

x2 + y2 + 2y � 24 = 0

(8� 2y)2 + y2 + 2y � 24 = 0

y2 � 6y + 8 = 0

y = 2 or y = 4

y = 2 =) x = 8� 2(2) = 4 and y = 4 =) x = 8� 2(4) = 0

The solutions are (4, 2) and (0, 4).

The graphs of both equations are circles. One has center (3, 5) and radiusp10, while the other has center (0,�1) and radius 5. The graphs with the points

of intersection are show below.

73


Solve the system, and graph the curves in one Cartesian plane showing thepoint(s) of intersection.

1.

8<

:x2 + y2 = 25

x2

18+

y2

32= 1

Answer: (3, 4), (�3, 4), (3,�4), and (�3,�4)

2.

8<

:x2 + 2y � 12 = 0

x2 + y2 = 36

Answer: (0, 6),�2p5,�4

�, and

��2

p5,�4

�

74

3.

8<

:(x� 1)2 + (y � 3)2 = 10

x2 + (y � 1)2 = 5

Answer: (�2, 2) and (2, 0)

1.6.4. Applications of Systems of Nonlinear Equations

As we expect, systems of equations are important in applications. In this session,we consider some of them.

?Example 1.6.6. The screen size of television sets is given in inches. Thisindicates the length of the diagonal. Screens of the same size can come in di↵erentshapes. Wide-screen TV’s usually have screens with aspect ratio 16 : 9, indicatingthe ratio of the width to the height. Older TV models often have aspect ratio4 : 3. A 40-inch LED TV has screen aspect ratio 16 : 9. Find the length and thewidth of the screen.

Solution. Let w represent the width and h the height of the screen. Then, byPythagorean Theorem, we have the system

8<

:w2 + h2 = 402 =) w2 + h2 = 1600w

h=

16

9=) h =

9w

16

75

w2 + h2 = 1600 =) w2 +

✓9w

16

◆2

= 1600

337w2

256= 1600

w =

r409 600

337⇡ 34.86

h =19x

16⇡ 19(34.86)

16= 19.61

Therefore, a 40-inch TV with aspect ratio 16 : 9 is about 35.86 inches wide and19.61 inches high. 2


1. From a circular piece of metal sheet with diameter 20 cm, a rectangularpiece with perimeter 28 cm is to be cut as shown. Find the dimensions ofthe rectangular piece. Answer: 6 cm⇥ 8 cm

Exercises 1.6

1. Solve the system, and graph the curves.

(a)

8<

:y = 2x+ 4

y = 2x2

Answer: (�1, 2) and (2, 8)

76

(b)

8<

:x2 + y2 = 25

2x� 3y = �6

Answer: (3, 4) and��63

13 ,�1613

�

(c)

8<

:x2 + y2 = 12

x2 � y2 = 4

Answer:�2p2, 2

�,��2

p2, 2

�,�2p2,�2

�, and

��2

p2,�2

�

77

(d)

8<

:x2 � 4y2 = 200

x+ 2y = 100

Answer:�51, 492

�

(e)

8<

:

14(x+ 1)2 � (y + 2)2 = 1

(y + 2)2 = �14(x� 1)

Answer: (1,�2),⇣�4,�2 +

p52

⌘, and

⇣�4,�2�

p52

⌘

78

?2. A laptop has screen size 13 inches with aspect ratio 5 : 4. Find the length andthe width of the screen. Answer: 10.15 in⇥ 8.12 in

?3. What are the dimensions of a rectangle whose perimeter is 50 cm and diagonal18 cm? Answer: 14.9 cm⇥ 10.1 cm

4. The graph of 2xy�y2+5x+20 = 0 is a rotated hyperbola. Find the intersectionof this hyperbola with the graph of 3x+ 2y = 3. (The graph is not required.)

Answer: (�1, 3),�7121 ,�

257

�

5. For what values of a will the system8<

:x2 + y2 + 2x� 1 = 0

x� y + a = 0

have only one solution? Answer: a = �1 or a = 3

4

79

Unit 2

Mathematical Induction

https://commons.wikimedia.org/wiki/File%3ABatad rice terraces in Ifugao.jpg

By Ericmontalban (Own work)[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)],

via Wikimedia Commons

Listed as one of the UNESCOWorld Heritage sites since 1995, the two-millennium-old Rice Terraces of the Philippine Cordilleras by the Ifugaos is a living testimonyof mankind’s creative engineering to adapt with physically-challenging environ-ment in nature. One of the five clusters of terraces inscribed in the UNESCO listis the majestic Batad terrace cluster (shown above), which is characterized by itsamphitheater-like semi-circular terraces with a village at its base.

https://commons.wikimedia.org/wiki/File%3ABatad_rice_terraces_in_Ifugao.jpg

Lesson 2.1. Review of Sequences and Series

Time Frame: 1 one-hour session



(1) illustrate a series; and

(2) di↵erentiate a series from a sequence.

Lesson Outline

(1) Sequences and series

(2) Di↵erent types of sequences and series (Fibonacci sequence, arithmetic andgeometric sequence and series, and harmonic series)

(3) Di↵erence between sequence and series

Introduction

Pose the following problem to the class:

Jason’s classroom is on the second floor of the school. He can takeone or two steps of the stairs in one leap. In how many ways canJason climb the stairs if it has 16 steps?

Get students to suggest strategies they can use to solve this problem. Leador encourage them to try out smaller number of steps and find a pattern. Workwith the class to complete the following table (on the board):

81

Teaching Notes

This is equivalentto the number ofways to express a

number (number ofsteps in the stairs)as a sum of 1’s and2’s. For example,

we can write 3 as asum of 1’s and 2’s

in three ways:2 + 1, 1 + 2, and

1 + 1 + 1. In 2 + 1,it means Jason

leaps 2 steps first,then 1 step to

finish thethree-step stairs.

Number of Stepsin the Stairs

Number of Waysto Climb the Stairs

1 1

2 2

3 3

4 5

5 8...

...

The students should be able to recognize the Fibonacci sequence. Ask thestudents to recall what Fibonacci sequences are and where they had encounteredthis sequence before.

In this lesson, we will review the definitions and di↵erent types of sequencesand series.

Lesson Proper

Recall the following definitions:

A sequence is a function whose domain is the set of positive integersor the set {1, 2, 3, . . . , n}.

A series represents the sum of the terms of a sequence.

If a sequence is finite, we will refer to the sum of the terms of thesequence as the series associated with the sequence. If the sequence hasinfinitely many terms, the sum is defined more precisely in calculus.

A sequence is a list of numbers (separated by commas), while a series is asum of numbers (separated by “+” or “�” sign). As an illustration, 1,�1

2 ,13 ,�

14

is a sequence, and 1� 12 +

13 �

14 = 7

12 is its associated series.

The sequence with nth term an

is usually denoted by {an

}, and the associatedseries is given by

S = a1 + a2 + a3 + · · ·+ an

.

82

Example 2.1.1. Determine the first five terms of each defined sequence, andgive their associated series.(1) {2� n}

(2) {1 + 2n+ 3n2}

(3) {(�1)n}

(4) {1 + 2 + 3 + · · ·+ n}

Solution. We denote the nth term of a sequence by an

, and S = a1 + a2 + a3 +a4 + a5.

(1) an

= 2� n

First five terms: a1 = 2� 1 = 1, a2 = 2� 2 = 0, a3 = �1, a4 = �2, a5 = �3

Associated series: S = a1 + a2 + a3 + a4 + a5 = 1 + 0� 1� 2� 3 = �5

(2) an

= 1 + 2n+ 3n2

First five terms: a1 = 1+2 · 1+3 · 12 = 6, a2 = 17, a3 = 34, a4 = 57, a5 = 86

Associated series: S = 6 + 17 + 34 + 57 + 86 = 200

(3) an

= (�1)n

First five terms: a1 = (�1)1 = �1, a2 = (�1)2 = 1, a3 = �1, a4 = 1,a5 = �1

Associated series: S = �1 + 1� 1 + 1� 1 = �1

(4) an

= 1 + 2 + 3 + · · ·+ n

First five terms: a1 = 1, a2 = 1+2 = 3, a3 = 1+2+3 = 6, a4 = 1+2+3+4 =10, a5 = 1 + 2 + 3 + 4 + 5 = 15

Associated series: S = 1 + 3 + 6 + 10 + 15 = 35 2

An arithmetic sequence is a sequence in which each term after the firstis obtained by adding a constant (called the common di↵erence) to thepreceding term.

If the nth term of an arithmetic sequence is an

and the common di↵erence isd, then

an

= a1 + (n� 1)d.

The associated arithmetic series with n terms is given by

Sn

=n(a1 + a

n

)

2=

n[2a1 + (n� 1)d]

2.

83

A geometric sequence is a sequence in which each term after the firstis obtained by multiplying the preceding term by a constant (calledthe common ratio).

If the nth term of a geometric sequence is an

and the common ratio is r, then

an

= a1rn�1.

The associated geometric series with n terms is given by

Sn

=

8><

>:

na1 if r = 1a1(1� rn)

(1� r)if r 6= 1.

The proof of this sum formula is an example in Lesson 2.3.

When �1 < r < 1, the infinite geometric seriesTeaching Notes

The proof of thefact that the

infinite geometricseries

a1 + a1r + · · · hasa sum when |r| < 1is beyond the scopeof Precalculus, and

can be found inuniversityCalculus.

a1 + a1r + a1r2 + · · ·+ a1r

n�1 + · · ·

has a sum, and is given by

S =a1

1� r.

If {an

} is an arithmetic sequence, then the sequence with nth termbn

= 1an

is a harmonic sequence.

Seatwork/Homework

1. Write SEQ if the given item is a sequence, and write SER if it is a series.

(a) 1, 2, 4, 8, . . . Answer: SEQ

(b) 2, 8, 10, 18, . . . Answer: SEQ

(c) �1 + 1� 1 + 1� 1 Answer: SER

(d) 12 ,

23 ,

34 ,

45 , . . . Answer: SEQ

(e) 1 + 2 + 22 + 23 + 24 Answer: SER

(f) 1 + 0.1 + 0.001 + 0.0001 Answer: SER

2. Write A if the sequence is arithmetic, G if it is geometric, F if Fibonacci, andO if it is not one of the mentioned types.

(a) 3, 5, 7, 9, 11, . . . Answer: A

84

(b) 2, 4, 9, 16, 25, . . . Answer: O

(c) 14 ,

116 ,

164 ,

1256 , . . . Answer: G

(d) 13 ,

29 ,

327 ,

481 , . . . Answer: O

(e) 15 ,

19 ,

113 ,

117 ,

121 , . . . Answer: A

(f) 4, 6, 10, 16, 26, . . . Answer: F

(g)p3,p4,p5,p6, . . . Answer: O

(h) 0.1, 0.01, 0.001, 0.0001, . . . Answer: G

3. Determine the first five terms of each defined sequence, and give their associ-ated series.

(a) {1 + n� n2}Answer: a1 = 1, a2 = �1, a3 = �5, a4 = �11, a5 = �19

Associated series: 1� 1� 5� 11� 19 = �35

(b) {1� (�1)n+1}Answer: a1 = 0, a2 = 2, a3 = 0, a4 = 2, a5 = 0

Associated series: 0 + 2 + 0 + 2 + 0 = 4

(c) a1 = 3 and an

= 2an�1 + 3 for n � 2

Answer: a1 = 3, a2 = 9, a3 = 21, a4 = 45, a5 = 93

Associated series: 1� 1� 5� 11� 19 = �35

(d) {1 · 2 · 3 · · ·n}Answer: a1 = 1, a2 = 1 · 2 = 2, a3 = 1 · 2 · 3 = 6, a4 = 24, a5 = 120

Associated series: 1 + 2 + 6 + 24 + 120 = 153

4. Identify the series (and write NAGIG if it is not arithmetic, geometric, andinfinite geometric series), and determine the sum (and write NO SUM if itcannot be summed up).

(a) 4 + 9 + 14 + · · ·+ 64 Answer: Arithmetic, 442

(b) 81 + 27 + 9 + · · ·+ 181 Answer: Geometric, 9841

81

(c) 1 + 3 + 6 + 10 + 15 + 21 + · · ·+ 55 Answer: NAGIG, 220

(d) �10� 2 + 6 + · · ·+ 46 Answer: Arithmetic, 144

(e) 10 + 2 + 0.4 + 0.08 + · · · Answer: Infinite geometric, 12.5

(f) 12 +

13 +

15 +

17 + · · · Answer: NAGIG, NO SUM

(g) 1� 0.1 + 0.01� 0.001 + · · · Answer: Infinite geometric, 1011

4

85

Lesson 2.2. Sigma Notation



At the end of the lesson, the student is able to use the sigma notation torepresent a series.

Lesson Outline

(1) Definition of and writing in sigma notation

(2) Evaluate sums written in sigma notation

(3) Properties of sigma notation

(4) Calculating sums using the properties of sigma notation

Introduction

The sigma notation is a shorthand for writing sums. In this lesson, we willsee the power of this notation in computing sums of numbers as well as algebraicexpressions.

2.2.1. Writing and Evaluating Sums in Sigma Notation

Mathematicians use the sigma notation to denote a sum. The uppercase Greekletter ⌃ (sigma) is used to indicate a “sum.” The notation consists of severalcomponents or parts.

Let f(i) be an expression involving an integer i. The expression

f(m) + f(m+ 1) + f(m+ 2) + · · ·+ f(n)

Teaching Notes

Emphasize thatthe value of i startsat m, increases by1, and ends at n.

can be compactly written in sigma notation, and we write it as

nX

i=m

f(i),

which is read “the summation of f(i) from i = m to n.” Here, mand n are integers with m n, f(i) is a term (or summand) of thesummation, and the letter i is the index, m the lower bound, and nthe upper bound.

Example 2.2.1. Expand each summation, and simplify if possible.

86

(1)4X

i=2

(2i+ 3)

(2)5X

i=0

2i

(3)nX

i=1

ai

(4)6X

n=1

pn

n+ 1

Solution. We apply the definition of sigma notation.

(1)4X

i=2

(2i+ 3) = [2(2) + 3] + [2(3) + 3] + [2(4) + 3] = 27

(2)5X

i=0

2i = 20 + 21 + 22 + 23 + 24 + 25 = 63

(3)nX

i=1

ai

= a1 + a2 + a3 + · · ·+ an

(4)6X

n=1

pn

n+ 1=

1

2+

p2

3+

p3

4+

2

5+

p5

6+

p6

72

Example 2.2.2. Write each expression in sigma notation.

(1) 1 +1

2+

1

3+

1

4+ · · ·+ 1

100

(2) �1 + 2� 3 + 4� 5 + 6� 7 + 8� 9 + · · ·� 25

(3) a2 + a4 + a6 + a8 + · · ·+ a20

(4) 1 +1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128

Solution. (1) 1 +1

2+

1

3+

1

4+ · · ·+ 1

100=

100X

n=1

1

n

(2) �1 + 2� 3 + 4� 5 + · · ·� 25

= (�1)1 1 + (�1)2 2 + (�1)3 3 + (�1)4 4

+ (�1)5 5 + · · ·+ (�1)25 25

=25X

j=1

(�1)jj

(3) a2 + a4 + a6 + a8 + · · ·+ a20= a2(1) + a2(2) + a2(3) + a2(4) + · · ·+ a2(10)

=10X

i=1

a2i

87

(4) 1 +1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128=

7X

k=0

1

2k2

The sigma notation of a sum expression is not necessarily unique. For ex-ample, the last item in the preceding example can also be expressed in sigmanotation as follows:

1 +1

2+

1

4+

1

8+

1

16+

1

32+

1

64+

1

128=

8X

k=1

1

2k�1.

However, this last sigma notation is equivalent to the one given in the example.


1. Expand each summation, and simplify if possible.

(a)5X

k=�1

(2� 3k) Answer: �28

(b)nX

j=1

xj Answer: x+ x2 + x3 + · · ·+ xn

(c)6X

j=3

(j2 � j) Answer: 68

(d)4X

k=1

(�1)k+1k Answer: �2

(e)3X

n=1

(an+1 � a

n

) Answer: a4 � a1

2. Write each expression in sigma notation.

(a) x+ 2x2 + 3x3 + 4x4 + 5x5 Answer:5X

k=1

kxk

(b) 1� 2 + 3� 4 + 5� 6 + · · ·� 10 Answer:10X

k=1

(�1)k+1 kTeaching Notes

Equivalent answer:1+3+5+· · ·+101 =

51X

k=1

(2k � 1)

(c) 1 + 3 + 5 + 7 + · · ·+ 101 Answer:50X

k=0

(2k + 1)

(d) a4 + a8 + a12 + a16 Answer:4X

k=1

a4k

(e) 1� 1

3+

1

5� 1

7+

1

9Answer:

4X

k=0

(�1)k

2k + 1

88

2.2.2. Properties of Sigma Notation

We start with finding a formula for the sum of

nX

i=1

i = 1 + 2 + 3 + · · ·+ n

in terms of n.

The sum can be evaluated in di↵erent ways. A simple, though informal,approach is pictorial.

Teaching Notes

This illustrationcan be done withmanipulatives, andallow the studentsto guess.

nX

i=1

i = 1 + 2 + 3 + · · ·+ n =n(n+ 1)

2

Another way is to use the formula for an arithmetic series with a1 = 1 andan

= n:

S =n(a1 + a

n

)

2=

n(n+ 1)

2.

We now derive some useful summation facts. They are based on the axiomsof arithmetic addition and multiplication.

nX

i=m

cf(i) = cnX

i=m

f(i), c any real number.

Proof. Teaching Notes

Some proofs can beskipped. However,it is helpful if theyare all discussed inclass.

nX

i=m

cf(i) = cf(m) + cf(m+ 1) + cf(m+ 2) + · · ·+ cf(n)

89

= c[f(m) + f(m+ 1) + · · ·+ f(n)]

= cnX

i=m

f(i) 2

nX

i=m

[f(i) + g(i)] =nX

i=m

f(i) +nX

i=m

g(i)

Proof.

nX

i=m

[f(i) + g(i)]

= [f(m) + g(m)] + · · ·+ [f(n) + g(n)]

= [f(m) + · · ·+ f(n)] + [g(m) + · · ·+ g(n)]

=nX

i=m

f(i) +nX

i=m

g(i) 2

nX

i=m

c = c(n�m+ 1)

Proof.

nX

i=m

c = c+ c+ c+ · · ·+ c| {z }n�m+1 terms

= c(n�m+ 1) 2

A special case of the above result which you might encounter more often isthe following:

nX

i=1

c = cn.

Telescoping Sum

nX

i=m

[f(i+ 1)� f(i)] = f(n+ 1)� f(m)

90

Proof.

nX

i=m

⇥f(i+ 1)� f(i)

⇤

= [f(m+ 1)� f(m)] + [f(m+ 2)� f(m+ 1)]

+ [f(m+ 3)� f(m+ 2)] + · · ·+ [f(n+ 1)� f(n)]

Note that the terms, f(m+1), f(m+2), . . . , f(n), all cancel out. Hence, we have

nX

i=m

[f(i+ 1)� f(i)] = f(n+ 1)� f(m). 2

Example 2.2.3. Evaluate:30X

i=1

(4i� 5).

Solution.

30X

i=1

(4i� 5) =30X

i=1

4i�30X

i=1

5

= 430X

i=1

i�30X

i=1

5

= 4(30)(31)

2� 5(30)

= 1710 2

Example 2.2.4. Evaluate:

1

1 · 2 +1

2 · 3 +1

3 · 4 + · · ·+ 1

99 · 100 .

Solution.

1

1 · 2 +1

2 · 3 +1

3 · 4 + · · ·+ 1

99 · 100

=99X

i=1

1

i(i+ 1)

=99X

i=1

i+ 1� i

i(i+ 1)

=99X

i=1

i+ 1

i(i+ 1)� i

i(i+ 1)

�

91

=99X

i=1

✓1

i� 1

i+ 1

◆

= �99X

i=1

✓1

i+ 1� 1

i

◆

Using f(i) =1

iand the telescoping-sum property, we get

99X

i=1

1

i(i+ 1)= �

✓1

100� 1

1

◆=

99

100. 2

Example 2.2.5. Derive a formula fornX

i=1

i2 using a telescoping sum with terms

f(i) = i3.

Solution. The telescoping sum property implies that

nX

i=1

⇥i3 � (i� 1)3

⇤= n3 � 03 = n3.

On the other hand, using expansion and the other properties of summation,we have

nX

i=1

⇥i3 � (i� 1)3

⇤=

nX

i=1

(i3 � i3 + 3i2 � 3i+ 1)

= 3nX

i=1

i2 � 3nX

i=1

i+nX

i=1

1

= 3nX

i=1

i2 � 3 · n(n+ 1)

2+ n.

Equating the two results above, we obtain

3nX

i=1

i2 � 3n(n+ 1)

2+ n = n3

6nX

i=1

i2 � 3n(n+ 1) + 2n = 2n3

6nX

i=1

i2 = 2n3 � 2n+ 3n(n+ 1)

92

= 2n(n2 � 1) + 3n(n+ 1)

= 2n(n� 1)(n+ 1) + 3n(n+ 1)

= n(n+ 1)[2(n� 1) + 3]

= n(n+ 1)(2n+ 1).

Finally, after dividing both sides of the equation by 6, we obtain the desiredformula

nX

i=1

i2 =n(n+ 1)(2n+ 1)

6. 2


1. Use the properties of sigma notation to evaluate the following summations.

(a)50X

k=1

(2� 3k) Answer: �3725

(b)nX

j=1

(1 + 2j) Answer: 2n+ n2

(c)99X

j=1

1pi+ 1 +

pi

Answer: 9

Solution:

99X

j=1

1pi+ 1 +

pi=

99X

j=1

1pi+ 1 +

pi·pi+ 1�

pip

i+ 1�pi

=99X

j=1

⇣pi+ 1�

pi⌘

=p99 + 1�

p1

= 9

2. IfnX

i=1

(i+ 1)2 = an3 + bn2 + cn+ d, what is a+ b+ c+ d? Answer: 4

Exercises 2.2

1. Expand each sum.

(a)9X

i=5

i

x+ iAnswer:

1

x+ 1+

2

x+ 2+

3

x+ 3+

4

x+ 4+

5

x+ 5

93

(b)6X

i=0

3p2i Answer: 0 + 3

p2 + 3

p4 + 3

p6 + 2 + 3

p10 + 3

p12

(c)3X

i=�2

3�i Answer: 9 + 3 + 1 + 1/3 + 1/9 + 1/27

2. Write each expression in sigma notation.

(a) 1 + 22 + 33 + 44 + · · ·+ 1212 Answer:12X

i=1

ii

(b) (x� 5) + (x� 3) + (x� 1) + (x+ 1) + (x+ 3) + · · ·+ (x+ 15)Teaching Notes

Another possibleanswer for (b) is11X

i=1

[x+ (2i� 7)]. Answer:7X

i=�3

[x+ (2i+ 1)]

(c) a1 + a4 + a9 + a16 + · · ·+ a81 Answer:9X

i=1

ai

2

3. Evaluate each sum.

(a)120X

i=1

(4i� 15) Answer: 27240

(b)50X

i=1

[(5i� 2)(i+ 3)] Answer: 230900

(c)nX

i=1

(3i� 1)2 Answer:6n3 + 3n2 � 3n+ 2

2

4. If30X

i=1

f(i) = 70 and30X

i=1

g(i) = 50, what is the value of30X

i=1

3g(i)� f(i) + 7

2?

Answer: 145

5. If s =100X

i=1

i, express200X

i=1

i in terms of s. Answer: 2s+ 100000

6. If s =nX

i=1

ai

, does it follow thatnX

i=1

a2i

= s2?

Answer: No. If s =2X

i=1

ai

= a1 + a2, thennX

i=1

a2i

= a21 + a22, while s2 =

a21 + 2a1a2 + a22.

94

7. Derive a formula fornX

i=1

i3 by using a telescoping sum with terms f(i) = i4.

Answer:n2(n+ 1)2

4

4

Lesson 2.3. Mathematical Induction




(1) illustrate the Principle of Mathematical Induction; and

(2) apply mathematical induction in proving identities.

Lesson Outline

(1) State the principle of mathematical induction

(2) Prove summation identities using mathematical induction

(3) Prove divisibility statements using mathematical induction

(4) Prove inequalities using mathematical induction

Introduction

We have derived and used formulas for the terms of arithmetic and geometricsequences and series. These formulas and many other theorems involving positiveintegers can be proven with the use of a technique called mathematical induction.

2.3.1. Proving Summation Identities

The Principle of Mathematical Induction

Let P (n) be a property or statement about an integer n. Supposethat the following conditions can be proven:

(1) P (n0) is true (that is, the statement is true when n = n0).

(2) If P (k) is true for some integer k � n0, then P (k + 1) is true(that is, if the statement is true for n = k, then it is also true forn = k + 1).

Then the statement P (n) is true for all integers n � n0.

95

The Principle of Mathematical Induction is often compared to climbing aninfinite staircase. First, you need to be able to climb up to the first step. Second,if you are on any step (n = k), you must be able to climb up to the next step(n = k + 1). If you can do these two things, then you will be able to climb upthe infinite staircase.

Part 1 Part 2

Another analogy of the Principle of Mathematical Induction that is used istoppling an infinite line of standing dominoes. You need to give the first dominoa push so that it falls down. Also, the dominoes must be arranged so that if thekth domino falls down, the next domino will also fall down. These two conditionswill ensure that the entire line of dominoes will fall down.

https://commons.wikimedia.org/wiki/File:Wallpaper kartu domino.png

By Nara Cute (Own work)[CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)],


96

https://commons.wikimedia.org/wiki/File:Wallpaper_kartu_domino.png

There are many mathematical results that can be proven using mathematicalinduction. In this lesson, we will focus on three main categories: summationidentities, divisibility statements, and inequalities.

We now consider some examples on the use of mathematical induction inproving summation identities.

Example 2.3.1. Using mathematical induction, prove that

1 + 2 + 3 + · · ·+ n =n(n+ 1)

2

for all positive integers n.

Solution. We need to establish the two conditions stated in the Principle of Math-ematical Induction.

Part 1. Prove that the identity is true for n = 1.

The left-hand side of the equation consists of one term equal to 1. The right-hand side becomes

1(1 + 1)

2=

2

2= 1.

Hence, the formula is true for n = 1.

Part 2. Assume that the formula is true for n = k � 1:

1 + 2 + 3 + · · ·+ k =k(k + 1)

2.

We want to show that the formula is true for n = k + 1; that is,

1 + 2 + 3 + · · ·+ k + (k + 1) =(k + 1)(k + 1 + 1)

2.

Using the formula for n = k and adding k + 1 to both sides of the equation,we get

1 + 2 + 3 + · · ·+ k + (k + 1) =k(k + 1)

2+ (k + 1)

=k(k + 1) + 2(k + 1)

2

=(k + 1)(k + 2)

2

=(k + 1) [(k + 1) + 1]

2

We have proven the two conditions required by the Principle of MathematicalInduction. Therefore, the formula is true for all positive integers n. 2

97

Example 2.3.2. Use mathematical induction to prove the formula for the sumof a geometric series with n terms:

Sn

=a1 (1� rn)

1� r, r 6= 1.

Solution. Let an

be the nth term of a geometric series. From Lesson 2.1, we knowthat a

n

= a1rn�1.Teaching Notes

The fact thatan = a1r

n�1 canalso be proven by

mathematicalinduction. Here,

however, we simplyrecall a formula inLesson 2.1 because

our focus in thisexample is the

proof of the sum.

Part 1. Prove that the formula is true for n = 1.

a1(1� r1)

1� r= a1 = S1

The formula is true for n = 1.

Part 2. Assume that the formula is true for n = k � 1: Sk

=a1(1� rk)

1� r. We

want to prove that it is also true for n = k + 1; that is,

Sk+1 =

a1(1� rk+1)

1� r.

We know that

Sk+1 = a1 + a2 + · · ·+ a

k| {z }Sk

+ak+1

= Sk

+ ak+1

=a1�1� rk

�

1� r+ a1r

k

=a1�1� rk

�+ a1rk (1� r)

1� r

=a1�1� rk + rk � rk+1

�

1� r

=a1�1� rk+1

�

1� r

By the Principle of Mathematical Induction, we have proven that

Sn

=a1(1� rn)

1� r

for all positive integers n. 2

Example 2.3.3. Using mathematical induction, prove that

12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all positive integers n.

98

Solution. We again establish the two conditions stated in the Principle of Math-ematical Induction.

Part 11(1 + 1)(2 · 1 + 1)

6=

1 · 2 · 36

= 1 = 12


Part 2

Assume: 12 + 22 + 32 + · · ·+ k2 =k(k + 1)(2k + 1)

6.

Prove: 12 + 22 + 32 + · · ·+ k2 + (k + 1)2

=(k + 1)(k + 2) [2(k + 1) + 1]

6

=(k + 1)(k + 2)(2k + 3)

6.

12 + 22 + 32 + · · ·+ k2 + (k + 1)2

=k(k + 1)(2k + 1)

6+ (k + 1)2

=k(k + 1)(2k + 1) + 6(k + 1)2

6

=(k + 1) [k(2k + 1) + 6(k + 1)]

6

=(k + 1) (2k2 + 7k + 6)

6

=(k + 1)(k + 2)(2k + 3)

6

Therefore, by the Principle of Mathematical Induction,

12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all positive integers n. 2


Using mathematical induction, prove that

1 · 3 + 2 · 4 + 3 · 5 + · · ·+ n(n+ 2) =n(n+ 1)(2n+ 7)

6.

99

Answer:

Part 11(1 + 1)[2(1) + 7]

6=

2 · 96

= 3 = 1 · 3


Part 2

Assume: 1 · 3 + 2 · 4 + 3 · 5 + · · ·+ k(k + 2) =k(k + 1)(2k + 7)

6

To show: 1 · 3 + 2 · 4 + · · ·+ k(k + 2) + (k + 1)(k + 3)

=(k + 1)(k + 2) [2(k + 1) + 7]

6

=(k + 1)(k + 2)(2k + 9)

6

1 · 3 + 2 · 4 + · · ·+ k(k + 2) + (k + 1)(k + 3)

=k(k + 1)(2k + 7)

6+ (k + 1)(k + 3)

=(k + 1)

6[k(2k + 7) + 6(k + 3)]

=(k + 1)

6

⇥2k2 + 13k + 18

⇤

=(k + 1)(k + 2)(2k + 9)

6

Therefore, by the Principle of Math Induction, the formula is true for all positiveintegers n.

Teaching Notes

Recall thedefinition of

divisibility: aninteger n is

divisible by aninteger k if n = kr

for some integer r.

2.3.2. Proving Divisibility Statements

We now prove some divisibility statements using mathematical induction.

Example 2.3.4. Use mathematical induction to prove that, for every positiveinteger n, 7n � 1 is divisible by 6.

Solution. Similar to what we did in the previous session, we establish the twoconditions stated in the Principle of Mathematical Induction.

Part 1

71 � 1 = 6 = 6 · 171 � 1 is divisible by 6.

100

Part 2

Assume: 7k � 1 is divisible by 6.

To show: 7k+1 � 1 is divisible by 6.

7k+1 � 1 = 7 · 7k � 1 = 6 · 7k + 7k � 1 = 6 · 7k + (7k � 1)

By definition of divisibility, 6 · 7k is divisible by 6. Also, by the hypothesis(assumption), 7k � 1 is divisible by 6. Hence, their sum (which is equal to7k+1 � 1) is also divisible by 6.

Therefore, by the Principle of Math Induction, 7n � 1 is divisible by 6 for allpositive integers n. 2

Note that 70 � 1 = 1� 1 = 0 = 6 · 0 is also divisible by 6. Hence, a strongerand more precise result in the preceding example is: 7n � 1 is divisible by 6 forevery nonnegative integer n. It does not make sense to substitute negative valuesof n since this will result in non-integer values for 7n � 1.

Example 2.3.5. Use mathematical induction to prove that, for every nonnega-tive integer n, n3 � n+ 3 is divisible by 3.

Solution. We again establish the two conditions in the Principle of MathematicalInduction.

Part 1 Note that claim of the statement is that it is true for every nonnegativeinteger n. This means that Part 1 should prove that the statement is true forn = 0.

03 � 0 + 3 = 3 = 3(1)

03 � 0 + 3 is divisible by 3.

Part 2. We assume that k3 � k+3 is divisible by 3. By definition of divisibility,we can write k3 � k + 3 = 3a for some integer a.

To show: (k + 1)3 � (k + 1) + 3 is divisible by 3.

(k + 1)3 � (k + 1) + 3 = k3 + 3k2 + 2k + 3

= (k3 � k + 3) + 3k2 + 3k

= 3a+ 3k2 + 3k

= 3(a+ k2 + k)

Since a+k2+k is also an integer, by definition of divisibility, (k+1)3�(k+1)+3is divisible by 3.

101

Therefore, by the Principle of Math Induction, n3 �n+3 is divisible by 3 forall positive integers n. 2


Use mathematical induction to prove each divisibility statement for all nonnega-tive integers n.

(1) 72n � 3 · 5n + 2 is divisible by 12.

Answer:

Part 1

72(0) � 3 · 50 + 2 = 1� 3(1) + 2 = 0 = 12(0)

72(0) � 3 · 50 + 2 is divisible by 12

Part 2

Assume: 72k � 3 · 5k + 2 is divisible by 12

To show: 72(k+1) � 3 · 5(k+1) + 2 is divisible by 12

72(k+1) � 3 · 5(k+1) + 2

= 7272k � 3 · 5 · 5k + 2

= 49 · 72k � 15 · 5k + 2

= 72k + 48 · 72k � 3 · 5k � 12 · 5k + 2

=�72k � 3 · 5k + 2

�+ 48 · 72k � 12 · 5k

=�72k � 3 · 5k + 2

�+ 12

�4 · 72k � 5k

�

By the hypothesis, 72k � 3 · 5k + 2 is divisible by 12. The second term,12�4 · 72k � 5k

�, is divisible by 12 because 4 · 72k � 5k is an integer. Hence

their sum, which is equal to 72(k+1) � 3 · 5(k+1) + 2, is divisible by 12.

Therefore, by the Principle of Math Induction, 72n � 3 · 5n + 2 is divisible by12 for every nonnegative integer n.

(2) n3 + 3n2 + 2n is divisible by 3.

Answer:

Part 1

03 + 3 · 02 + 2(0) = 0 = 3(0)

Thus, 03 + 3 · 02 + 2(0) is divisible by 3.

102

Part 2

Assume: k3 + 3k2 + 2k is divisible by 3.

=) k3 + 3k2 + 2k = 3a, a integer

To show: (k + 1)3 + 3(k + 1)2 + 2(k + 1) is divisible by 3.

(k + 1)3 + 3(k + 1)2 + 2(k + 1)

= k3 + 6k2 + 11k + 6

= (k3 + 3k2 + 2k) + 3k2 + 9k + 6

= 3a+ 3k2 + 9k + 6

= 3(a+ k2 + 3k + 2)

Since a+ k2+3k+2 is also an integer, by definition of divisibility, (k+1)3+3(k + 1)2 + 2(k + 1) is divisible by 3.

Therefore, by the Principle of Math Induction, n3 + 3n2 + 2n is divisible by3 for all positive integers n.

?2.3.3. Proving Inequalities

Finally, we now apply the Principle of Mathematical Induction in proving someinequalities involving integers.

Example 2.3.6. Use mathematical induction to prove that 2n > 2n for everyinteger n � 3.

Solution. Just like the previous example, we establish the two conditions in thePrinciple of Mathematical Induction.

Part 1

23 = 8 > 6 = 2(3)

This confirms that 23 > 2(3).

Part 2

Assume: 2k > 2k, where k is an integer with k � 3

To show: 2k+1 > 2(k + 1) = 2k + 2

We compare the components of the assumption and the inequality we need toprove. On the left-hand side, the expression is doubled. On the right-hand side,the expression is increased by 2. We choose which operation we want to apply toboth sides of the assumed inequality.

103

Alternative 1. We double both sides.

Since 2k > 2k, by the multiplication property of inequality, we have 2 · 2k >2 · 2k.

2k+1 > 2(2k) = 2k + 2k > 2k + 2 if k � 3.

Hence, 2k+1 > 2(k + 1).

Alternative 2. We increase both sides by 2.

Since 2k > 2k, by the addition property of inequality, we have 2k+2 > 2k+2.

2(k + 1) = 2k + 2 < 2k + 2 < 2k + 2k if k � 3.

The right-most expression above, 2k + 2k, is equal to 2�2k�= 2k+1.

Hence, 2(k + 1) < 2k+1.

Therefore, by the Principle of Math Induction, 2n > 2n for every integern � 3. 2

We test the above inequality for integers less than 3.

20 = 1 > 0 = 2(0) True

21 = 2 = 2(1) False

22 = 4 = 2(2) False

The inequality is not always true for nonnegative integers less than 3. Thisillustrates the necessity of Part 1 of the proof to establish the result. However,the result above can be modified to: 2n � 2n for all nonnegative integers n.

Before we discuss the next example, we review the factorial notation. Recallthat 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · ·n. The factorial alsosatisfies the property that (n+ 1)! = (n+ 1) · n!.

Example 2.3.7. Use mathematical induction to prove that 3n < (n + 2)! forevery positive integer n. Can you refine or improve the result?

Solution. We proceed with the usual two-part proof.

Part 1

31 = 3 < 6 = 3! = (1 + 2)! =) 31 < (1 + 2)!

Thus, the desired inequality is true for n = 1.

Part 2

Assume: 3k < (k + 2)!

To show: 3k+1 < (k + 3)!

104

Given that 3k < (k + 2)!, we multiply both sides of the inequality by 3 andobtain

3�3k�< 3 [(k + 2)!] .

This implies that

3�3k�< 3 [(k + 2)!] < (k + 3) [(k + 2)!] , since k > 0,

and so3k+1 < (k + 3)!.

Therefore, by the Principle of Math Induction, we conclude that 3n < (n+2)!for every positive integer n.

The left-hand side of the inequality is defined for any integer n. The right-hand side makes sense only if n+ 2 � 0, or n � �2.

When n = �2: 3�2 =1

9< 1 = 0! = (�2 + 2)!

When n = �1: 3�1 =1

3< 1 = 1! = (�1 + 2)!

When n = 0: 30 = 1 < 2 = 2! = (0 + 2)!

Therefore, 3n < (n+ 2)! for any integer n � �2. 2


Use mathematical induction to prove that 2n+ 3 < 2n for n � 4.

Answer:

Part 1

2(4) + 3 = 11 < 16 = 24

Thus, 2(4) + 3 <= 24.

Part 2

Assume: 2k + 3 < 2k, k � 4

To show: 2(k + 1) + 3 < 2k+1

2(k + 1) + 3 = 2k + 5 = (2k + 3) + 2

< 2k + 2 < 2k + 2k = 2k+1

Therefore, by the Principle of Math Induction, 2n+ 3 < 2n for n � 4.

105

Exercises 2.3

Prove the following statements by mathematical induction.

(1)nX

i=1

(3i� 1) =3n2 + n

2

(2)1

1 · 2 +1

2 · 3 +1

3 · 4 + · · ·+ 1

n(n+ 1)=

n

n+ 1

(3)nX

i=1

2 · 3i�1 = 3n � 1

Hint:k+1X

i=1

2·3i�1 =kX

i=1

2·3i�1+2·3(k+1)�1 = 3k�1+2·3k = 3·3k�1 = 3k+1�1

(4)nX

i=1

i3 =n2(n+ 1)2

4

(5) a1 + (a1 + d) + (a1 + 2d) + · · ·+ [a1 + (n� 1)d] =n [2a1 + (n� 1)d]

2

(6) 1 (1!) + 2 (2!) + · · ·+ n (n!) = (n+ 1)!� 1

Hint:k+1X

i=1

i · i! =kX

i=1

i · i! + (k + 1)(k + 1)! = (k + 1)!� 1 + (k + 1)(k + 1)! =

(k + 1)!(1 + k + 1)� 1 = (k + 2)!� 1

(7) 7n � 4n is divisible by 3

Hint: 7k+1 � 4k+1 = 7 · 7k � 4 · 4k = (3 + 4)7k � 4 · 4k = 3 · 7k + (7k � 4k)

(8) 10n + 3 · 4n+2 + 5 is divisible by 9

Hint: 10k+1+3·4k+3+5 = 10·10k+3·4·4k+2+5 = (9+1)10k+(9+3)4k+2+5 =9(10k + 4k+2) + 10k + 3 · 4k+2 + 5

(9) 11n+2 + 122n+1 is divisible by 133

Hint: 11k+3+122k+3 = 11·11k+2+122 ·122k+1 = 11·11k+2+(133+11)122k+1 =11(11k+2 + 122k+1) + 133 · 122k+1

(10) xn � yn is divisible by x� y for any positive integer n

Hint: xk+1 � yk+1 = x · xk � y · xk + y · xk � y · yk = (x� y)xk + y(xk � yk)

(11) xn + yn is divisible by x+ y for any odd positive integer n

Hint: xk+2+ yk+2 = x2xk + y2yk = x2xk +x2yk �x2yk + y2yk = x2(xk + yk)�yk(x� y)(x+ y)

106

(12) If 0 < a < 1, then 0 < an < 1 for any positive integer n

Hint: 0 < ak < 1 =) 0 · a < ak · a < 1 · a =) 0 < ak+1 < a < 1

(13) (1 + a)n > 1 + na for a > �1, a 6= 0 and n an integer greater than 1

Hint: (1 + a)k+1 > (1 + ka)(1 + a) = 1 + (k + 1)a+ ka2 > 1 + (k + 1)a

(14) 2n > n2 for every integer n > 4

Hint: 2k+1 = 2 · 2k > 2k2 = k2 + k2 > k2 + 2k + 1 = (k + 1)2. The lastinequality follows from (k�1)2 > 2 for k > 4, which implies that k2 > 2k+1.

For k > 4, (k � 1)2 > 2

(15) 2n < n! for every integer n > 3

Hint: 2k+1 = 2 · 2k = 2k! < (k + 1)k! = (k + 1)!

4

Lesson 2.4. The Binomial Theorem




(1) illustrate Pascal’s Triangle in the expansion of (x + y)n for small positiveintegral values of n;

(2) prove the Binomial Theorem;

(3) determine any term in (x + y)n, where n is a positive integer, without ex-panding; and

(4) solve problems using mathematical induction and the Binomial Theorem.

Lesson Outline

(1) Expand (x+ y)n for small values of n using Pascal’s Triangle

(2) Review the definition of and formula for combinationTeaching Notes

The concept ofcombination wasintroduced inGrade 10. Inparticular, theconcept wasdiscussed withcompetency codesfrom M10SP-IIIc-1to M10SP-IIId-e-1.

(3) State and prove the Binomial Theorem

(4) Compute all or specified terms of a binomial expansion

(5) Prove some combination identities using the Binomial Theorem

Introduction

In this lesson, we study two ways to expand (a + b)n, where n is a positiveinteger. The first, which uses Pascal’s Triangle, is applicable if n is not too big,

107

and if we want to determine all the terms in the expansion. The second methodgives a general formula for the expansion of (a + b)n for any positive integer n.This formula is useful especially when n is large because it avoids the process ofgoing through all the coe�cients for lower values of n obtained through Pascal’sTriangle.

Teaching Notes

Calculations withbig numbers arerequired in manyof the examplesand exercises in

this section. Theuse of scientific

calculators isdesirable.

Moreover, if only a specific term is required, it can be computed directlyusing a simple formula. Lastly, the theorem can be used to derive and prove someuseful and interesting results about sums of combinations.

2.4.1. Pascal’s Triangle and the Concept of Combination

Consider the following powers of a+ b:Teaching Notes

You may ask thestudents to expandthese powers usinglong multiplication.

(a+ b)1 = a+ b

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

(a+ b)5 = a5 + 5a4b+ 10a3b2 + 10a2b3 + 5ab4 + b5

(a+ b)6 = a6 + 6a5b+ 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

We now list down the coe�cients of each expansion in a triangular array asfollows:

n = 1 : 1 1

n = 2 : 1 2 1

n = 3 : 1 3 3 1

n = 4 : 1 4 6 4 1

n = 5 : 1 5 10 10 5 1

n = 6 : 1 6 15 20 15 6 1

This is part of the Pascal’s Triangle.

Named after the French mathematician Blaise Pascal (1623-1662), some prop-erties of the Pascal’s Triangle are the following:

(1) Each row begins and ends with 1.

(2) Each row has n+ 1 numbers.

(3) The second and second to the last number of each row correspond to therow number.

(4) There is symmetry of the numbers in each row.

108

(5) The number of entries in a row is one more than the row number (or onemore than the number of entries in the preceding row).

(6) Every middle number after first row is the sum of the two numbers aboveit.

It is the last statement which is useful in constructing the succeeding rows of thetriangle.

Example 2.4.1. Use Pascal’s Triangle to expand the expression (2x� 3y)5.

Solution. We use the coe�cients in the fifth row of the Pascal’s Triangle.

(2x� 3y)5 = (2x)5 + 5(2x)4(�3y) + 10(2x)3(�3y)2

+ 10(2x)2(�3y)3 + 5(2x)(�3y)4

+ (�3y)5

= 32x5 � 240x4y + 720x3y2 � 1080x2y3

+ 810xy4 � 243y5 2

Example 2.4.2. Use Pascal’s Triangle to expand (a+ b)8.

Solution. We start with the sixth row (or any row of the Pascal’s Triangle thatwe remember).

n = 6 : 1 6 15 20 15 6 1

n = 7 : 1 7 21 35 35 21 7 1

n = 8 : 1 8 28 56 70 56 28 8 1

Therefore, we get

(a+ b)8 = a8 + 8a7b+ 28a6b2 + 56a5b3

+ 70a4b4 + 56a3b5 + 28a2b6

+ 8ab7 + b8 2

We observe that, for each n, the expansion of (a + b)n starts with an

and the exponent of a in the succeeding terms decreases by 1, whilethe exponent of b increases by 1. This observation will be shown tobe true in general.

Let us review the concept of combination. Recall that C(n, k) or�n

k

�counts

the number of ways of choosing k objects from a set of n objects. It is also usefulto know some properties of C(n, k):

109

(1) C(n, 0) = C(n, n) = 1,

(2) C(n, 1) = C(n, n� 1) = n, and

(3) C(n, k) = C(n, n� k).

These properties can explain some of the observations we made on the num-bers in the Pascal’s Triangle. Recall also the general formula for the number ofcombinations of n objects taken k at a time:

C(n, k) =

✓n

k

◆=

n!

k!(n� k)!,

where 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · ·n.

Example 2.4.3. Compute

✓5

3

◆and

✓8

5

◆.

Solution. ✓5

3

◆=

5!

(5� 3)!3!=

5!

2!3!= 10

✓8

5

◆=

8!

(8� 5)!5!=

10!

3!5!= 56 2

You may observe that the value of�53

�and the fourth coe�cient in the fifth

row of Pascal’s Triangle are the same. In the same manner,�85

�is equal to the

sixth coe�cient in the expansion of (a+ b)8 (see Example 2.4.2). These observedequalities are not coincidental, and they are, in fact, the essence embodied in theBinomial Theorem, as you will see in the succeeding sessions.


1. Use Pascal’s Triangle to expand each expression.

(a) (x� 2y)4 Answer: x4 � 8x3y + 24x2y2 � 32xy3 + 16y4

(b) (2a� b2)3 Answer: 8a3 � 12a2b2 + 6ab4 � b6

(c) (a+ b)9

Answer: a9+9a8b+36a7b2+84a6b3+126a5b4+126a4b5+84a3b6+36a2b7+9ab8 + b9

2. Compute.

(a)

✓5

2

◆Answer: 10

110

(b)

✓9

7

◆Answer: 36

(c)

✓12

10

◆Answer: 66

(d)

✓20

5

◆Answer: 15504

3. Prove:

✓n

2

◆=

n(n� 1)

2.

Answer: ✓n

2

◆=

n!

(n� 2)!2!=

n(n� 1)(n� 2)!

(n� 2)!2!=

n(n� 1)

2

2.4.2. The Binomial Theorem

As the power n gets larger, the more laborious it would be to use Pascal’s Triangle(and impractical to use long multiplication) to expand (a + b)n. For example,using Pascal’s Triangle, we need to compute row by row up to the thirtieth rowto know the coe�cients of (a+ b)30. It is, therefore, delightful to know that it ispossible to compute the terms of a binomial expansion of degree n without goingthrough the expansion of all the powers less than n.

We now explain how the concept of combination is used in the expansion of(a+ b)n.

(a+ b)n = (a+ b)(a+ b)(a+ b) · · · (a+ b)| {z }n factors

When the distributive law is applied, the expansion of (a + b)n consists ofterms of the form ambi, where 0 m, i n. This term is obtained by choosinga for m of the factors and b for the rest of the factors. Hence, m + i = n, orm = n � i. This means that the number of times the term an�ibi will appearin the expansion of (a + b)n equals the number of ways of choosing (n � i) or ifactors from the n factors, which is exactly C(n, i). Therefore, we have

(a+ b)n =nX

i=0

✓n

i

◆an�ibi.

To explain the reasoning above, consider the case n = 3.

(a+ b)3 = (a+ b)(a+ b)(a+ b)

= aaa+ aab+ aba+ abb+ baa+ bab+ bba+ bbb

= a3 + 3a2b+ 3ab2 + b3

111

That is, each term in the expansion is obtained by choosing either a or b in eachfactor. The term a3 is obtained when a is chosen each time, while a2b is obtainedwhen a is selected 2 times, or equivalently, b is selected exactly once.

We will give another proof of this result using mathematical induction. Butfirst, we need to prove a result about combinations.

Pascal’s Identity

If n and k are positive integers with k n, then✓n+ 1

k

◆=

✓n

k

◆+

✓n

k � 1

◆.

Proof. The result follows from the combination formula.Teaching Notes

The formula canalso be proved

using the fact that�nk

�is the number

of ways to choose k

from n distinctobjects. Suppose a

is one of the n

objects. Then, inselecting k objects,either a is selected

or not. If a isincluded in the k

objects, then thereare

� nk�1

�ways to

complete theselection of the k

objects; if a is notincluded, then

there are�nk

�ways.

✓n

k

◆+

✓n

k � 1

◆=

n!

k!(n� k)!+

n!

(k � 1)!(n� k + 1)!

=n!(n� k + 1) + n!(k)

k!(n� k + 1)!

=n!(n� k + 1 + k)

k!(n+ 1� k)!

=n!(n+ 1)

k!(n+ 1� k)!

=(n+ 1)!

k!(n+ 1� k)!

=

✓n+ 1

k

◆2

Pascal’s identity explains the method of constructing Pascal’s Triangle, inwhich an entry is obtained by adding the two numbers above it. This identityis also an essential part of the second proof of the Binomial Theorem, which wenow state.

The Binomial Theorem

For any positive integer n,

(a+ b)n =nX

i=0

✓n

i

◆an�ibi.

Proof. We use mathematical induction.

112

Part 11X

i=0

✓1

i

◆a1�ibi =

✓1

0

◆a1b0 +

✓1

1

◆a0b1 = a+ b

Hence, the formula is true for n = 1.

Part 2. Assume that

(a+ b)k =kX

i=0

✓k

i

◆ak�ibi.

We want to show that

(a+ b)k+1 =k+1X

i=0

✓k + 1

i

◆ak+1�ibi.

(a+ b)k+1 = (a+ b)(a+ b)k

= (a+ b)kX

i=0

✓k

i

◆ak�ibi

= akX

i=0

✓k

i

◆ak�ibi + b

kX

i=0

✓k

i

◆ak�ibi

=kX

i=0

✓k

i

◆ak�i+1bi +

kX

i=0

✓k

i

◆ak�ibi+1

=

✓k

0

◆ak+1b0 +

kX

i=1

✓k

i

◆ak+1�ibi

+

✓k

0

◆akb1 +

✓k

1

◆ak�1b2 +

✓k

2

◆ak�2b3

+ · · ·+✓

k

k � 1

◆a1bk +

✓k

k

◆a0bk+1

= ak+1 +kX

i=1

✓k

i

◆ak+1�ibi

+kX

i=1

✓k

i� 1

◆ak+1�ibi + bk+1

=

✓k + 1

0

◆ak+1b0

+kX

i=1

✓k

i

◆+

✓k

i� 1

◆�ak+1�ibi

+

✓k + 1

k + 1

◆a0bk+1

113

=k+1X

i=0

✓k + 1

i

◆ak+1�ibi

The last expression above follows from Pascal’s Identity.

Therefore, by the Principle of Mathematical Induction,

(a+ b)n =nX

i=1

✓n

i

◆an�ibi

for any positive integer n. 2

2.4.3. Terms of a Binomial Expansion

We now apply the Binomial Theorem in di↵erent examples.

Example 2.4.4. Use the Binomial Theorem to expand (x+ y)6.

Solution.

(x+ y)6 =6X

k=0

✓6

k

◆x6�kyk

=

✓6

0

◆x6y0 +

✓6

1

◆x5y1 +

✓6

2

◆x4y2

+

✓6

3

◆x3y3 +

✓6

4

◆x2y4 +

✓6

5

◆x1y5

+

✓6

6

◆x0y6

= x6 + 6x5y + 15x4y2 + 20x3y3

+ 15x2y2 + 6xy5 + y6 2

Since the expansion of (a + b)n begins with k = 0 and ends with k = n, theexpansion has n + 1 terms. The first term in the expansion is

�n

0

�an = an, the

second term is�n

1

�an�1b = nan=1b, the second to the last term is

�n

n�1

�abn�1 =

nabn�1, and the last term is�n

n

�bn = bn.

The kth term of the expansion is�

n

k�1

�an�k+1bk�1. If n is even, there is a

middle term, which is the�n

2 + 1�th term. If n is odd, there are two middle

terms, the�n+12

�th and

�n+12 + 1

�th terms.

The general term is often represented by�n

k

�an�kbk. Notice that, in any term,

the sum of the exponents of a and b is n. The combination�n

k

�is the coe�cient

of the term involving bk. This allows us to compute any particular term withoutneeding to expand (a+ b)n and without listing all the other terms.

114

Example 2.4.5. Find the fifth term in the expansion of�2x�p

y�20

.Teaching Notes

To find a specificterm in theexpansion of(a+ b)n, it isimportant to findthe value of k.

Solution. The fifth term in the expansion of a fifth power corresponds to k = 4.✓20

4

◆(2x)20�4 (�p

y)4 = 4845�65536x16

�y2

= 317521920x16y2 2

Example 2.4.6. Find the middle term in the expansion of⇣x2+ 3y

⌘6

.

Solution. Since there are seven terms in the expansion, the middle term is thefourth term (k = 3), which is

✓6

3

◆⇣x2

⌘3

(3y)3 = 20

✓x3

8

◆�27y3

�=

135x3y3

2. 2

Example 2.4.7. Find the term involving x (with exponent 1) in the expansion

of

✓x2 � 2y

x

◆8

.

Solution. The general term in the expansion is

✓8

k

◆�x2�8�k

✓�2y

x

◆k

=

✓8

k

◆x16�2k · (�2)kyk

xk

=

✓8

k

◆(�2)kx16�2k�kyk

=

✓8

k

◆(�2)kx16�3kyk.

The term involves x if the exponent of x is 1, which means 16 � 3k = 1, ork = 5. Hence, the term is

✓8

5

◆(�2)5xy5 = �1792xy5. 2


1. Use the Binomial Theorem to expand (2a� b2)5.

Answer:

�2a� b2

�5=

✓5

0

◆(2a)5 +

✓5

1

◆(2a)4b2

115

+

✓5

2

◆(2a)3

�b2�2

+

✓5

3

◆(2a)2

�b2�3

+

✓5

4

◆(2a)

�b2�4

+

✓5

5

◆�b2�5

= 32a5 � 80a4b2 + 80a3b4 � 40a2b6

+ 10ab8 � b10

2. Find the two middle terms in the expansion of

✓x1/3 +

2

y

◆11

.

Answer: There are 12 terms in the expansion, so the two middle terms are the6th (corresponding to k = 5) and the 7th (corresponding to k = 6) terms.

✓11

5

◆�x1/3

�11�5✓2

y

◆5

= 462x2

✓32

y5

◆=

14784x2

y5

✓11

6

◆�x1/3

�11�6✓2

y

◆6

= 462x5/3

✓64

y6

◆=

29568x5/3

y6

3. Find the constant term in the expansion of

✓x3

2+

3

x2

◆10

.

Answer: The general term is

✓10

k

◆✓x3

2

◆10�k

✓3

x2

◆k

=

✓10

k

◆✓x30�3k

210�k

◆✓3k

x2k

◆

=

✓10

k

◆3k

210�k

x30�5k

The constant term contains x0, which means 30� 5k = 0, or k = 6.

✓10

6

◆36

24x0 =

76545

8

?2.4.4. Approximation and Combination Identities

We continue applying the Binomial Theorem.

?Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in theexpansion of (1� 0.2)8. Compare your answer with the calculator value.

(2) Use 5 terms in the binomial expansion to approximate (0.8)8. Is there animprovement in the approximation?

116

Solution.

(0.8)8 = (1� 0.2)8 =8X

k=0

✓8

k

◆(1)8�k(�0.2)k

=8X

k=0

✓8

k

◆(�0.2)k

(1)2X

k=0

✓8

k

◆(�0.2)k =

✓8

0

◆+

✓8

1

◆(�0.2) +

✓8

2

◆(�0.2)2

= 1� 1.6 + 1.12 = 0.52

The calculator value is 0.16777216, so the error is 0.35222784.

(2)4X

k=0

✓8

k

◆(�0.2)k =

✓8

0

◆+

✓8

1

◆(�0.2) +

✓8

2

◆(�0.2)2

+

✓8

3

◆(�0.2)3 +

✓8

4

◆(�0.2)4

= 0.52� 0.448 + 0.112 = 0.184

The error is 0.01622784, which is an improvement on the previous estimate.2

Example 2.4.9. Use the Binomial Theorem to prove that, for any positive in-teger n,

nX

k=0

✓n

k

◆= 2n.

Solution. Set a = b = 1 in the expansion of (a+ b)n. Then

2n = (1 + 1)n =nX

k=0

✓n

k

◆(1)n�k(1)k =

nX

k=0

✓n

k

◆. 2

Example 2.4.10. Use the Binomial Theorem to prove that✓100

0

◆+

✓100

2

◆+

✓100

4

◆+ · · ·+

✓100

100

◆

=

✓100

1

◆+

✓100

3

◆+

✓100

5

◆+ · · ·+

✓100

99

◆

Solution. Let a = 1 and b = �1 in the expansion of (a+ b)100. Then

⇥1 + (�1)

⇤100=

100X

k=0

✓100

k

◆(1)100�k(�1)k.

117

0 =

✓100

0

◆+

✓100

1

◆(�1) +

✓100

2

◆(�1)2 +

✓100

3

◆(�1)3

+ · · ·+✓100

99

◆(�1)99 +

✓100

100

◆(�1)100

If k is even, then (�1)k = 1. If k is odd, then (�1)k = �1. Hence, we have

0 =

✓100

0

◆�✓100

1

◆+

✓100

2

◆�✓100

3

◆

+ · · ·�✓100

99

◆+

✓100

100

◆

Therefore, after transposing the negative terms to other side of the equation, weobtain

✓100

0

◆+

✓100

2

◆+

✓100

4

◆+ · · ·+

✓100

100

◆

=

✓100

1

◆+

✓100

3

◆+

✓100

5

◆+ · · ·+

✓100

99

◆2


?1. Approximate (1.9)10 using the first three terms in the expansion of(2� 0.1)10, and find its error compared to the calculator value.

Answer:

(1.9)10 = (2� 0.1)10 ⇡2X

k=0

✓10

k

◆210�k(�0.1)k

= 210 � 10 · 29 · 0.1 + 45 · 28 · 0.12

= 627.2

Calculator value = 613.1066258

Error from the calculator value = 14.09337422

2. Prove that, for any positive integer n,

nX

k=0

✓n

k

◆3k = 4n.

Answer: 4n = (1 + 3)n =nX

k=0

✓n

k

◆1n�k3k =

nX

k=0

✓n

k

◆3k

118

Exercises 2.4

1. Use the Binomial Theorem to expand each expression.

(a) (x� 2)5 Answer: x5 � 10x4 + 40x3 � 80x2 + 80x� 32

(b)

✓x+

1

y

◆7

Answer: x7 +7x6

y+

21x5

y2+

35x4

y3+

35x3

y4+

21x2

y5+

7x

y6+

1

y7

(c)

✓3� 1

2

◆4

Answer: 81� 3(27)

✓1

2

◆+ 3(9)

✓1

4

◆� 1

8=

377

8

2. Without expanding completely, compute the indicated term(s) in the expan-sion of the given expression.

(a)

✓x3 +

1

2x

◆15

, first 3 terms Answer: x45 +15x41

2+

105x37

4

(b) (4� 3x)6, last 3 terms Answer: 19440x4 � 5832x5 + 729x6

(c)

✓x+

3

2

◆12

, 9th term Answer:3247695

256x4

(d)

✓px� 1

y

◆25

, 6th term Answer: �53130x10

y5

(e)

✓1

2p+

1

q

◆18

, middle term Answer:12155p9

128q9

(f)

✓2

a+

a2

3

◆11

, two middle terms Answer:9856

81a4 +

4928

243a7

(g)�p

y + x�9, term involving y3 Answer: 84x3y3

(h)

✓1

x3� 2x

◆16

, constant term Answer:366080

729

(i) (xy � 2y�2)21, term that does not contain y Answer: �14883840x14

(j)

✓px

y2� y

x

◆18

, term in which the exponents of x and y are equal

Answer:43758

x6y6

?3. Approximate (1.1)10 by using the first 4 terms in the expansion of(1 + 0.1)10. Compare your answer with the calculator result.

Answer: 2.57, with an error of 0.0237424601 from the calculator value of2.59374246

119

4. Use the Binomial Theorem to prove that

nX

k=0

✓n

k

◆2k = 3n.

Hint: Expand (1 + 2)n.

5. Use the Binomial Theorem to prove that

50X

k=0

✓50

k

◆(�2)k = 1.

Hint: Expand (1� 2)50.

4

120

Unit 3

Trigonometry

https://commons.wikimedia.org/wiki/File%3AUnderground River.jpg

By Giovanni G. Navata (Own work)[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)],


Named as one of the New Seven Wonders of Nature in 2012, the PuertoPrincesa Subterranean River National Park is world-famous for its limestonekarst mountain landscape with an underground river. The Park was also listedas UNESCO World Heritage Site in 1999. The underground river stretches about8.2 km long, making it one of the world’s longest rivers of its kind.

121

https://commons.wikimedia.org/wiki/File%3AUnderground_River.jpg

Lesson 3.1. Angles in a Unit Circle




(1) illustrate the unit circle and the relationship between the linear and angularmeasures of arcs in a unit circle.

(2) convert degree measure to radian measure, and vice versa.

(3) illustrate angles in standard position and coterminal angles.

Lesson Outline

(1) Linear and angular measure of arcs

(2) Conversion of degree to radian, and vice versa

(3) Arc length and area of the sector

(4) Angle in standard position and coterminal angles

Introduction

There are many problems involving angles in several fields like engineering,medical imaging, electronics, astronomy, geography and many more. Survey-ors, pilots, landscapers, designers, soldiers, and people in many other professionsheavily use angles and trigonometry to accomplish a variety of practical tasks.In this lesson, we will deal with the basics of angle measures together with arclength and sectors.

3.1.1. Angle Measure

An angle is formed by rotating a ray about its endpoint.Teaching Notes

Angles intrigonometry di↵er

from angles inEuclidean

geometry in thesense of motion.

An angle ingeometry is definedas a union of rays

(that is, static)and has measurebetween 0� and

180�. An angle intrigonometry is arotation of a ray,

and, therefore, hasno limit. It has

positive andnegative directions

and measures.

In the figure shownbelow, the initial side of \AOB is OA, while its terminal side is OB. An angleis said to be positive if the ray rotates in a counterclockwise direction, and theangle is negative if it rotates in a clockwise direction.

122

An angle is in standard position if it is drawn in the xy-plane with its vertexat the origin and its initial side on the positive x-axis. The angles ↵, �, and ✓ inthe following figure are angles in standard position.

To measure angles, we use degrees, minutes, seconds, and radians.

A central angle of a circle measures one degree, written 1�, if it inter-cepts 1

360 of the circumference of the circle. One minute, written 10, is160 of 1�, while one second, written 100, is 1

60 of 10.

For example, in degrees, minutes, and seconds,

10�3001800 = 10�✓30 +

18

60

◆0

= 10�30.30

=

✓10 +

30.3

60

◆�

= 10.505�

and

79.251� = 79�(0.251⇥ 60)0

= 79�15.060

= 79�150(0.06⇥ 60)00

= 79�1503.600.

Recall that the unit circle is the circle with center at the origin and radius 1unit.

123

A central angle of the unit circle that intercepts an arc of the circlewith length 1 unit is said to have a measure of one radian, written 1rad. See Figure 3.1.

Figure 3.1

In trigonometry, as it was studied in Grade 9, the degree measure is often used.On the other hand, in some fields of mathematics like calculus, radian measure ofangles is preferred. Radian measure allows us to treat the trigonometric functionsas functions with the set of real numbers as domains, rather than angles.

Example 3.1.1. In the following figure, identify the terminal side of an angle instandard position with given measure.

(1) degree measure: 135�, �135�, �90�, 405�

(2) radian measure: ⇡

4 rad, �3⇡4 rad, 3⇡

2 rad, �⇡

2 rad

124

Solution. (1) 135�:�!OC; �135�:

��!OD; �90�:

��!OE; and 405�:

��!OB

(2) radian measure: ⇡

4 rad:��!OB; �3⇡

4 rad:��!OD; 3⇡

2 rad:��!OE; and �⇡

2 rad:��!OE 2

Since a unit circle has circumference 2⇡, a central angle that measures 360�

has measure equivalent to 2⇡ radians. Thus, we obtain the following conversionrules.

Converting degree to radian,and vice versa

1. To convert a degree measure to radian, multiply it by ⇡

180 .

2. To convert a radian measure to degree, multiply it by 180⇡

.

Figure 3.2 shows some special angles in standard position with the indicatedterminal sides. The degree and radian measures are also given.

Figure 3.2

Example 3.1.2. Express 75� and 240� in radians.

Solution.

75⇣ ⇡

180

⌘=

5⇡

12=) 75� =

5⇡

12rad

240⇣ ⇡

180

⌘=

4⇡

3=) 240� =

4⇡

3rad 2

125

Example 3.1.3. Express ⇡

8 rad and 11⇡6 rad in degrees.

Solution.⇡

8

✓180

⇡

◆= 22.5 =) ⇡

8rad = 22.5�

11⇡

6

✓180

⇡

◆= 330 =) 11⇡

6rad = 330� 2


1. Convert the following degree measures to radian measure.

(a) 60� Answer: ⇡

3 rad

(b) 90� Answer: ⇡

2 rad

(c) 150� Answer: 5⇡6 rad

2. Convert the following radian measures to degree measure.

(a) ⇡

9 rad Answer: 20�

(b) 3⇡4 rad Answer: 135�

3.1.2. Coterminal Angles

Two angles in standard position that have a common terminal side are calledcoterminal angles. Observe that the degree measures of coterminal angles di↵erby multiples of 360�.

Two angles are coterminal if and only if their degree measures di↵erby 360k, where k 2 Z.

Similarly, two angles are coterminal if and only if their radian mea-sures di↵er by 2⇡k, where k 2 Z.

As a quick illustration, to find one coterminal angle with an angle that mea-sures 410�, just subtract 360�, resulting in 50�. See Figure 3.3.

126

Figure 3.3

Example 3.1.4. Find the angle coterminal with �380� that has measure

(1) between 0� and 360�, and

(2) between �360� and 0�.

Solution. A negative angle moves in a clockwise direction, and the angle �380�

lies in Quadrant IV.

(1) �380� + 2 · 360� = 340�

(2) �380� + 360� = �20� 2


1. Find the angle between 0� and 360� (if in degrees) or between 0 rad and 2⇡ rad(if in radians) that is coterminal with the given angle.

(a) 736� Answer: 16�

(b) �28�4806500 Answer: 331�1005500

(c) 13⇡2 rad Answer: ⇡

2 rad?(d) 10 rad Answer: 3.72 rad

2. Find the angle between �360� and 0� (if in degrees) or between �2⇡ rad and0 rad (if in radians) that is coterminal with the given angle.

(a) 142� Answer: �218�

(b) �400�102300 Answer: �40�102300

(c) ⇡

6 rad Answer: �11⇡6 rad

?(d) �20 rad Answer: �1.15 rad

127

3.1.3. Arc Length and Area of a Sector

In a circle, a central angle whose radian measure is ✓ subtends an arc that is thefraction ✓

2⇡ of the circumference of the circle.Teaching Notes

Review how arcswere measured inGrade 10. Whatunit of measure

was used? For twocircles with

di↵erent radii, doequal central

angles interceptarcs of the same

measure?Conclude that

previous notion ofarc measure is notthe same as length.

Arcs are nowmeasured in terms

of length andmeasure changes

with the radius ofthe circle.

Thus, in a circle of radius r (seeFigure 3.4), the length s of an arc that subtends the angle ✓ is

s =✓

2⇡⇥ circumference of circle =

✓

2⇡(2⇡r) = r✓.

Figure 3.4

In a circle of radius r, the length s of an arc intercepted by a centralangle with measure ✓ radians is given by

s = r✓.

Example 3.1.5. Find the length of an arc of a circle with radius 10 m thatsubtends a central angle of 30�.

Solution. Since the given central angle is in degrees, we have to convert it intoradian measure. Then apply the formula for an arc length.

30⇣ ⇡

180

⌘=

⇡

6rad

s = 10⇣⇡6

⌘=

5⇡

3m 2

Example 3.1.6. A central angle ✓ in a circle of radius 4 m is subtended by anarc of length 6 m. Find the measure of ✓ in radians.

Solution.

✓ =s

r=

6

4=

3

2rad 2

128

A sector of a circle is the portion of the interior of a circle bounded by theinitial and terminal sides of a central angle and its intercepted arc. It is like a“slice of pizza.” Note that an angle with measure 2⇡ radians will define a sectorthat corresponds to the whole “pizza.” Therefore, if a central angle of a sectorhas measure ✓ radians, then the sector makes up the fraction ✓

2⇡ of a completecircle. See Figure 3.5. Since the area of a complete circle with radius r is ⇡r2, wehave

Area of a sector =✓

2⇡(⇡r2) =

1

2✓r2.

Figure 3.5

In a circle of radius r, the area A of a sector with a central anglemeasuring ✓ radians is

A =1

2r2✓.

Example 3.1.7. Find the area of a sector of a circle with central angle 60� ifthe radius of the circle is 3 m.

Solution. First, we have to convert 60� into radians. Then apply the formula forcomputing the area of a sector.

60⇣ ⇡

180

⌘=

⇡

3rad

A =1

2(32)

⇡

3=

3⇡

2m2 2

Example 3.1.8. A sprinkler on a golf course fairway is set to spray water overa distance of 70 feet and rotates through an angle of 120�. Find the area of thefairway watered by the sprinkler.

129

Solution.

120⇣ ⇡

180

⌘=

2⇡

3rad

A =1

2(702)

2⇡

3=

4900⇡

3⇡ 5131 ft2 2


1. In a circle of radius 7 feet, find the length of the arc that subtends a centralangle of 5 radians. Answer: 35 ft

2. A central angle ✓ in a circle of radius 20 m is subtended by an arc of length15⇡ m. Find the measure of ✓ in degrees. Answer: 135�

3. Find the area of a sector of a circle with central angle that measures 75� if theradius of the circle is 6 m. Answer: 7.5 m2

Exercises 3.1

1. Give the degree/radian measure of the following special angles.

2. Convert each degree measure to radians. Leave answers in terms of ⇡.

(a) 330� Answer: 11⇡6 rad

(b) 480� Answer: 8⇡3 rad

130

(c) 15� Answer: ⇡

12 rad

(d) 105� Answer: 7⇡12 rad

(e) 265� Answer: 53⇡36 rad

(f) �120� Answer: �2⇡3 rad

(g) �315� Answer: �7⇡4 rad

3. Convert each radian measure to degree-minute-second measure (approximateif necessary).

(a) 5⇡6 rad Answer: 150�

(b) 8⇡3 rad Answer: 480�

(c) 15⇡4 rad Answer: 675�

(d) �⇡

6 rad Answer: �30�

(e) �7⇡20 rad Answer: �63�

?(f) 20 rad Answer: 1145�54056.1200

?(g) �35 rad Answer: �2005�2108.2200

?(h) �5 rad Answer: �286�28044.0300

4. Find the angle between 0� and 360� (if in degrees) or between 0 rad and 2⇡ rad(if in radians) that is coterminal with the given angle.

(a) 685� Answer: 325�

(b) 451� Answer: 91�

(c) �1400� Answer: 40�

(d) 960�4503400 Answer: 240�4503400

(e) �728�1504300 Answer: 352�1504300

(f) 29⇡6 rad Answer: 5⇡

6 rad

(g) �3⇡2 rad Answer: ⇡

2 rad?(h) 16 rad Answer: 3.43 rad?(i) �20 rad Answer: 5.13 rad

5. Find the angle between �360� and 0� (if in degrees) or between �2⇡ rad and0 rad (if in radians) that is coterminal with the given angle.

(a) 685� Answer: �35�

(b) 451� Answer: �269�

(c) �1400� Answer: �320�

(d) 960�4503400 Answer: �120�4503400

131

(e) �728�1504300 Answer: �8�1504300

(f) 29⇡6 rad Answer: �7⇡

6 rad

(g) �3⇡2 rad Answer: �3⇡

2 rad?(h) 16 rad Answer: ⇡ �2.850 rad?(i) �20 rad Answer: ⇡ �1.150 rad

6. Find the length of an arc of a circle with radius 21 m that subtends a centralangle of 15�. Answer: 7⇡

4 m

7. A central angle ✓ in a circle of radius 9 m is subtended by an arc of length 12m. Find the measure of ✓ in radians. Answer: 4

3 rad

8. Find the radius of a circle in which a central angle of ⇡

6 rad determines a sectorof area 64 m2. Answer: 16 m

9. If the radius of a circle is doubled, how is the length of the arc intercepted bya fixed central angle changed? Answer: The length is doubled.

10. Radian measure simplifies many formulas, such as the formula for arc length,s = r✓. Give the corresponding formula when ✓ is measured in degrees insteadof radians. Answer: s = ⇡r✓

180

?11. As shown below, find the radius of the pulley if a rotation of 51.6� raises theweight by 11.4 cm. Answer: 12.7 cm

?12. How many inches will the weight rise if the pulley whose radius is 9.27 inchesis rotated through an angle of 71�500? Answer: 11.6 in

?13. Continuing with the previous item, through what angle (to the nearest minute)must the pulley be rotated to raise the weight 6 in? Answer: 37�50

?14. Given a circle of radius 3 in, find the measure (in radians) of the central angleof a sector of area 16 in2. Answer: 3.6 rad

132

?15. An automatic lawn sprinkler sprays up to a distance of 20 feet while rotating30�. What is the area of the sector the sprinkler covers? Answer: 104.72 ft2

?16. A jeepney has a windshield wiper on the driver’s side that has total arm andblade 10 inches long and rotates back and forth through an angle of 95�. Theshaded region in the figure is the portion of the windshield cleaned by the7-inch wiper blade. What is the area of the region cleaned? Answer: 75.4 in2

17. If the radius of a circle is doubled and the central angle of a sector is unchanged,how is the area of the sector changed? Answer: The area is quadrupled.

18. Give the corresponding formula for the area of a sector when the angle ismeasured in degrees. Answer: A = ⇡r

2✓

360

?19. A frequent problem in surveying city lots and rural lands adjacent to curvesof highways and railways is that of finding the area when one or more of theboundary lines is the arc of a circle. Approximate the total area of the lotshown in the figure. Answer: 1909.0 m2

133

20. Two gears of radii 2.5 cm and 4.8 cm are adjusted so that the smaller geardrives the larger one, as shown. If the smaller gear rotates counterclockwisethrough 225�, through how many degrees will the larger gear rotate?

Answer: 117�

4

Lesson 3.2. Circular Functions




(1) illustrate the di↵erent circular functions; and

(2) use reference angles to find exact values of circular functions.

Lesson Outline

(1) Circular functions

(2) Reference angles

Introduction

We define the six trigonometric functionTeaching Notes

The teacher cangive a review of

trigonometricratios as discussed

in Grade 9.

in such a way that the domain ofeach function is the set of angles in standard position. The angles are measuredeither in degrees or radians. In this lesson, we will modify these trigonometricfunctions so that the domain will be real numbers rather than set of angles.

134

3.2.1. Circular Functions on Real Numbers

Recall that the sine and cosine functions (and four others: tangent, cosecant,secant, and cotangent) of angles measuring between 0� and 90� were defined inthe last quarter of Grade 9 as ratios of sides of a right triangle. It can be verifiedthat these definitions are special cases of the following definition.

Let ✓ be an angle in standard position and P (✓) = P (x, y) the pointon its terminal side on the unit circle. Define

sin ✓ = y csc ✓ =1

y, y 6= 0

cos ✓ = x sec ✓ =1

x, x 6= 0

tan ✓ =y

x, x 6= 0 cot ✓ =

x

y, y 6= 0

Example 3.2.1. Find the values of cos 135�, tan 135�, sin(�60�), and sec(�60�).

Solution. Refer to Figure 3.6(a).

(a) (b)

Figure 3.6

From properties Teaching Notes

A 45�-45� righttriangle is isosceles.Moreover, theopposite side of the30�-angle in a30�-60� righttriangle is half thelength of itshypotenuse.

of 45�-45� and 30�-60� right triangles (with hypotenuse 1unit), we obtain the lengths of the legs as in Figure 3.6(b). Thus, the coordinatesof A and B are

A =

�p2

2,

p2

2

!and B =

1

2,�

p3

2

!.

135

Therefore, we get

cos 135� = �p2

2, tan 135� = �1,

sin(�60�) = �p3

2, and sec(�60�) = 2. 2

From the last example, we may then also say that

cos⇣⇡4rad

⌘=

p2

2, sin

⇣�⇡

3rad

⌘= �

p3

2,

and so on.

From the above definitions, we define the same six functions on real numbers.These functions are called trigonometric functions.

Let s be any real number. Suppose ✓ is the angle in standard positionwith measure s rad. Then we define

sin s = sin ✓ csc s = csc ✓

cos s = cos ✓ sec s = sec ✓

tan s = tan ✓ cot s = cot ✓

From the last example, we then have

cos⇣⇡4

⌘= cos

⇣⇡4rad

⌘= cos 45� =

p2

2

and

sin⇣�⇡

3

⌘= sin

⇣�⇡

3rad

⌘= sin(�60�) = �

p3

2.

In the same way, we have

tan 0 = tan(0 rad) = tan 0� = 0.

Example 3.2.2. Find the exact values of sin 3⇡2 , cos

3⇡2 , and tan 3⇡

2 .

Solution. Let P�3⇡2

�be the point on the unit circle and on the terminal side of

the angle in the standard position with measure 3⇡2 rad. Then P

�3⇡2

�= (0,�1),

and so

sin3⇡

2= �1, cos

3⇡

2= 0,

but tan 3⇡2 is undefined. 2

136

Example 3.2.3. Suppose s is a real number such that sin s = �34 and cos s > 0.

Find cos s.

Solution. We may consider s as the angle with measure s rad. Let P (s) = (x, y)be the point on the unit circle and on the terminal side of angle s.

Since P (s) is on the unit circle, we know that x2 + y2 = 1. Since sin s = y =�3

4 , we get

x2 = 1� y2 = 1�✓�3

4

◆2

=7

16=) x = ±

p7

4.

Since cos s = x > 0, we have cos s =p74 . 2

Let P (x1, y1) and Q(x, y) be points on the terminal side of an angle ✓ instandard position, where P is on the unit circle and Q on the circle of radius r(not necessarily 1) with center also at the origin, as shown above. Observe thatwe can use similar triangles to obtain

cos ✓ = x1 =x1

1=

x

rand sin ✓ = y1 =

y11

=y

r.

We may then further generalize the definitions of the six circular functions.

137

Let ✓ be an angle in standard position, Q(x, y) any point on the ter-minal side of ✓, and r =

px2 + y2 > 0. Then

sin ✓ =y

rcsc ✓ =

r

y, y 6= 0

cos ✓ =x

rsec ✓ =

r

x, x 6= 0

tan ✓ =y

x, x 6= 0 cot ✓ =

x

y, y 6= 0

We then have a second solution for Example 3.2.3 as follows. With sin s = �34

and sin s = y

r

, we may choose y = �3 and r = 4 (which is always positive). Inthis case, we can solve for x, which is positive since cos s = x

4 is given to bepositive.

4 =p

x2 + (�3)2 =) x =p7 =) cos s =

p7

4


1. Given ✓, find the exact values of the six circular functions.

(a) ✓ = 30�

Answer: sin 30� = 12 , cos 30

� =p32 , tan 30� =

p33 , csc 30� = 2, sec 30� =

2p3

3 , cot 30� =p3

(b) ✓ = 3⇡4

Answer: sin 3⇡4 =

p22 , cos 3⇡

4 = �p22 , tan 3⇡

4 = �1, csc 3⇡4 =

p2, sec 3⇡

4 =

�p2, cot 3⇡

4 = �1

(c) ✓ = �150�

Answer: sin(�150�) = �12 , cos(�150�) = �

p32 , tan(�150�) =

p33 , csc(�150�) =

�2, sec(�150�) = �2p3

3 , cot(�150�) =p3

(d) ✓ = �4⇡3

Answer: sin(�4⇡3 ) =

p32 , cos(�4⇡

3 ) = �12 , tan(�

4⇡3 ) = �

p3, csc(�4⇡

3 ) =2p3

3 , sec(�4⇡3 ) = �2, cot(�4⇡

3 ) = �p33

2. Given a value of one circular function and sign of another function (or thequadrant where the angle lies), find the value of the indicated function.

(a) sin ✓ = 12 , ✓ in QI; cos ✓ Answer:

p32

(b) cos ✓ = 35 , ✓ in QIV; csc ✓ Answer: 5

4

(c) sin ✓ = �37 , sec ✓ < 0; tan ✓ Answer: 3

p10

20

(d) cot ✓ = �29 , cos ✓ > 0; csc ✓ Answer: �

p859

138

3.2.2. Reference Angle

We observe that if ✓1 and ✓2 are coterminal angles, the values of the six circularor trigonometric functions at ✓1 agree with the values at ✓2. Therefore, in findingthe value of a circular function at a number ✓, we can always reduce ✓ to a numberbetween 0 and 2⇡. For example, sin 14⇡

3 = sin�14⇡3 � 4⇡

�= sin 2⇡

3 . Also, observefrom Figure 3.7 that sin 2⇡

3 = sin ⇡

3 .

Figure 3.7

In general, if ✓1, ✓2, ✓3, and ✓4 are as shown in Figure 3.8 with P (✓1) =(x1, y1), then each of the x-coordinates of P (✓2), P (✓3), and P (✓4) is ±x1, whilethe y-coordinate is ±y1. The correct sign is determined by the location of theangle. Therefore, together with the correct sign, the value of a particular circularfunction at an angle ✓ can be determined by its value at an angle ✓1 with radianmeasure between 0 and ⇡

2 . The angle ✓1 is called the reference angle of ✓.

Figure 3.8

139

The signs of the coordinates of P (✓) depends on the quadrant or axis whereit terminates. It is important to know the sign of each circular function in eachquadrant. See Figure 3.9. It is not necessary to memorize the table, since thesign of each function for each quadrant is easily determined from its definition.We note that the signs of cosecant, secant, and cotangent are the same as sine,cosine, and tangent, respectively.

Figure 3.9

Using the fact that the unit circle is symmetric with respect to the x-axis, they-axis, and the origin, we can identify the coordinates of all the points using thecoordinates of corresponding points in the Quadrant I, as shown in Figure 3.10for the special angles.

Figure 3.10

140

Example 3.2.4. Use reference angle and appropriate sign to find the exact valueof each expression.(1) sin 11⇡

6 and cos 11⇡6

(2) cos��7⇡

6

�(3) sin 150�

(4) tan 8⇡3

Solution. (1) The reference angle of 11⇡6 is ⇡

6 , and it lies in Quadrant IV whereinsine and cosine are negative and positive, respectively.

sin11⇡

6= � sin

⇡

6= �1

2

cos11⇡

6= cos

⇡

6=

p3

2

(2) The angle �7⇡6 lies in Quadrant II wherein cosine is negative, and its refer-

ence angle is ⇡

6 .

cos

✓�7⇡

6

◆= � cos

⇡

6= �

p3

2

(3) sin 150� = sin 30� = 12

(4) tan 8⇡3 = � tan ⇡

3 = � sin⇡

3cos

⇡

3= �

p3212

= �p3 2


Use reference angle and appropriate sign to find the exact value of each expression.

(1) sin 510� Answer: 12

(2) tan(�225�) Answer: �1

(3) sec 13⇡3 Answer: 2

(4) cot��10⇡

3

�Answer: �

p33

Exercises 3.2

1. Find the exact value.

(a) sin 600� Answer: �p32

(b) tan(�810�) Answer: Undefined

(c) sec 585� Answer: �p2

(d) cos(�420�) Answer: 12

141

(e) sin 7⇡6 Answer: �1

2

(f) cos 5⇡3 Answer: 1

2

(g) tan 3⇡4 Answer: �1

(h) sec 2⇡3 Answer: �2

(i) csc 11⇡6 Answer: �2

(j) cot 35⇡6 Answer: �

p3

(k) cos��4⇡

3

�Answer: �1

2

(l) tan 17⇡3 Answer: �

p33

(m) cos 7⇡4 Answer:

p22

(n) sec 19⇡4 Answer: �

p2

(o) sin��4⇡

3

�Answer:

p32

(p) sec 23⇡6 Answer: 2

p3

3

(q) csc 13⇡3 Answer: 2

p3

3

(r) tan 5⇡6 Answer: �

p33

2. Find the exact value of each expression.Teaching Notes

(sinx)2 is denotedby sin2 x.

Similarly, thisnotation is usedwith the othertrigonometricfunctions. Ingeneral, for a

positive integer n,sinn x = (sinx)n.

(a) sin2 150� + cos2 150� Answer: 1

(b) cos(�30�) + sin 420� Answer:p3

(c) tan(�225�) + tan 405� Answer: 0

(d) sec 750� � csc(�300�) Answer: 0

(e) cos2 2⇡3 + sin2 2⇡

3 Answer: 1

(f) sin 11⇡6 + cos 5⇡

3 Answer: 0

(g) 2 cos 5⇡3 � sin 5⇡

2 Answer: 0

(h) tan2 ⇡

4 + 2 cos 8⇡3 � sin 13⇡

6 Answer: �12

(i)tan 2⇡

3 �tan 5⇡6

1+tan 2⇡3 tan 5⇡

6

Answer: �p33

(j)sin 11⇡

6 �cos ⇡6

sin(�⇡6 )+cos 5⇡

6

Answer: 1

3. Compute P (✓), and find the exact values of the six circular functions.

(a) ✓ = 19⇡6

Answer: P (✓) =⇣�

p32 ,�1

2

⌘, sin 19⇡

6 = �12 , cos

19⇡6 = �

p32 , tan 19⇡

6 =p33 ,

csc 19⇡6 = �2, sec 19⇡

6 = �2p3

3 , cot 19⇡6 =

p3

142

(b) ✓ = 32⇡3

Answer: P (✓) =⇣�1

2 ,p32

⌘, sin 32⇡

3 =p32 , cos 32⇡

3 = �12 , tan

32⇡3 = �

p3,

csc 32⇡3 = 2

p3

3 , sec 32⇡3 = �2, cot 32⇡

3 = �p33

4. Given the value of a particular circular function and an information about theangle ✓, find the values of the other circular functions.

(a) cos ✓ = 12 and 3⇡

2 < ✓ < 2⇡

Answer: sec ✓ = 2, sin ✓ = �p32 , tan ✓ = �

p3, csc ✓ = �2

p3

3 , cot ✓ = �p33

(b) sin ✓ = 817 and 0 < ✓ < ⇡

2

Answer: csc ✓ = 178 , cos ✓ = 15

17 , tan ✓ = 815 , sec ✓ = 17

15 , cot ✓ = 158

(c) cos ✓ = 2p13

13 and 3⇡2 < ✓ < 2⇡

Answer: sec ✓ =p132 , sin ✓ = �3

p13

13 , tan ✓ = �32 , csc ✓ = �

p133 , cot ✓ =

�23

4

Lesson 3.3. Graphs of Circular Functions and SituationalProblems




(1) determine the domain and range of the di↵erent circular functions;

(2) graph the six circular functions with its amplitude, period, and phase shift;and

(3) solve situational problems involving circular functions.

Lesson Outline

(1) Domain and range of circular functions

(2) Graphs of circular functions

(3) Amplitude, period, and phase shift

Introduction

There are many things that occur periodically. Phenomena like rotation ofthe planets and comets, high and low tides, and yearly change of the seasons

143

follow a definite pattern. In this lesson, we will graph the six circular functionsand we will see that they are periodic in nature.

3.3.1. Graphs of y = sinx and y = cosx

Recall that, for a real number x, sin x = sin ✓ for an angle ✓ with measure xradians, and that sin ✓ is the second coordinate of the point P (✓) on the unitcircle. Since each x corresponds to an angle ✓, we can conclude that

(1) sin x is defined for any real number x or the domain of the sine function isR, and

(2) the range of sine is the set of all real numbers between �1 and 1 (inclusive).

From the definition, it also follows that sin(x+2⇡) = sin x for any real numberx. This means that the values of the sine function repeat every 2⇡ units. In thiscase, we say that the sine function is a periodic function with period 2⇡.

Table 3.11 below shows the values of y = sin x, where x is the equivalent radianmeasure of the special angles and their multiples from 0 to 2⇡. As commentedabove, these values determine the behavior of the function on R.

x 0 ⇡

6⇡

4⇡

3⇡

22⇡3

3⇡4

5⇡6 ⇡

y 0 12

p22

p32 1

p32

p22

12 0

0 0.5 0.71 0.87 1 0.87 0.71 0.5 0

x 7⇡6

5⇡4

4⇡3

3⇡2

5⇡3

7⇡4

11⇡6 2⇡

y �12 �

p22 �

p32 �1 �

p32 �

p22 �1

2 0

�0.5 �0.71 �0.87 �1 �0.87 �0.71 �0.5 0

Table 3.11

From the table, we can observe that as x increases from 0 to ⇡

2 , sin x alsoincreases from 0 to 1. Similarly, as x increases from 3⇡

2 to 2⇡, sin x also increasesfrom �1 to 0. On the other hand, notice that as x increases from ⇡

2 to ⇡, sin xdecreases from 1 to 0. Similarly, as x increases from ⇡ to 3⇡

2 , sin x decreases from0 to �1.

To sketch the graph of y = sin x, we plot the points presented in Table 3.11,and join them with a smooth curve. See Figure 3.12. Since the graph repeatsevery 2⇡ units, Figure 3.13 shows periodic graph over a longer interval.

144

Teaching Notes

It is a goodexercise toconstruct thegraph of the sinefunction using theheight of P (✓).Put the unit circleside-by-side withthe coordinateplane for thegraph, and tracethe height for eachvalue of x onto thegraph of y = sinx.

Figure 3.12

Figure 3.13

We can make observations about the cosine function that are similar to thesine function.

• y = cosx has domain R and range [�1, 1].

• y = cos x is periodic with period 2⇡. The graph of y = cos x is shown inFigure 3.14.

Figure 3.14

From the graphs of y = sin x and y = cosx in Figures 3.13 and 3.14, re-spectively, we observe that sin(�x) = � sin x and cos(�x) = cosx for any realnumber x. In other words, the graphs of y = cos(�x) and y = cosx are the same,while the graph of y = sin(�x) is the same as that of y = � sin x.

In general, if a function f satisfies the property that f(�x) = f(x) for all xin its domain, we say that such function is even. On the other hand, we say thata function f is odd if f(�x) = �f(x) for all x in its domain. For example, thefunctions x2 and cos x are even, while the functions x3 � 3x and sin x are odd.

145

3.3.2. Graphs of y = a sin bx and y = a cos bx

Using a table of values from 0 to 2⇡, we can sketch the graph of y = 3 sin x, andcompare it to the graph of y = sin x. See Figure 3.15 wherein the solid curvebelongs to y = 3 sin x, while the dashed curve to y = sin x. For instance, if x = ⇡

2 ,then y = 1 when y = sin x, and y = 3 when y = 3 sin x. The period, x-intercepts,and domains are the same for both graphs, while they di↵er in the range. Therange of y = 3 sin x is [�3, 3].

Figure 3.15

In general, the graphs of y = a sin x and y = a cos x with a > 0 have the sameshape as the graphs of y = sin x and y = cosx, respectively. If a < 0, there isa reflection across the x-axis.

Teaching Notes

Review or teachthe reflection

across the x-axiswhen the sign ofthe function is

changed.

The range of both y = a sin x and y = a cos x is[�|a|, |a|].

In the graphs of y = a sin x and y = a cos x, the number |a| is calledits amplitude. It dictates the height of the curve. When |a| < 1,the graphs are shrunk vertically, and when |a| > 1, the graphs arestretched vertically.

Now, in Table 3.16, we consider the values of y = sin 2x on [0, 2⇡].

x 0 ⇡

6⇡

4⇡

3⇡

22⇡3

3⇡4

5⇡6 ⇡

y 0p32 1

p32 0 �

p32 �1 �

p32 0

0 0.87 1 0.87 0 �0.87 �1 �0.87 0

x 7⇡6

5⇡4

4⇡3

3⇡2

5⇡3

7⇡4

11⇡6 2⇡

yp32 1

p32 0 �

p32 �1 �

p32 0

0.87 1 0.87 0 �0.87 �1 �0.87 0

Table 3.16

146

Figure 3.17

Figure 3.17 shows the graphs of y = sin 2x (solid curve) and y = sin x (dashedcurve) over the interval [0, 2⇡]. Notice that, for sin 2x to generate periodic valuessimilar to [0, 2⇡] for y = sin x, we just need values of x from 0 to ⇡. We thenexpect the values of sin 2x to repeat every ⇡ units thereafter. The period ofy = sin 2x is ⇡.

If b 6= 0, then both y = sin bx and y = cos bx have period given by2⇡

|b| .

If 0 < |b| < 1, the graphs are stretched horizontally, and if |b| > 1, thegraphs are shrunk horizontally.

To sketch the graphs of y = a sin bx and y = a cos bx, a, b 6= 0, we may proceedwith the following steps:

(1) Determine the amplitude |a|, and find the period 2⇡|b| . To draw one cycle

of the graph (that is, one complete graph for one period), we just need tocomplete the graph from 0 to 2⇡

|b| .

(2) Divide the interval into four equal parts, and get five division points: x1 = 0,x2, x3, x4, and x5 =

2⇡|b| , where x3 is the midpoint between x1 and x5 (that

is, 12(x1 + x5) = x3), x2 is the midpoint between x1 and x3, and x4 is the

midpoint between x3 and x5.

(3) Evaluate the function at each of the five x-values identified in Step 2. Thepoints will correspond to the highest point, lowest point, and x-interceptsof the graph.

(4) Plot the points found in Step 3, and join them with a smooth curve similarto the graph of the basic sine curve.

(5) Extend the graph to the right and to the left, as needed.

Example 3.3.1. Sketch the graph of one cycle of y = 2 sin 4x.

147

Solution. (1) The period is 2⇡4 = ⇡

2 , and the amplitude is 2.

(2) Dividing the interval [0, ⇡2 ] into 4 equal parts, we get the following x-coordinates: 0, ⇡

8 ,⇡

4 ,3⇡8 , and

⇡

2 .

(3) When x = 0, ⇡

4 , and⇡

2 , we get y = 0. On the other hand, when x = ⇡

8 , wehave y = 2 (the amplitude), and y = �2 when x = 3⇡

8 .

(4) Draw a smooth curve by connecting the points. There is no need to proceedto Step 5 because the problem only asks for one cycle.

Example 3.3.2. Sketch the graph of y = �3 cos x

2 .

Solution. (1) The amplitude is |� 3| = 3, and the period is2⇡12

= 4⇡.

(2) We divide the interval [0, 4⇡] into four equal parts, and we get the followingx-values: 0, ⇡, 2⇡, 3⇡, and 4⇡.

(3) We have y = 0 when x = ⇡ and 3⇡, y = �3 when x = 0 and 4⇡, and y = 3when x = 2⇡.

(4) We trace the points in Step 3 by a smooth curve.

(5) We extend the pattern in Step 4 to the left and to the right.

Example 3.3.3. Sketch the graph of two cycles of y = 12 sin

��2x

3

�.

148

Solution. Since the sine function is odd, the graph of y = 12 sin

��2x

3

�is the same

as that of y = �12 sin

2x3 .

(1) The amplitude is 12 , and the period is

2⇡23

= 3⇡.

(2) Dividing the interval [0, 3⇡] into four equal parts, we get the x-coordinatesof the five important points:

0 + 3⇡

2=

3⇡

2,

0 + 3⇡2

2=

3⇡

4,

3⇡2 + 3⇡

2=

9⇡

4.

(3) We get y = 0 when x = 0, 3⇡2 , and 3⇡, y = �1

2 when 3⇡4 , and y = 1

2 when9⇡4 .

(4) We trace the points in Step 3 by a smooth curve.

(5) We extend the pattern in Step 4 by one more period to the right.


(1) Sketch the graph of one cycle of y = 12 sin 3x.

Answer:

(2) Sketch the graph of two cycles of y = �2 cos��x

2

�.

Answer:

149

(3) Sketch the graph of y = �2 cos 4x.

Answer:

(4) Sketch the graph of one cycle of y = 3 sin��x

3

�.

Answer:

3.3.3. Graphs of y = a sin b(x � c) + d and y = a cos b(x � c) + d

We first compare the graphs of y = sin x and y = sin�x� ⇡

3

�using a table of

values and the 5-step procedure discussed earlier.Teaching Notes

Review or teachthe horizontal

translation rule: ifx is replaced by

x� h in theequation, the

graph is translated|h| units to the

right if h > 0 andto the left if h < 0.

As x runs from ⇡

3 to 7⇡3 , the value of the expression x� ⇡

3 runs from 0 to 2⇡. Sofor one cycle of the graph of y = sin

�x� ⇡

3

�, we then expect to have the graph of

y = sin x starting from x = ⇡

3 . This is confirmed by the values in Table 3.18. Wethen apply a similar procedure to complete one cycle of the graph; that is, dividethe interval [⇡3 ,

7⇡3 ] into four equal parts, and then determine the key values of

x in sketching the graphs as discussed earlier. The one-cycle graph of y = sin x(dashed curve) and the corresponding one-cycle graph of y = sin

�x� ⇡

3

�(solid

curve) are shown in Figure 3.19.

x ⇡

35⇡6

4⇡3

11⇡6

7⇡3

x� ⇡

3 0 ⇡

2 ⇡ 3⇡2 2⇡

sin�x� ⇡

3

�0 1 0 �1 0

Table 3.18

150

Figure 3.19

Observe that the graph of y = sin�x� ⇡

3

�shifts ⇡

3 units to the right ofy = sin x. Thus, they have the same period, amplitude, domain, and range.

The graphs of

y = a sin b(x� c) and y = a cos b(x� c)

have the same shape as y = a sin bx and y = a cos bx, respectively, butshifted c units to the right when c > 0 and shifted |c| units to the leftif c < 0. The number c is called the phase shift of the sine or cosinegraph.

Example 3.3.4. In the same Cartesian plane, sketch one cycle of the graphs ofy = 3 sin x and y = 3 sin

�x+ ⇡

4

�.

Solution. We have sketched the graph of y = 3 sin x earlier at the start of thelesson. We consider y = 3 sin

�x+ ⇡

4

�. We expect that it has the same shape as

that of y = 3 sin x, but shifted some units.

Here, we have a = 3, b = 1, and c = �⇡

4 . From these constants, we getthe amplitude, the period, and the phase shift, and these are 3, 2⇡, and �⇡

4 ,respectively.

One cycle starts at x = �⇡

4 and ends at x = �⇡

4 +2⇡ = 7⇡4 . We now compute

the important values of x.

�⇡

4 + 7⇡4

2=

3⇡

4,

�⇡

4 + 3⇡4

2=

⇡

4,

3⇡4 + 7⇡

4

2=

5⇡

4

x �⇡

4⇡

43⇡4

5⇡4

7⇡4

y = 3 sin�x+ ⇡

4

�0 3 0 �3 0

151

WhileTeaching Notes

Review or teachthe vertical

translation rule: ifthe equationy = f(x) ischanged to

y = f(x) + k, thegraph is translated|k| units upward if

k > 0 anddownward if k < 0.

the e↵ect of c in y = a sin b(x � c) and y = a cos b(x � c) isa horizontal shift of their graphs from the corresponding graphs ofy = a sin bx and y = a cos bx, the e↵ect of d in the equations y =a sin b(x� c) + d and y = a cos b(x� c) + d is a vertical shift. That is,the graph of y = a sin b(x�c)+d has the same amplitude, period, andphase shift as that of y = a sin b(x � c), but shifted d units upwardwhen d > 0 and |d| units downward when d < 0.

Example 3.3.5. Sketch the graph of

y = �2 cos 2⇣x� ⇡

6

⌘� 3.

Solution. Here, a = �2, b = 2, c = ⇡

6 , and d = �3. We first sketch one cycle ofthe graph of y = �2 cos 2

�x� ⇡

6

�, and then extend this graph to the left and to

the right, and then move the resulting graph 3 units downward.

The graph of y = �2 cos 2�x� ⇡

6

�has amplitude 2, period ⇡, and phase shift

⇡

6 .

Start of one cycle: ⇡

6

End of the cycle: ⇡

6 + ⇡ = 7⇡6

⇡

6 + 7⇡6

2=

2⇡

3,

⇡

6 + 2⇡3

2=

5⇡

12,

2⇡3 + 7⇡

6

2=

11⇡

12

x ⇡

65⇡12

2⇡3

11⇡12

7⇡6

y = �2 cos 2�x� ⇡

6

��2 0 2 0 �2

y = �2 cos 2�x� ⇡

6

�� 3 �5 �3 �1 �3 �5

152

Before we end this sub-lesson, we make the following observation, which willbe used in the discussion on simple harmonic motion (Sub-Lesson 3.3.6).

Di↵erent Equations, The Same Graph

1. The graphs of y = sin x and y = sin(x + 2⇡k), k any integer, arethe same.

2. The graphs of y = sin x, y = � sin(x + ⇡), y = cos(x � ⇡

2 ), andy = � cos(x+ ⇡

2 ) are the same.

3. In general, the graphs of

y = a sin b(x� c) + d,

y = �a sin[b(x� c) + ⇡ + 2⇡k] + d,

y = a cos[b(x� c)� ⇡

2 + 2⇡k] + d,

andy = �a cos[b(x� c) + ⇡

2 + 2⇡k] + d,

where k is any integer, are all the same.

Similar observations are true for cosine.


(1) In the same Cartesian plane, sketch one cycle of the graphs of y = 3 cosxand y = 3 cos

�x+ ⇡

3

�� 1.

Answer:

153

(2) In the same Cartesian plane, sketch one cycle of the graphs of y = �14 sin 2x

and y = 2� 14 sin 2

�x� ⇡

4

�.

Answer:

(3) Sketch the graph of y = 2 sin�⇡

2 � x�� 2.

Answer:

3.3.4. Graphs of Cosecant and Secant Functions

We know that cscx = 1sinx

if sin x 6= 0. Using this relationship, we can sketch thegraph of y = cscx.

154

First, we observe that the domain of the cosecant function is

{x 2 R : sin x 6= 0} = {x 2 R : x 6= k⇡, k 2 Z}.

Table 3.20 shows the key numbers (that is, numbers where y = sin x crosses thex-axis, attain its maximum and minimum values) and some neighboring points,where “und” stands for “undefined,” while Figure 3.21 shows one cycle of thegraphs of y = sin x (dashed curve) and y = cscx (solid curve). Notice theasymptotes of the graph y = cscx.

x 0 ⇡

6⇡

25⇡6 ⇡ 7⇡

63⇡2

11⇡6 2⇡

y = sin x 0 12 1 1

2 0 �12 �1 �1

2 0

y = cscx und 2 1 2 und �2 �1 �2 und

Table 3.20

Figure 3.21

We could also sketch the graph of cscx directly from the graph of y = sin xby observing the following facts:

(1) If sin x = 1 (or �1), then csc x = 1 (or �1).

(2) At each x-intercept of y = sin x, y = cscx is undefined; but a verticalasymptote is formed because, when sin x is close to 0, the value of cscx willhave a big magnitude with the same sign as sinx.

Refer to Figure 3.22 for the graphs of y = sin x (dashed curve) and y = cscx(solid curve) over a larger interval.

155

Figure 3.22

Like the sine and cosecant functions, the cosine and secant functions are alsoreciprocals of each other. Therefore, y = secx has domain

{x 2 R : cosx 6= 0} = {x 2 R : x 6= k⇡

2, k odd integer}.

Similarly, the graph of y = secx can be obtained from the graph of y = cosx.These graphs are shown in Figure 3.23.

Figure 3.23

Example 3.3.6. Sketch the graph of y = 2 csc x

2 .

Solution. First, we sketch the graph of y = 2 sin x

2 , and use the technique dis-cussed above to sketch the graph of y = 2 csc x

2 .

156

The vertical asymptotes of y = 2 csc x

2 are the x-intercepts of y = 2 sin x

2 :x = 0,±2⇡,±4⇡, . . .. After setting up the asymptotes, we now sketch the graphof y = 2 csc x

2 as shown below.

Example 3.3.7. Sketch the graph of y = 2� sec 2x.

Solution. Sketch the graph of y = � cos 2x (note that it has period ⇡), then sketchthe graph of y = � sec 2x (as illustrated above), and then move the resultinggraph 2 units upward to obtain the graph of y = 2� sec 2x.

157


(1) Sketch the graph of y = � sec x on the interval [0, 2⇡].

Answer:

(2) Sketch the graph of y = 2 csc 4x� 1 on the interval⇥�⇡

2 ,⇡

2

⇤.

Answer:

158

3.3.5. Graphs of Tangent and Cotangent Functions

We know that tanx = sinx

cosx , where cosx 6= 0. From this definition of the tangentfunction, it follows that its domain is the same as that of the secant function,which is

{x 2 R : cosx 6= 0} = {x 2 R : x 6= k⇡

2, k odd integer}.

We note that tan x = 0 when sin x = 0 (that is, when x = k⇡, k any integer), andthat the graph of y = tan x has asymptotes x = k⇡

2 , k odd integer. Furthermore,by recalling the signs of tangent from Quadrant I to Quadrant IV and its values,we observe that the tangent function is periodic with period ⇡.

To sketch the graph of y = tan x, it will be enough to know its one-cyclegraph on the open interval

��⇡

2 ,⇡

2

�. See Table 3.24 and Figure 3.25.

Teaching Notes

There is also a wayof sketching thegraph of y = tanx

based on thetangent segment tothe unit circle,similar to theconstructiondescribed insketching thegraph of y = sinx.But we do not goanymore into thedetails of thisapproach.

x �⇡

2 �⇡

3 �⇡

4 �⇡

6 0

y = tan x und �p3 �1 �

p33 0

x ⇡

6⇡

4⇡

3⇡

2

y = tan xp33 1

p3 und

Table 3.24

Figure 3.25

In the same manner, the domain of y = cotx = cosxsinx

is

{x 2 R : sin x 6= 0} = {x 2 R : x 6= k⇡, k 2 Z},

and its period is also ⇡. The graph of y = cotx is shown in Figure 3.26.

159

Figure 3.26

In general, to sketch the graphs of y = a tan bx and y = a cot bx, a 6= 0 andb > 0, we may proceed with the following steps:

(1) Determine the period ⇡

b

. Then we draw one cycle of the graph on�� ⇡

2b ,⇡

2b

�

for y = a tan bx, and on�0, ⇡

b

�for y = a cot bx.

(2) Determine the two adjacent vertical asymptotes. For y = a tan bx, thesevertical asymptotes are given by x = ± ⇡

2b . For y = a cot bx, the verticalasymptotes are given by x = 0 and x = ⇡

b

.

(3) Divide the interval formed by the vertical asymptotes in Step 2 into fourequal parts, and get three division points exclusively between the asymp-totes.

(4) Evaluate the function at each of these x-values identified in Step 3. Thepoints will correspond to the signs and x-intercept of the graph.

(5) Plot the points found in Step 3, and join them with a smooth curve ap-proaching to the vertical asymptotes. Extend the graph to the right and tothe left, as needed.

Example 3.3.8. Sketch the graph of y = 12 tan 2x.

Solution. The period of the function is ⇡

2 , and the adjacent asymptotes are x =±⇡

4 ,±3⇡4 , . . .. Dividing the interval

��⇡

4 ,⇡

4

�into four equal parts, the key x-values

are �⇡

8 , 0, and⇡

8 .

x �⇡

8 0 ⇡

8

y = 12 tan 2x �1

2 0 12

160

Example 3.3.9. Sketch the graph of y = 2 cot x

3 on the interval (0, 3⇡).

Solution. The period of the function is 3⇡, and the adjacent asymptotes are x = 0and x = 3⇡. We now divide the interval (0, 3⇡) into four equal parts, and thekey x-values are 3⇡

4 ,3⇡2 , and

9⇡4 .

x 3⇡4

3⇡2

9⇡4

y = 2 cot x

3 2 0 �2

161


(1) Sketch the graph of y = cot(�x) on the interval [�⇡, ⇡].

Answer:

(2) Sketch the graph of y = 2 tan x

4 on the interval [�2⇡, 2⇡].

Answer:

3.3.6. Simple Harmonic Motion

Repetitive or periodic behavior is common in nature. The time-telling deviceknown as sundial is a result of the predictable rising and setting of the suneveryday. It consists of a flat plate and a gnomon. As the sun moves across thesky, the gnomon casts a shadow on the plate, which is calibrated to tell the timeof the day.

162

https://commons.wikimedia.org/wiki/File:Sundial 2r.jpg

By liz west (Sundial)[CC BY 2.0 (http://creativecommons.org/licenses/by/2.0)],


Some motions are also periodic. When a weight is suspended on a spring,pulled down, and released, the weight oscillates up and down. Neglecting resis-tance, this oscillatory motion of the weight will continue on and on, and its heightis periodic with respect to time.

t = 0 sec t = 2.8 sec

163

https://commons.wikimedia.org/wiki/File:Sundial_2r.jpg

t = 6.1 sec t = 9 sec

Periodic motions are usually modeled by either sine or cosine function, and arecalled simple harmonic motions. Unimpeded movements of objects like oscilla-tion, vibration, rotation, and motion due to water waves are real-life occurrencesthat behave in simple harmonic motion.

Equations of Simple Harmonic Motion

The displacement y (directed height or length) of an object behavingin a simple harmonic motion with respect to time t is given by one ofthe following equations:

y = a sin b(t� c) + d

ory = a cos b(t� c) + d.

In both equations, we have the following information:

• amplitude = |a| = 12(M � m) - the maximum displacement above

and below the rest position or central position or equilibrium, whereM is the maximum height and m is the minimum height;

• period = 2⇡|b| - the time required to complete one cycle (from one

highest or lowest point to the next);

• frequency = |b|2⇡ - the number of cycles per unit of time;

• c - responsible for the horizontal shift in time; and

• d - responsible for the vertical shift in displacement.

Example 3.3.10. A weight is suspended from a spring and is moving up anddown in a simple harmonic motion. At start, the weight is pulled down 5 cm belowthe resting position, and then released. After 8 seconds, the weight reaches its

164

highest location for the first time. Find the equation of the motion.

Solution. We are given that the weight is located at its lowest position at t = 0;that is, y = �5 when t = 0. Therefore, the equation is y = �5 cos bt.

Because it took the weight 8 seconds from the lowest point to its immediatehighest point, half the period is 8 seconds.

1

2· 2⇡b

= 8 =) b =⇡

8=) y = �5 cos

⇡t

82

?Example 3.3.11. Suppose you ride a Ferris wheel. The lowest point of thewheel is 3 meters o↵ the ground, and its diameter is 20 m. After it started, theFerris wheel revolves at a constant speed, and it takes 32 seconds to bring youback again to the riding point. After riding for 150 seconds, find your approximateheight above the ground.

Solution. We ignore first the fixed value of 3 m o↵ the ground, and assume thatthe central position passes through the center of the wheel and is parallel to theground.

Let t be the time (in seconds) elapsed that you have been riding the Ferriswheel, and y is he directed distance of your location with respect to the assumedcentral position at time t. Because y = �10 when t = 0, the appropriate modelis y = �10 cos bt for t � 0.

Given that the Ferris wheel takes 32 seconds to move from the lowest pointto the next, the period is 32.

2⇡

b= 32 =) b =

⇡

16=) y = �10 cos

⇡t

16

When t = 150, we get y = 10 cos 150⇡16 ⇡ 3.83.

Bringing back the original condition given in the problem that the riding pointis 3 m o↵ the ground, after riding for 150 seconds, you are approximately located3.83 + 13 = 16.83 m o↵ the ground. 2

In the last example, the central position or equilibrium may be verticallyshifted from the ground or sea level (the role of the constant d). In the same way,the starting point may also be horizontally shifted (the role of the constant c).Moreover, as observed in Sub-Lesson 3.3.3 (see page 153), to find the functionthat describes a particular simple harmonic motion, we can either choose

y = a sin b(t� c) + d

ory = a cos b(t� c) + d,

and determine the appropriate values of a, b, c, and d. In fact, we can assumethat a and b are positive numbers, and c is the smallest such nonnegative number.

165

Example 3.3.12. A signal buoy in Laguna Bay bobs up and down with theheight h of its transmitter (in feet) above sea level modeled by h(t) = a sin bt+ dat time t (in seconds). During a small squall, its height varies from 1 ft to 9 ftabove sea level, and it takes 3.5 seconds from one 9-ft height to the next. Findthe values of the constants a, b, and d.

Solution. We solve the constants step by step.

• The minimum and maximum values of h(t) are 1 ft and 9 ft, respectively.Thus, the amplitude is a = 1

2(M �m) = 12(9� 1) = 4.

• Because it takes 3.5 seconds from one 9-ft height to the next, the period is3.5. Thus, we have 2⇡

b

= 3.5, which gives b = 4⇡7 .

• Because the lowest point is 1 ft above the sea level and the amplitude is 4,it follows that d = 5. 2

Example 3.3.13. A variable star is a star whose brightness fluctuates as ob-served from Earth. The magnitude of visual brightness of one variable star rangesfrom 2.0 to 10.1, and it takes 332 days to observe one maximum brightness tothe next. Assuming that the visual brightness of the star can be modeled by theequation y = a sin b(t � c) + d, t in days, and putting t = 0 at a time when thestar is at its maximum brightness, find the constants a, b, c, and d, where a, b > 0and c the least nonnegative number possible.

Solution.

a =M �m

2=

10.1� 2.0

2= 4.05

2⇡

b= 332 =) b =

⇡

166d = a+m = 4.05 + 2.0 = 6.05

For the (ordinary) sine function to start at the highest point at t = 0, the leastpossible horizontal movement to the right (positive value) is 3⇡

2 units.

bc =3⇡

2=) c =

3⇡

2b=

3⇡

2 · ⇡

166

= 249 2

?Example 3.3.14. The path of a fast-moving particle traces a circle with equa-tion

(x+ 7)2 + (y � 5)2 = 36.

It starts at point (�1, 5), moves clockwise, and passes the point (�7, 11) for thefirst time after traveling 6 microseconds. Where is the particle after traveling 15microseconds?

166

Solution. As described above, we may choose sine or cosine function. Here, wechoose the sine function to describe both x and y in terms of time t in microsec-onds; that is, we let

x = a sin b(t� c) + d and y = e sin f(t� g) + h,

where we appropriately choose the positive values for a, b, e, and f , and the leastnonnegative values for c and g.

The given circle has radius 6 and center (�7, 5). Defining the central positionof the values of x as the line x = �7 and that of the values of y as the line y = 5,we get a = e = 6, d = �7, and h = 5.

From the point (�1, 5) to the point (�7, 11) (moving clockwise), the particlehas traveled three-fourths of the complete cycle; that is, three-fourths of theperiod must be 2.

3

4· 2⇡b

=3

4· 2⇡f

= 6 =) b = f =⇡

4

As Teaching Notes

Here, we need anequation with thesame graph asy = a sin(bt+ ⇡

2 )+d

that will fit in theequationy = a sin b(t�c)+d,where c is the leastnonnegativepossible number.Recall theobservation madeon page 153.

the particle starts at (�1, 5) and moves clockwise, the values of x startat its highest value (x = �1) and move downward toward its central position(x = �7) and continue to its lowest value (x = �13). Therefore, the graph ofa sin bt+ d has to move 3⇡

2b = 6 units to the right, and so we get c = 6.

As to the value of g, we observe the values of y start at its central position(y = 5) and go downward to its lowest value (y = �1). Similar to the argumentused in determining c, the graph of y = e sin ft + h has to move ⇡

b

= 4 units tothe right, implying that g = 4.

Hence, We have the following equations of x and y in terms of t:

x = 6 sin ⇡

4 (t� 6)� 7 and y = 6 sin ⇡

4 (t� 4) + 5.

When t = 15, we get

x = 6 sin ⇡

4 (15� 6)� 7 = �7 + 3p2 ⇡ �2.76

andy = 6 sin ⇡

4 (15� 4) + 5 = 5 + 3p2 ⇡ 9.24.

That is, after traveling for 15 microseconds, the particle is located near the point(�2.76, 9.24). 2


?1. A weight is suspended from a spring and is moving up and down in a simpleharmonic motion. At start, the weight is pushed up 6 cm above the resting

167

position, and then released. After 14 seconds, the weight reaches again to itshighest position. Find the equation of the motion, and locate the weight withrespect to the resting position after 20 seconds since it was released.

Answer: y = 6 cos ⇡

7 t or y = 6 sin pi

7 (t +72), location of the weight after 20

seconds: about 5.4 cm below the resting position

2. Suppose the lowest point of a Ferris wheel is 1.5 meters o↵ the ground, and itsradius is 15 m. It makes one complete revolution every 30 seconds. Starting atthe lowest point, find a cosine function that gives the height above the groundof a riding child in terms of the time t in seconds.

Answer: y = 15 cos ⇡

15

�t� 15

2

�+ 16.5

Exercises 3.3

1. Sketch two cycles of the graph (starting from x = 0) of the given function.Indicate the amplitude, period, phase shift, domain, and range for each func-tion.

(a) y = 4 sin x

Answer: amplitude = 4, period = 2⇡, phase shift = 0, domain = R,range = [�4, 4]

(b) y = 3 cosx


(c) y = cos x

4


(d) y = � sin 2x

Answer: amplitude = 1, period = ⇡, phase shift = 0, domain = R,range = [�1, 1]

168

(e) y = 2 + sin 4x

Answer: amplitude = 1, period = ⇡

2 , phase shift = 0, domain = R,range = [1, 3]

(f) y = �1 + cos x


(g) y = �12 sin 3x

Answer: amplitude = 12 , period = 2⇡

3 , phase shift = 0, domain = R,range = [�1

2 ,12 ]

(h) y = 3 sin(�x)


(i) y = 3� 2 cos x

2

Answer: amplitude = 2, period = 4⇡, phase shift = 0, domain = R,range = [1, 5]

(j) y = sin�x� ⇡

4

�

Answer: amplitude = 1, period = 2⇡, phase shift = ⇡

4 , domain = R,range = [�1, 1]

169

(k) y = 2 cos�x+ ⇡

3

�

Answer: amplitude = 2, period = 2⇡, phase shift = �⇡


(l) y = 3 sin(x� 4⇡)

Answer: amplitude = 3, period = 2⇡, phase shift = 4⇡, domain = R,range = [�3, 3]

(m) y = 2� 23 cos

�x� ⇡

2

�

Answer: amplitude = 23 , period = 2⇡, phase shift = ⇡

2 , domain = R,range = [43 ,

83 ]

(n) y = �4 cos�x� ⇡

3

�+ 2

Answer: amplitude = 4, period = 2⇡, phase shift = ⇡


2. Sketch the graph of the following functions.

(a) y = | sin x|Answer:

170

(b) y = |4 cosx|+ 2

Answer:

(c) y = |2 sin 2(x+ ⇡)|� 1

Answer:

3. Sketch the graph of each function over two periods, starting from x = 0.Indicate the period, phase shift, domain, and range of each function.

(a) y = csc(�x)

Answer: period = 2⇡, phase shift = 0, domain = {x|x 6= k⇡, k 2 Z},range = (�1,�1] [ [1,1)

(b) y = cot(�x)

Answer: period = ⇡, phase shift = 0, domain = {x|x 6= k⇡, k 2 Z},range = R

(c) y = � tan x

Answer: period = ⇡, phase shift = 0, domain = {x|x 6= (2k + 1)⇡2 , k 2Z}, range = R

171

(d) y = � sec x

Answer: period = 2⇡, phase shift = 0, domain = {x|x 6= (2k + 1)⇡2 , k 2Z}, range = (�1,�1] [ [1,1)

(e) y = � sec 3x

Answer: period = 2⇡3 , phase shift = 0, domain = {x|x 6= (2k + 1)⇡6 , k 2

Z}, range = (�1,�1] [ [1,1)

(f) y = 3 cscx


(g) y = 4 sec 2x3

Answer: period = 3⇡, phase shift = 0, domain = {x|x 6= (2k + 1)3⇡4 , k 2Z}, range = (�1,�4] [ [4,1)

(h) y = tan(x+ ⇡)

172

Answer: period = ⇡, phase shift = �⇡, domain = {x|x 6= (2k+ 1)⇡2 , k 2Z}, range = R

(i) y = tan�x� ⇡

2

�

Answer: period = ⇡, phase shift = ⇡

2 , domain = {x|x 6= k⇡, k 2 Z},range = R

(j) y = cot�x+ ⇡

4

�

Answer: period = ⇡, phase shift = ⇡

4 , domain = {x|x 6= (2k�1)⇡4 , k even integer},range = R

(k) y = 2� 3 cscx


(l) y = 4 + sec 3x

Answer: period = 2⇡3 , phase shift = 0, domain = {x|x 6= (2k + 1)⇡6 , k 2

Z}, range = (�1, 3] [ [6,1)

(m) y = 2 sec�x� ⇡

3

�

Answer: period = 2⇡, phase shift = ⇡

3 , domain = {x|x 6= (2k+1)3⇡4 , k 2Z}, range = (�1,�2] [ [2,1)

(n) y = 2� 3 sec 2x3

Answer: period = 3⇡, phase shift = 0, domain = {x|x 6= (2k + 1)3⇡4 , k 2Z}, range = (�1,�2] [ [5,1)

173

(o) y = 3 csc�x� 3⇡

2

�

Answer: period = 2⇡, phase shift = 3⇡2 , domain = {x|x 6= (2k+1)⇡2 , k 2

Z}, range = (�1,�3] [ [3,1)

4. Assuming that there is no vertical shift, find a function that describes a simpleharmonic motion with the following properties.

(a) sine function; displacement zero at time t = 0; moving up initially;amplitude = 6 cm; period = 4 sec Answer: y = 6 sin ⇡

2 t

(b) cosine function; highest point 4 cm above the equilibrium at time t = 0;period = 10 sec Answer: y = 5 cos ⇡

5 t

174

(c) cosine function; lowest point 9 cm below the equilibrium at time t = 0;period = 5 sec Answer: y = 9 cos 2⇡

5 (t�52)

5. A point P moving in a simple harmonic motion makes 10 complete revolutionsevery 1 second. The amplitude of the motion is 3 m. Assuming that P is atits minimum displacement with respect to the equilibrium when t = 0 andthere is a vertical shift of 2 m downward, find a sine function that describesthe path traced by P in terms of time t. Answer: y = 3 sin 20⇡(t� 1

40)� 2?6. The path of a fast-moving particle (assuming constant speed) traces a circle

with equation(x� 3)2 + (y � 4)2 = 25.

It starts at point (3,�1), moves counterclockwise, and passes the point (8, 4)for the first time after traveling 7 microseconds. Where is the particle aftertraveling for 20 microseconds? Answer: about the point (�1.87, 5.11)

Hint. The coordinates (x, y) of the location of the particle at time t (in mi-croseconds) are given by x = 5 cos ⇡

14(t� 7) + 3 and y = 5 sin ⇡

14(t� 7) + 4.

7. A wooden ball is tied on a string 30 cm long, and is oscillating like a pendulum.See figure below. It is initially pulled back at 90� angle with the vertical, andis released with a push so that the ball reaches its maximum height back andforth. If it reaches its maximum height again after 3 seconds, find its height10 seconds after it was released. Answer: 27 cm

Hint. The height h(t) (in cm) of the ball at time t (in seconds) is given byh(t) = |30 sin ⇡

3 (t�32)|+ 12.

8. For what values of k do y = cotx and y = cot(x� k⇡) have the same graph?

Answer: any integer

9. For what values of k do y = secx and y = sec(x� k⇡) have the same graph?

Answer: any even integer

175

10. Find the least positive value of c such that the graph of y = �2 sin 2(x + c)coincide with that of y = 2 cos 2x. Answer: ⇡

4

11. Find the largest positive value of c such that the graph of y = 2 cos 3(x � c)coincide with that of y = �2 cos 3(x� 2). Answer: 2 + ⇡

3

12. For what values of a do the graphs of y = a cos b(x�c) and y = �2 sec ⇡

6 (x�6)never intersect for any values of b and c? Answer: �2 < a < 2

4

Lesson 3.4. Fundamental Trigonometric Identities




(1) determine whether an equation is an identity or a conditional equation;

(2) derive the fundamental trigonometric identities;

(3) simplify trigonometric expressions using fundamental trigonometric identi-ties; and

(4) prove other trigonometric identities using fundamental trigonometric identi-ties.

Lesson Outline

(1) Domain of an equation

(2) Identity and conditional equation

(3) Fundamental trigonometric identities

(4) Proving trigonometric identities

Introduction

In previous lessons, we have defined trigonometric functions using the unitcircle and also investigated the graphs of the six trigonometric functions. Thislesson builds on the understanding of the di↵erent trigonometric functions bydiscovery, deriving, and working with trigonometric identities.

3.4.1. Domain of an Expression or Equation

Consider the following expressions:

2x+ 1,px2 � 1,

x

x2 � 3x� 4,

xpx� 1

.

176

What are the real values of the variable x that make the expressions defined inthe set of real numbers?

In the first expression, every real value of x when substituted to the expressionmakes it defined in the set of real numbers; that is, the value of the expression isreal when x is real.

In the second expression, not every real value of xmakes the expression definedin R. For example, when x = 0, the expression becomes

p�1, which is not a real

number.px2 � 1 2 R () x2 � 1 � 0 () x �1 or x � 1

Here, forpx2 � 1 to be defined in R, x must be in (�1,�1] [ [1,1).

In the third expression, the values of x that make the denominator zero makethe entire expression undefined.

x2 � 3x� 4 = (x� 4)(x+ 1) = 0 () x = 4 or x = �1

Hence, the expressionx

x2 � 3x� 4is real when x 6= 4 and x 6= �1.

In the fourth expression, because the expressionpx� 1 is in the denominator,

x must be greater than 1. Although the value of the entire expression is 0 whenx = 0, we do not include 0 as allowed value of x because part of the expressionis not real when x = 0.

In the expressions above, the allowed values of the variable x constitute thedomain of the expression.

The domain of an expression (or equation) is the set of all real values ofthe variable for which every term (or part) of the expression (equation)is defined in R.

In the expressions above, the domains of the first, second, third, and fourthexpressions are R, (�1,�1] [ [1,1), R \ {�1, 4}, and (1,1), respectively.

Example 3.4.1. Determine the domain of the expression/equation.

(a)x2 � 1

x3 + 2x2 � 8x�

px+ 1

1� x

(b) tan ✓ � sin ✓ � cos 2✓

(c) x2 �p1 + x2 =

23px2 � 1

(d) z � cos2 z

1 + sin z= 4 sin z � 1

177

Solution. (a) x3 + 2x2 � 8x = x(x + 4)(x � 2) = 0 () x = 0, x =�4, or x = 2px+ 1 2 R () x+ 1 � 0 () x � �1

1� x = 0 () x = 1

Domain = [�1,1) \ {�4, 0, 1, 2}= [�1, 0) [ (0, 1) [ (1, 2) [ (2,1)

(b) tan ✓ � sin ✓ � cos 2✓ = sin ✓

cos ✓ � sin ✓ � cos 2✓

cos ✓ = 0 () ✓ = k⇡

2 , k odd integer

Domain = R \ {k⇡

2 | k odd integer}

(c) The expression 1+x2 is always positive, and sop1 + x2 is defined in R. On

the other hand, the expression 3px2 � 1 is also defined in R, but it cannot

be zero because it is in the denominator. Therefore, x should not be �1and 1.

Domain = R \ {�1, 1}

(d) 1 + sin z = 0 () z = 3⇡2 + 2k⇡, k 2 Z

Domain = R \ {3⇡2 + 2k⇡|k 2 Z} 2


Find the domain of the expression/equation.

(1)x

x+ 1� x+ 1

x2 + 2x+ 1� 2

x2 � 1Answer: R \ {�1, 1}

(2) 2 sec2 ✓ =1

1� sin ✓+

1

1 + sin ✓Answer: R \

�(2k + 1)⇡2 |k 2 Z

(3) 2 tan2 x = 2 cot x+ 1 Answer: R \�

k⇡

2 |k 2 Z

(4)1p

1� x2� tan x = sin x Answer: R \

�{�1, 1} [

�(2k + 1)⇡2 |k 2 Z

3.4.2. Identity and Conditional Equation

Consider the following two groups of equations:

Group A Group B

(A1) x2 � 1 = 0 (B1) x2 � 1 = (x� 1)(x+ 1)

(A2) (x+ 7)2 = x2 + 49 (B2) (x+ 7)2 = x2 + 14x+ 49

(A3)x2 � 4

x� 2= 2x� 1 (B3)

x2 � 4

x� 2= x+ 2

178

In each equation in Group A, some values of the variable that are in thedomain of the equation do not satisfy the equation (that is, do not make theequation true). On the other hand, in each equation in Group B, every elementin the domain of the equation satisfies the given equation. The equations inGroup A are called conditional equations, while those in Group B are calledidentities.

An identity is an equation that is true for all values of the variablein the domain of the equation. An equation that is not an identity iscalled a conditional equation. (In other words, if some values of thevariable in the domain of the equation do not satisfy the equation,then the equation is a conditional equation.)

Example 3.4.2. Identify whether the given equation is an identity or a condi-tional equation. For each conditional equation, provide a value of the variable inthe domain that does not satisfy the equation.

(1) x3 � 2 =�x� 3

p2� �

x2 + 3p2x+ 3

p4�

(2) sin2 ✓ = cos2 ✓ + 1

(3) sin ✓ = cos ✓ � 1

(4)1�

px

1 +px=

1� 2px+ x

1� x

Solution. (1) This is an identity because this is simply factoring of di↵erence oftwo cubes.

(2) This is a conditional equation. If ✓ = 0, then the left-hand side of the equationis 0, while the right-hand side is 2.

(3) This is also a conditional equation. If ✓ = 0, then both sides of the equationare equal to 0. But if ✓ = ⇡, then the left-hand side of the equation is 0,while the right-hand side is �2.

(4) This is an identity because the right-hand side of the equation is obtained byrationalizing the denominator of the left-hand side. 2


Identify whether the given equation is an identity or a conditional equation. Foreach conditional equation, provide a value of the variable in the domain that doesnot satisfy the equation.

(1) 1 + x+x2

1� x=

1

1� xAnswer: identity

179

(2)cos2 ✓ � sin2 ✓

cos ✓ + sin ✓= cos ✓ � sin ✓ Answer: identity

(3) tan ✓ = cot ✓ Answer: conditional equation, ✓ = ⇡

2

(4) cos2 x = 2 cosx+ 3 Answer: conditional equation, x = 0

3.4.3. The Fundamental Trigonometric Identities

Recall that if P (x, y) is the terminal point on the unit circle corresponding to ✓,then we have

sin ✓ = y csc ✓ =1

ytan ✓ =

y

x

cos ✓ = x sec ✓ =1

xcot ✓ =

x

y.

From the definitions, the following reciprocal and quotient identities immedi-ately follow. Note that these identities hold if ✓ is taken either as a real numberor as an angle.

Reciprocal Identities

csc ✓ =1

sin ✓sec ✓ =

1

cos ✓cot ✓ =

1

tan ✓

Quotient Identities

tan ✓ =sin ✓

cos ✓cot ✓ =

cos ✓

sin ✓

We can use these identities to simplify trigonometric expressions.

Example 3.4.3. Simplify:

(1)tan ✓ cos ✓

sin ✓(2)

cos ✓

cot ✓

Solution. (1)tan ✓ cos ✓

sin ✓=

sin ✓

cos ✓ cos ✓

sin ✓= 1

(2)cos ✓

cot ✓=

cos ✓cos ✓sin ✓

= sin ✓ 2

If P (x, y) is the terminal point on the unit circle corresponding to ✓, thenx2 + y2 = 1. Since sin ✓ = y and cos ✓ = x, we get

sin2 ✓ + cos2 ✓ = 1.

180

By dividingTeaching Notes

The assumption inthe division is thatthe divisor isnonzero.

both sides of this identity by cos2 ✓ and sin2 ✓, respectively, we obtain

tan2 ✓ + 1 = sec2 ✓ and 1 + cot2 ✓ = csc2 ✓.

Pythagorean Identities

sin2 ✓ + cos2 ✓ = 1

tan2 ✓ + 1 = sec2 ✓ 1 + cot2 ✓ = csc2 ✓

Example 3.4.4. Simplify:

(1) cos2 ✓ + cos2 ✓ tan2 ✓ (2)1 + tan2 ✓

1 + cot2 ✓

Solution. (1) cos2 ✓ + cos2 ✓ tan2 ✓ = (cos2 ✓)(1 + tan2 ✓)

= cos2 ✓ sec2 ✓

= 1

(2)1 + tan2 ✓

1 + cot2 ✓=

sec2 ✓

csc2 ✓=

1cos2 ✓

1sin2 ✓

=sin2 ✓

cos2 ✓= tan2 ✓ 2

In addition to the eight identities presented above, we also have the followingidentities.

Even-Odd Identities

sin(�✓) = � sin ✓ cos(�✓) = cos ✓

tan(�✓) = � tan ✓

The first two of the negative identities can be obtained from the graphs ofthe sine and cosine functions, respectively. (Please review the discussion on page145.) The third identity can be derived as follows:

Teaching Notes

The correspondingreciprocalfunctions followthe sameEven-OddIdentities:csc(�✓) = � csc ✓sec(�✓) = sec ✓cot(�✓) = � cot ✓.

tan(�✓) =sin(�✓)

cos(�✓)=

� sin ✓

cos ✓= � tan ✓.

The reciprocal, quotient, Pythagorean, and even-odd identities constitutewhat we call the fundamental trigonometric identities.

We now solve Example 3.2.3 in a di↵erent way.

Example 3.4.5. If sin ✓ = �34 and cos ✓ > 0. Find cos ✓.

181

Solution. Using the identity sin2 ✓ + cos2 ✓ = 1 with cos ✓ > 0, we have

cos ✓ =p

1� sin2 ✓ =

s

1�✓�3

4

◆2

=

p7

4. 2

Example 3.4.6. If sec ✓ = 52 and tan ✓ < 0, use the identities to find the values

of the remaining trigonometric functions of ✓.

Solution. Note that ✓ lies in QIV.

cos ✓ =1

sec ✓=

2

5

sin ✓ = �p1� cos2 ✓ = �

s

1�✓2

5

◆2

= �p21

5

csc ✓ =1

sin ✓= �5

p21

21

tan ✓ =sin ✓

cos ✓=

�p215

25

= �p21

2

cot ✓ =1

tan ✓= �2

p21

212


1. Use the identities presented in this lesson to simplify each trigonometric ex-pression.

(a)1 + tan x

1 + cot xAnswer: tan x

Solution.1 + tan x

1 + cot x=

1 + tan x

1 + 1tanx

= tan x

(b)sin ✓

1 + cos ✓+

1 + cos ✓

sin ✓Answer: 2 csc ✓

Solution.sin ✓

1 + cos ✓+

1 + cos ✓

sin ✓=

sin2 ✓

sin ✓(1 + cos ✓)+

(1 + cos ✓)(1 + cos ✓)

sin ✓(1 + cos ✓)

=sin2 ✓ + (1 + 2 cos ✓ + cos2 ✓)

sin ✓(1 + cos ✓)

=2 + 2 cos ✓

sin ✓(1 + cos ✓)=

2

sin ✓= 2 csc ✓

182

(c)tan y + cot y

sec y csc yAnswer: 1

Solution.tan y + cot y

sec y csc y=

sin y

cos y +cos ysin y

1cos y ·

1sin y

=sin2 y+cos2 ycos y sin y

1cos y sin y

= sin2 y + cos2 y = 1

(d) 1� cos2 ✓

1 + sin ✓Answer: sin ✓

Solution. 1� cos2 ✓

1 + sin ✓=

1 + sin ✓ � cos2 ✓

1 + sin ✓

=sin2 ✓ + sin ✓

1 + sin ✓=

sin ✓(1 + sin ✓)

1 + sin ✓= sin ✓

2. Given some initial values, use the identities to find the values of the remainingtrigonometric functions of ✓.

(a) sin ✓ = 25 and sec ✓ > 0

Answer: ✓ in QI; csc ✓ = 52 , cos ✓ =

p1� sin2 ✓ =

p215 , sec ✓ = 5

p21

21 ,

tan ✓ = 2p21

21 , cot ✓ =p212

(b) sec ✓ = �83 and tan ✓ > 0

Answer: ✓ in QIII; cos ✓ = �38 , sin ✓ = �

p1� cos2 ✓ = �

p74 , csc ✓ =

�4p7

7 , tan ✓ = 2p7

3 , cot ✓ = 3p7

14

(c) tan ✓ = 2 and csc ✓ < 0

Answer: ✓ in QIII; cot ✓ = 12 , sec ✓ = �

p5, cos ✓ = �

p55 , csc ✓ = �

p54 ,

sin ✓ = �4p5

5

(d) csc ✓ = 32 and sec ✓ < 0

Answer: ✓ in QII; sin ✓ = 23 , cos ✓ = �

p1� sin2 ✓ = �

p53 , sec ✓ = �3

p5

5 ,

tan ✓ = �2p5

5 , cot ✓ = �p52

3.4.4. Proving Trigonometric Identities

We can use the eleven fundamental trigonometric identities to establish otheridentities. For example, suppose we want to establish the identity

csc ✓ � cot ✓ =sin ✓

1 + cos ✓.

To verify that it is an identity, recall that we need to establish the truth of theequation for all values of the variable in the domain of the equation. It is notenough to verify its truth for some selected values of the variable. To prove it, weuse the fundamental trigonometric identities and valid algebraic manipulationslike performing the fundamental operations, factoring, canceling, and multiplyingthe numerator and denominator by the same quantity.

183

Start on the expression on one side of the proposed identity (preferably thecomplicated side), use and apply some of the fundamental trigonometric identitiesand algebraic manipulations, and arrive at the expression on the other side of theproposed identity.

Expression Explanation

csc ✓ � cot ✓ Start on one side.

=1

sin ✓� cos ✓

sin ✓Apply some reciprocal andquotient identities.

=1� cos ✓

sin ✓Add the quotients.

=1� cos ✓

sin ✓· 1 + cos ✓

1 + cos ✓Multiply the numeratorand denominator by1 + cos ✓.

=1� cos2 ✓

(sin ✓)(1 + cos ✓)Multiply.

=sin2 ✓

(sin ✓)(1 + cos ✓)Apply a Pythagoreanidentity.

=sin ✓

1 + cos ✓Reduce to lowest terms.

Upon arriving at the expression of the other side, the identity has been estab-lished. There is no unique technique to prove all identities, but familiarity withthe di↵erent techniques may help.

Example 3.4.7. Prove: secx� cos x = sin x tan x.

Solution.

sec x� cos x =1

cos x� cos x

=1� cos2 x

cos x=

sin2 x

cos x= sin x · sin x

cos x= sin x tan x 2

Example 3.4.8. Prove:1 + sin ✓

1� sin ✓� 1� sin ✓

1 + sin ✓= 4 sin ✓ sec2 ✓

Solution.

1 + sin ✓

1� sin ✓� 1� sin ✓

1 + sin ✓=

(1 + sin ✓)2 � (1� sin ✓)2

(1� sin ✓)(1 + sin ✓)

=1 + 2 sin ✓ + sin2 ✓ � 1 + 2 sin ✓ � sin2 ✓

1� sin2 ✓

184

=4 sin ✓

cos2 ✓= 4 sin ✓ sec2 ✓ 2


Prove each identity.

1. tan x+ cot x = cscx sec x

Answer: tan x+ cot x =sin x

cos x+

cos x

sin x

=sin2 x+ cos2 x

sin x cos x

=1

sin x cos x=

1

cos x· 1

sin x= cscx sec x

2. sec ✓ + tan ✓ =1

sec ✓ � tan ✓

Answer:1

sec ✓ � tan ✓=

11

cos ✓ �sin ✓

cos ✓

=cos ✓

1� sin ✓

=cos ✓

1� sin ✓· 1 + sin ✓

1 + sin ✓

=cos ✓(1 + sin ✓)

1� sin2 ✓

=cos ✓(1 + sin ✓)

cos2 ✓

=1 + sin ✓

cos ✓=

1

cos ✓+

sin ✓

cos ✓= sec ✓ + tan ✓

3.sec y + tan y

csc y + 1= tan y

Answer:sec y + tan y

csc y + 1=

1cos y +

sin y

cos y1

sin y

+ 1

=1+sin y

cos y1+sin y

sin y

=sin y

cos y= tan y

4. 2 csc2 ✓ =1

1� cos ✓+

1

1 + cos ✓

Answer:1

1� cos ✓+

1

1 + cos ✓=

1 + cos ✓ + 1� cos ✓

1� cos2 ✓=

2

sin2 ✓= 2 csc2 ✓

185

Exercises 3.4

1. Find the domain of the equation.

(a) 3px+ 2�

px = 2x Answer: {x|x � 0}

(b) sin3 x = sin x+ 1 Answer: R(c) tan x+ cot x = sin x Answer: R \

�k⇡

2 |k 2 Z

(d)x+ 1

x2 � 1+ cosx = cscx Answer: R \ {{�1, 1} [ {k⇡|k 2 Z}}

2. Simplify each expression using the fundamental identities.

(a)sin2 ✓

sec2 ✓ � 1Answer: cos2 ✓

Solution.sin2 ✓

sec2 ✓ � 1=

sin2 ✓

tan2 ✓=

sin2 ✓

sin2 ✓

cos2 ✓

= cos2 ✓

(b)1

1 + tan2 x+

1

1 + cot2 xAnswer: 1

Solution.1

1 + tan2 x+

1

1 + cot2 x=

1

sec2 x+

1

csc2 x= cos2 x+ sin2 x = 1

(c) 1� cos2 x

1 + sin xAnswer: � sin x

Solution. 1� cos2 x

1 + sin x= 1� 1� sin2 x

1 + sin x

= 1� (1� sin x)(1 + sin x)

1 + sin x= 1� 1� sin x = � sin x

(d)sin ✓

cos ✓ tan ✓Answer: 1

Solution.sin ✓

cos ✓ tan ✓= tan ✓ · 1

tan ✓= 1

3. Given some initial information, use the identities to find the values of thetrigonometric functions of ✓.

(a) csc ✓ = 53 and tan ✓ > 0

Answer: ✓ in QI; sin ✓ = 35 , cos ✓ =

p1� sin2 ✓ = 4

5 , sec ✓ = 54 , tan ✓ = 3

4 ,cot ✓ = 4

3

(b) tan ✓ = �125 and cos ✓ < 0

Answer: ✓ in QII; cot ✓ = � 512 , sec ✓ = �

ptan2 ✓ + 1 = �13

5 , cos ✓ = � 513 ,

sin ✓ =p1� cos2 ✓ = 12

13 , csc ✓ = 1312

186

(c) csc ✓ = �32 and ⇡ < x < 3⇡

2

Answer: ✓ in QIII; sin ✓ = �23 , cos ✓ = �

p1� sin2 ✓ = �

p53 , sec ✓ =

�3p5

5 , tan ✓ = 2p5

5 , cot ✓ =p52

(d) cot ✓ = �75 and 3⇡

2 < ✓ < 2⇡

Answer: ✓ in QIV; tan ✓ = �57 , sec ✓ =

ptan2 ✓ + 1 =

p747 , cos ✓ = 7

p74

74 ,

csc ✓ = �pcot2✓ + 1 = �

p745 , sin ✓ = �5

p74

74

(e) sin ✓ = �1

Answer: ✓ coterminal with 3⇡2 ; csc ✓ = �1, cos ✓ = 0, sec ✓ undefined,

tan ✓ undefined, cot ✓ = 0

(f) cot ✓ = �1

Answer: ✓ either in QII or QIV

✓ in QII: tan ✓ = �1, sin ✓ =p22 , csc ✓ =

p2, cos ✓ = �

p22 , sec ✓ = �

p2

✓ in QIV: tan ✓ = �1, sin ✓ = �p22 , csc ✓ = �

p2, cos ✓ =

p22 , sec ✓ =

p2

4. Determine whether the given equation is an identity or a conditional equation.If it is an identity, prove it; otherwise, provide a value of the variable in thedomain that does not satisfy the equation.

(a) sin x cos x = 1 Answer: conditional equation, x = 0

(b) sin3 x = cosx� 1 Answer: conditional equation, x = ⇡

2

(c) (sin x� cos x)2 + (sin x+ cosx)2 = 2 Answer: identity

Proof. (sin x� cos x)2 + (sin x+ cosx)2

= (cos2 x� 2 cosx sin x+ sin2 x) + (cos2 x+ 2 cosx sin x+ sin2 x)

= 1 + 1 = 2

(d) tan(�x) cot x = �1 Answer: conditional equation, x = ⇡

2

(e) 2� sin2 x = secx+ cosx Answer: conditional equation, x = ⇡

2

5. Prove the following identities.

(a) sin3 x = sin x� sin x cos2 x

Solution. sin3 x = sin2 x · sin x = (1� cos x) sin x = sin x� sin x cos2 x

(b) sin4 x� cos4 x = sin2 x� cos2 x

Solution. sin4 x�cos4 x = (sin2 x�cos2 x)(sin2 x+cos2 x) = sin2 x�cos2 x

(c) tan(�✓) sin(�✓) + cos(�✓) = sec(�✓)

187

Solution. tan(�✓) sin(�✓) + cos(�✓) = tan ✓ sin ✓ + cos ✓

=sin2 ✓

cos ✓+ cos ✓

=sin2 ✓ + cos2 ✓

cos ✓

=1

cos ✓= sec ✓ = sec(�✓)

(d)1 + sin u+ cosu

1 + sin u� cos u=

1 + cosu

sin u

Solution

1 + sin u+ cosu

1 + sin u� cos u=

1 + sin u+ cosu

sin u+ 1� cos u· 1 + cosu

1 + cosu

=(1 + sin u+ cosu)(1 + cosu)

sin u+ sin u cos u+ 1� cos2 uTeaching Notes

Since you need

1 + cosu to retainin the numerator

at the end, do not

expand thenumerator.


sin u+ sin u cos u+ sin2 u


(sin u)(1 + cosu+ sin u)

=1 + cosu

sin u

6. Express1� sec2 x

sec2 xin terms of sin x. Answer: � sin2 x

Solution.1� sec2 x

sec2 x=

1

sec2 x� 1 = cos2 x� 1 = � sin2 x

7. Express tan x sec x in terms of cosx. Answer:±p1� cos2 x

cos2 x

Solution. tan x sec x = sinx

cosx · 1cosx = sinx

cos2 x =±p1� cos2 x

cos2 x8. Express all other five trigonometric functions in terms of tanx (allowing ± in

the expression).

Answer: sin x =tan x

±ptan2 x+ 1

; cos x =1

±ptan2x+ 1

; cot x =1

tan x; sec x =

±ptan2 x+ 1; cscx =

±ptan2 x+ 1

tan x9. If sec ✓ � tan ✓ = 3, what is sec ✓ + tan ✓? Answer: 1

3

Solution

tan2 ✓ + 1 = sec2 ✓

sec2 ✓ � tan2 ✓ = 1

188

(sec ✓ � tan ✓)(sec ✓ + tan ✓) = 1

3(sec ✓ + tan ✓) = 1

sec ✓ + tan ✓ =1

3

4

Lesson 3.5. Sum and Di↵erence Identities




(1) derive trigonometric identities involving sum and di↵erence of two angles;

(2) simplify trigonometric expressions using fundamental trigonometric identitiesand sum and di↵erence identities;

(3) prove other trigonometric identities using fundamental trigonometric identi-ties and sum and di↵erence identities; and

(4) solve situational problems involving trigonometric identities.

Lesson Outline

(1) The sum and di↵erence identities for cosine, sine, and tangent functions

(2) Cofunction identities

(3) More trigonometric identities

Introduction

In previous lesson, we introduced the concept of trigonometric identity, pre-sented the fundamental identities, and proved some identities. In this lesson, wederive the sum and di↵erence identities for cosine, sine, and tangent functions,establish the cofunction identities, and prove more trigonometric identities.

3.5.1. The Cosine Di↵erence and Sum Identities

Let u and v be any real numbers with 0 < v u < 2⇡. Consider the unit circlewith points A = (1, 0), P1, P2, P3, and u and v with corresponding angles shownin Figure 3.27. Then P1P2 = AP3.

189

Figure 3.27

Recall that P1 = P (u) = (cos u, sin u), P2 = P (v) = (cos v, sin v), and P3 =P (u� v) = (cos(u� v), sin(u� v)), so that

P1P2 =p

(cosu� cos v)2 + (sin u� sin v)2,

whileAP3 =

p[cos(u� v)� 1]2 + [sin(u� v)� 0]2.

Equating these two expressions and expanding the squares, we get

(cosu� cos v)2 + (sin u� sin v)2 = [cos(u� v)� 1]2 + sin2(u� v)

cos2 u� 2 cosu cos v + cos2 v + sin2 u� 2 sin u sin v + sin2 v

= cos2(u� v)� 2 cos(u� v) + 1 + sin2(u� v)

Applying the Pythagorean identity cos2 ✓+sin2 ✓ = 1 and simplifying the resultingequations, we obtain

(cos2 u+ sin2 u) + (cos2 v + sin2 v)� 2 cosu cos v � 2 sin u sin v

= [cos2(u� v) + sin2(u� v)]� 2 cos(u� v) + 1

1 + 1� 2 cosu cos v � 2 sin u sin v = 1� 2 cos(u� v) + 1

cos(u� v) = cos u cos v + sin u sin v.

We have thus proved another identity.

Although we assumed at the start that 0 < v u < 2⇡, but becausecos(�✓) = cos ✓ (one of the even-odd identities), this new identity is true forany real numbers u and v. As before, the variables can take any real values orangle measures.

190

Cosine Di↵erence Identity

cos(A� B) = cosA cosB + sinA sinB

Replacing B with �B, and applying the even-odd identities, we immediatelyget another identity.

Cosine Sum Identity

cos(A+B) = cosA cosB � sinA sinB

Example 3.5.1. Find the exact values of cos 105� and cos ⇡

12 .

Solution.

cos 105� = cos(60� + 45�)

= cos 60� cos 45� � sin 60� sin 45�

=1

2·p2

2�

p3

2·p2

2

=

p2�

p6

4

cos⇡

12= cos

⇣⇡4� ⇡

6

⌘

= cos⇡

4cos

⇡

6+ sin

⇡

4sin

⇡

6

=

p2

2·p3

2+

p2

2· 12

=

p6 +

p2

42

Example 3.5.2. Given cos↵ = 35 and sin � = 12

13 , where ↵ lies in QIV and � inQI, find cos(↵ + �).

Solution. We will be needing sin↵ and cos �.

sin↵ = �p1� cos2 ↵ = �

s

1�✓3

5

◆2

= �4

5

cos � =q

1� sin2 � =

s

1�✓12

13

◆2

=5

13

191

cos(↵ + �) = cos↵ cos � � sin↵ sin �

=3

5· 5

13�✓�4

5

◆12

13

=63

652


1. Find the exact value of cos 7⇡12 .

2. Expresscos(5x) cos(2x) + sin(5x) sin(2x)

as a single cosine expression.

3. Given cos↵ = 13 and cos � = �1

4 , where ↵ lies in QI and � in QIII, findcos(↵� �).

3.5.2. The Cofunction Identities and the Sine Sum and Di↵erenceIdentities

In the Cosine Di↵erence Identity, if we let A = ⇡

2 , we get

cos⇣⇡2� B

⌘= cos

⇣⇡2

⌘cosB + sin

⇣⇡2

⌘sinB

= (0) cosB + (1) sinB

= sinB.

From this identity, if we replace B with ⇡

2 � B, we have

cosh⇡2�⇣⇡2� B

⌘i= sin

⇣⇡2� B

⌘

cosB = sin⇣⇡2� B

⌘.

As for the tangent function, we have

tan⇣⇡2� B

⌘=

sin�⇡

2 � B�

cos�⇡

2 � B�

=cosB

sinB= cotB.

We have just derived another set of identities.

192

Cofunction Identities

Teaching Notes

The CofunctionIdentities for thereciprocalfunctions willfollow:csc

�⇡2 �B

�=

secBsec

�⇡2 �B

�=

cscBcot

�⇡2 �B

�=

tanB.

cos⇣⇡2� B

⌘= sinB sin

⇣⇡2� B

⌘= cosB

tan⇣⇡2� B

⌘= cotB

Using the first two cofunction identities, we now derive the identity for sin(A+B).

sin(A+B) = cosh⇡2� (A+B)

i

= cosh⇣⇡

2� A

⌘� B)

i

= cos⇣⇡2� A

⌘cosB + sin

⇣⇡2� A

⌘sinB

= sinA cosB + cosA sinB

Sine Sum Identity

sin(A+B) = sinA cosB + cosA sinB

In the last identity, replacing B with �B and applying the even-odd identitiesyield

sin(A� B) = sin[A+ (�B)]

= sinA cos(�B) + cosA sin(�B)

= sinA cosB � cosA sinB.

Sine Di↵erence Identity

sin(A� B) = sinA cosB � cosA sinB

Example 3.5.3. Find the exact value of sin�5⇡12

�.

Solution.

sin

✓5⇡

12

◆= sin

⇣⇡4+

⇡

6

⌘

= sin⇣⇡4

⌘cos

⇣⇡6

⌘+ cos

⇣⇡4

⌘sin

⇣⇡6

⌘

193

=

p2

2·p3

2+

p2

2· 12

=

p6 +

p2

42

Example 3.5.4. If sin↵ = 313 and sin � = 1

2 , where 0 < ↵ < ⇡

2 and ⇡

2 < � < ⇡,find sin(↵ + �) and sin(� � ↵).

Solution. We first compute cos↵ and cos �.

cos↵ =p

1� sin2 ↵ =

s

1�✓

3

13

◆2

=4p10

13

cos � = �q

1� sin2 � = �

s

1�✓1

2

◆2

= �p3

2

sin(↵ + �) = sin↵ cos � + cos↵ sin �

=3

13

�p3

2

!+

4p10

13· 12

=4p10� 3

p3

26

sin(� � ↵) = sin � cos↵� cos � sin↵

=1

2· 4

p10

13� �p3

2

!3

13

=4p10 + 3

p3

262

Example 3.5.5. Prove:

sin(x+ y) = (1 + cot x tan y) sin x cos y.

Solution.

(1 + cot x tan y) sin x cos y

= sin x cos y + cot x tan y sin x cos y

= sin x cos y +cos x

sin x

sin y

cos ysin x cos y

= sin x cos y + cosx sin y

= sin(x+ y) 2

194


1. Find the exact value of sin�

⇡

12

�.

2. Find the exact value of sin 20� cos 80� � sin 80� cos 20�.

3. Prove:sin(x+ y)

sin(x� y)=

tan x+ tan y

tan x� tan y.

3.5.3. The Tangent Sum and Di↵erence Identities

Recall that tan x is the ratio of sinx over cos x. When we replace x with A+B,we obtain

tan(A+B) =sin(A+B)

cos(A+B).

Using the sum identities for sine and cosine, and then dividing the numeratorand denominator by cosA cosB, we have

tan(A+B) =sinA cosB + cosA sinB

cosA cosB � sinA sinB

=sinA cosBcosA cosB + cosA sinB

cosA cosBcosA cosBcosA cosB � sinA sinB

cosA cosB

=tanA+ tanB

1� tanA tanB.

We have just established the tangent sum identity.

In the above identity, if we replace B with �B and use the even-odd identitytan(�✓) = � tan ✓, we get

tan(A� B) = tan[A+ (�B)]

=tanA+ tan(�B)

1� tanA tan(�B)

=tanA� tanB

1 + tanA tanB.

This is the tangent di↵erence identity.

Tangent Sum and Di↵erence Identities

tan(A+B) =tanA+ tanB

1� tanA tanB

tan(A� B) =tanA� tanB

1 + tanA tanB

195


1. Find the exact values of tan�5⇡12

�, tan

�⇡

12

�, and tan

��7⇡

12

�.

2. Express tan�⇡

4 + ✓�and tan(2⇡ � x) in terms of tan ✓.

3. Prove: cot(A+B) =cotA cotB � 1

cotA+ cotB.

Exercises 3.5

1. Find the exact value.

(a) cos 255�

(b) tan ⇡

12

(c) sin 735�

(d) cot 285�

(e) cos�⇡

9

�cos

�2⇡9

�� sin

�⇡

9

�sin

�2⇡9

�

(f)tan 20� + tan 25�

1� tan 20� tan 25�

2. Given some information about a and b, find sin(a+b), cos(a�b), and tan(a+b).

(a) sin a = �35 , cos b = 12

13 , a lies in the third quadrant, and b in the firstquadrant

(b) cos a = 12 , tan b = �3

2 , 0 < a < ⇡

2 , and⇡

2 < b < ⇡

(c) sec a = �52 , cot b =

35 , a in QII, and b in QIII

3. Simplify the following expressions.

(a) cos(⇡ � x)

(b) tan(x+ ⇡)

(c) sin�3⇡2 + x

�

(d) cos(x� ⇡)

4. Prove each identity.

(a) sin(x� y) sin(x+ y) = sin2 x� sin2 y

(b) cos(x� y) = (cot x+ cot y) sin x cos y

(c) sec(x+ y) =csc x csc y

cot x cot y � 1

(d)cos(x+ y)

cos(x� y)=

1� tan x tan y

1 + tan x tan y

196

5. Let n be an integer. Prove that cos(n⇡ + ✓) = (�1)n cos ✓ and sin(n⇡ + ✓) =(�1)n sin ✓.

6. In an alternating current circuit, the instantaneous power P (t) at time t isgiven by

P (t) = Im

Vm

cos' sin2(!t)� Im

Vm

sin' sin(!t) cos(!t),

where Im

and Vm

are the maximum current (in amperes) and voltage (in volts),respectively. Express this function as a product of two sine functions.

?7. The force (in pounds) on the back of a person when he or she bends over atan acute angle ✓ (in degrees) is given by F (✓) = 0.6W sin(✓+90)

sin 12 , where W is theweight (in pounds) of the person.

(a) Simplify the formula F (✓).

(b) Find the force on the back of a person whose weight is 154.32 lbs if hebends an angle of 40�.

(c) How many pounds should a person weigh for his back to endure a forceof 275 lbs if he bends 38�?

8. (a) Prove: sinx+ sin y = 2 sin�x+y

2

�cos

�x�y

2

�.

(b) A particle is moving according to the equation of motion

s(t) = sin⇣4t+

⇡

3

⌘+ sin

⇣4t+

⇡

6

⌘,

where s(t) centimeters is the directed distance of the particle from theorigin at t seconds.

(i) Express s(t) in the form s(t) = a sin(bt+ c).

(ii) Find the amplitude and frequency of the motion. (Here, frequency isdefined as the reciprocal of the period.)

9. The dual tone multi frequency is the signal information used in touch-tonephones to identify which digit you touched on the keypad. It works by addinga pair of sounds, one with a lower frequency and one with a higher frequency.Refer to the chart below. For example, the sound created by touching 6 isproduced by adding a 770-hertz sound to a 1477-hertz sound. (Note that“hertz” is a unit of frequency and is equal to 1 cycle per second.) This soundis modeled by the equation

s(t) = sin(2⇡ · 770t) + sin(2⇡ · 1477t),

where t is time in seconds.

197

http://cnx.org/contents/XGjYtByD@4/Lab-6-Analog-to-Digital-Conver

(a) Write the equation of the sound created by touching the * key as a productof sine and cosine functions.

(b) In (a), what is the maximum value of s(t)?

10. (a) Prove: cosx+ cos y = 2 cos�x+y

2

�cos

�x�y

2

�.

(b) Two atmospheric waves in space produce pressures of F (t) and G(t) pas-cals at t seconds, where

F (t) = 0.04 cos(2⇡t)

and

G(t) = 0.04 cos

✓2⇡t� 3⇡

4

◆.

Express the total pressure of F (t) and G(t) in the form P (t) = a cos(bt+c).

11. (a) In the figure, two intersecting lines have equations y = m1x + b1 andy = m2x + b2, respectively. Let ✓ be the acute angle between them, asshown. Prove that

tan ✓ =m2 �m1

1 +m1m2.

198

http://cnx.org/contents/XGjYtByD@4/Lab-6-Analog-to-Digital-Conver

(b) Two non-vertical lines intersect at the point (�3, 2), and one angle be-tween them measures 30�. If one line is 2y = x + 7, find the equation ofthe other line.

12. The length s(✓) of the shadow cast by a vertical pole when the angle of thesun with the horizontal is given by

s(✓) =h sin(90� � ✓)

sin ✓,

where h is the height of the pole.

(a) Express s(✓) as a single trigonometric expression.

(b) At what angle ✓ will give the shortest shadow of the pole? Longestshadow?

13. In 4ABC, prove that

tanA+ tanB + tanC = tanA tanB tanC.

4

Lesson 3.6. Double-Angle and Half-Angle Identities




(1) derive the double-angle and half-angle identities;

199

(2) simplify trigonometric expressions using known identities;

(3) prove other trigonometric identities using known identities; and

(4) solve situational problems involving trigonometric identities.

Lesson Outline

(1) The double-angle and half-angle identities for cosine, sine, and tangent

(2) More trigonometric identities

Introduction

Trigonometric identities simplify the computations of trigonometric expres-sions. In this lesson, we continue on establishing more trigonometric identities.In particular, we derive the formulas for f(2✓) and f

�12✓�, where f is the sine,

cosine, or tangent function.

3.6.1. Double-Angle Identities

Recall the sum identities for sine and cosine.

sin(A+B) = sinA cosB + cosA sinB

cos(A+B) = cosA cosB � sinA sinB

When A = B, these identities becomes

sin 2A = sinA cosA+ cosA sinA = 2 sinA cosA

andcos 2A = cosA cosA� sinA sinA = cos2 A� sin2 A.

Double-Angle Identities for Sine and Cosine

sin 2A = 2 sinA cosA cos 2A = cos2 A� sin2 A

The double-identity for cosine has other forms. We use the Pythagoreanidentity sin2 ✓ + cos2 ✓ = 1.

cos 2A = cos2 A� sin2 A

= cos2 A� (1� cos2 A)

= 2 cos2 A� 1

cos 2A = cos2 A� sin2 A

= (1� sin2 A)� sin2 A

= 1� 2 sin2 A

200

Other Double-Angle Identities for Cosine

cos 2A = 2 cos2 A� 1 cos 2A = 1� 2 sin2 A

Example 3.6.1. Given sin t = 35 and ⇡

2 < t < ⇡, find sin 2t and cos 2t.

Solution. We first find cos t using the Pythagorean identity. Since t lies in QII,we have

cos t = �p

1� sin2 t = �

s

1�✓3

5

◆2

= �4

5.

sin 2t = 2 sin t cos t

= 2

✓3

5

◆✓�4

5

◆

= �24

25

cos 2t = 1� 2 sin2 t

= 1� 2

✓3

5

◆2

=7

252

In the last example, we may compute cos 2t using one of the other two double-angle identities for cosine. For the sake of answering the curious minds, we includethe computations here.

cos 2t = cos2 t� sin2 t

=

✓�4

5

◆2

�✓3

5

◆2

=7

25

cos 2t = 2 cos2 t� 1

= 2

✓�4

5

◆2

� 1

=7

25

In the three cosine double-angle identities, which formula to use depends onthe convenience, what is given, and what is asked.

Example 3.6.2. Derive an identity for sin 3x in terms of sin x.

Solution. We use the sum identity for sine, the double-angle identities for sineand cosine, and the Pythagorean identity.

sin 3x = sin(2x+ x)

= sin 2x cos x+ cos 2x sin x

= (2 sin x cos x) cosx+ (1� 2 sin2 x) sin x

= 2 sin x cos2 x+ sin x� 2 sin3 x

= 2(sin x)(1� sin2 x) + sin x� 2 sin3 x

= 3 sin x� 4 sin3 x 2

201

For the double-angle formula for tangent, we recall the tangent sum identity:

tan(A+B) =tanA+ tanB

1� tanA tanB.

When A = B, we obtain

tan(A+ A) =tanA+ tanA

1� tanA tanA=

2 tanA

1� tan2 A.

Tangent Double-Angle Identity

tan 2A =2 tanA

1� tan2 A

Example 3.6.3. If tan ✓ = �13 and sec ✓ > 0, find sin 2✓, cos 2✓, and tan 2✓.

Solution. We can compute immediately tan 2✓.

tan 2✓ =2 tan ✓

1� tan2 ✓=

2��1

3

�

1��1

3

�2 = �3

4

From the given information, we deduce that ✓ lies in QIV. Using one Pythagoreanidentity, we compute cos ✓ through sec ✓. (We may also use the technique dis-cussed in Lesson 3.2 by solving for x, y, and r.) Then we proceed to find cos 2✓.

sec ✓ =p

1 + tan2 ✓ =

s

1 +

✓�1

3

◆2

=

p10

3

cos ✓ =1

sec ✓=

1p103

=3p10

10

cos 2✓ = 2 cos2 ✓ � 1 = 2

3p10

10

!2

� 1 =4

5

tan 2✓ =sin 2✓

cos 2✓=) sin 2✓ = tan 2✓ cos 2✓ = �3

52


1. If cos ✓ = 23 and 3⇡

2 < ✓ < 2⇡, find sin 2✓, cos 2✓, and tan 2✓.

2. Express tan 3✓ in terms of tan ✓.

3. Prove:2 tan ✓

1 + tan2 ✓= sin 2✓.

202

3.6.2. Half-Angle Identities

Recall two of the three double-angle identities for cosine:

cos 2A = 2 cos2 A� 1 and cos 2A = 1� 2 sin2 A.

From these identities, we obtain two useful identities expressing sin2 A and cos2 Ain terms of cos 2A as follows:

cos2 A =1 + cos 2A

2and sin2 A =

1� cos 2A

2.

Some Useful Identities

cos2 A =1 + cos 2A

2sin2 A =

1� cos 2A

2

From these identities, replacing A with A

2 , we get

cos2A

2=

1 + cos 2�A

2

�

2=

1 + cosA

2and

sin2 A

2=

1� cos 2�A

2

�

2=

1� cosA

2.

These are the half-angle identities for sine and cosine.

Half-Angle Identities for Sine and Cosine

cos2✓A

2

◆=

1 + cosA

2sin2

✓A

2

◆=

1� cosA

2

Because of the “square” in the formulas, we get

cosA

2= ±

r1 + cosA

2and sin

A

2= ±

r1� cosA

2.

The appropriate signs of cos A

2 and sin A

2 depend on which quadrant A

2 lies.

Example 3.6.4. Find the exact values of sin 22.5� and cos 22.5�.

Solution. Clearly, 22.5� lies in QI (and so sin 22.5� and cos 22.5� are both posi-tive), and 22.5� is the half-angle of 45�.

sin 22.5� =

r1� cos 45�

2=

s1�

p22

2=

p2�

p2

2

cos 22.5� =

r1 + cos 45�

2=

s1 +

p22

2=

p2 +

p2

22

203

Example 3.6.5. Prove: cos2✓✓

2

◆=

tan ✓ + sin ✓

2 tan ✓.

Solution.

cos2✓✓

2

◆=

1 + cos ✓

2

=1 + cos ✓

2

✓tan ✓

tan ✓

◆

=tan ✓ + cos ✓ tan ✓

2 tan ✓

=tan ✓ + cos ✓ · sin ✓

cos ✓

2 tan ✓

=tan ✓ + sin ✓

2 tan ✓2

We now derive the first version of the half-angle formula for tangent.

tanA

2=

sin A

2

cos A

2

=sin A

2

cos A

2

2 sin A

2

2 sin A

2

!

=2 sin2

�A

2

�

2 sin A

2 cos A

2

=2 · 1�cosA

2

sin�2 · A

2

�

=1� cosA

sinA

There is another version of the tangent half-angle formula, and we can deriveit from the first version.

tanA

2=

1� cosA

sinA

=1� cosA

sinA

✓1 + cosA

1 + cosA

◆

=1� cos2 A

(sinA)(1 + cosA)

=sin2 A

(sinA)(1 + cosA)

=sinA

1 + cosA

204

Tangent Half-Angle Identities

tanA

2=

1� cosA

sinAtan

A

2=

sinA

1 + cosA

tanA

2=

sin A

2

cos A

2

tan2

✓A

2

◆=

1� cosA

1 + cosA

Example 3.6.6. Find the exact value of tan ⇡

12 .

Solution.

tan⇡

12=

1� cos ⇡

6

sin ⇡

6

=1�

p32

12

= 2�p3 2

Example 3.6.7. If sin ✓ = �25 , cot ✓ > 0, and 0 ✓ < 2⇡, find sin ✓

2 , cos✓

2 , andtan ✓

2 .

Solution. Since sin ✓ < 0 and cot ✓ > 0, we conclude the ⇡ < ✓ < 3⇡2 . It follows

that⇡

2<

✓

2<

3⇡

4,

which means that ✓

2 lies in QII.

cos ✓ = �p

1� sin2 ✓ = �

s

1�✓�2

5

◆2

= �p21

5

sin✓

2=

r1� cos ✓

2=

vuut1�⇣�

p215

⌘

2=

p50 + 10

p21

10

cos✓

2= �

r1 + cos ✓

2= �

vuut1 +⇣�

p215

⌘

2= �

p50� 10

p21

10

tan✓

2=

1� cos ✓

sin ✓=

1�⇣�

p215

⌘

�25

= �5 +p21

22


1. Find the exact value of tan ⇡

8 .

2. If cos ✓ = 35 and 3⇡

2 < ✓ < 2⇡, find sin ✓

2 , cos✓

2 , and tan ✓

2 .

3. Prove: sec2✓A

2

◆=

2� 2 cosA

sin2 A.

205

Exercises 3.6

1. Given some information about ✓, find sin 2✓, cos 2✓, and tan 2✓.

(a) cos ✓ = �14 and ⇡

2 < ✓ < ⇡

(b) sec ✓ = 52 and sin ✓ > 0

(c) tan ✓ = �2 and 3⇡2 < ✓ < 2⇡

(d) sin ✓ = 35 and tan ✓ < 0

2. Given the same information as in Item (1), where 0 ✓ < 2⇡, find sin ✓

2 , cos✓

2 ,and tan ✓

2 .

3. Express each expression as one trigonometric expression, but do not find theexact value.

(a) 2 sin 10� cos 10�

(b)

r1� cos

�7⇡6

�

2

(c) 1� 2 sin2�3⇡10

�

(d)1 + cos 8

2


(a) tan2�✓

2

�= (csc ✓ � cot ✓)2

(b) tan ✓

2 + cot ✓

2 = 2 csc ✓

(c) sec2�✓

2

�= (csc2 ✓)(2� 2 cos ✓)

5. If a = 2 tan ✓, express sin 2✓ and cos 2✓ in terms of a.

6. Find the exact value of cos 36� � cos 72�.

7. The range R of a projectile fired at an angle ✓ with the horizontal and withan initial velocity of v meters per second is given by

R =v2

gsin(2✓),

where g is the acceleration due to gravity, which is 9.81 m/sec2 near the Earth’ssurface.

(a) An archer targets an object 100 meters away from her position. If shepositions her arrow at an angle of 32� and releases the arrow at the speedof 30 m/sec, will she hit her target?

(b) If sin ✓ = 25 , solve for v when R = 50.

206

(c) Given v, what is ✓ to reach largest possible range? At this ✓, what is therange?

8. The figure shows a laser scanner projection system. The optical angle ✓, throwdistance D, and projected image width W are related by the equation

D =W

2

csc ✓ � cot ✓.

Solve for W in terms D and ✓

2 .

https://pangolin.com/userhelp/scanangles.htm

9. The slope of a mountain makes an angle of 45� with the horizontal. At thebase of the mountain, a cannon is fired at an angle ✓ with the horizontal, where45� < ✓ < 90�, and with initial velocity of v m/sec. Neglecting air resistance,the distance R (in meters) it drops on the slope of the mountain from the baseis given by

R =2p2v2

g(sin ✓ � cos ✓) cos ✓,

where g is the acceleration due to gravity in m/sec2. Express this formula forR in terms of 2✓.

4

207

https://pangolin.com/userhelp/scanangles.htm

Lesson 3.7. Inverse Trigonometric Functions




(1) graph the six basic inverse trigonometric functions;

(2) illustrate the domain and range of the inverse trigonometric functions;

(3) evaluate inverse trigonometric expressions; and

(4) solve situational problems involving inverse trigonometric functions.

Lesson Outline

(1) Definitions of the six inverse trigonometric functions

(2) Graphs of inverse trigonometric functions

(3) Domain and range of inverse trigonometric functions

(4) Evaluation of inverse trigonometric expressions

Introduction

In the previous lessons on functions (algebraic and trigonometric), given anumber in the domain of a function, we computed for the value of the functionat that number. Now, given a value in the range of the function, we reverse thisprocess by finding a number in the domain whose function value is the givenone. Observe that, in this process, the function involved may or may not give aunique number in the domain. For example, each of the functions f(x) = x2 andg(x) = cos x do not give a unique number in their respective domains for somevalues of each function. Given f(x) = 1, the function gives x = ±1. If g(x) = 1,then x = 2k⇡, where k is an integer. Because of this possibility, in order thatthe reverse process produces a function, we restrict this process to one-to-onefunctions or at least restrict the domain of a non-one-to-one function to make itone-to-one so that the process works. Loosely speaking, a function that reverseswhat a given function f does is called its inverse function, and is usually denotedby f�1.

More formally, two functions f and g are inverse functionsTeaching Notes

The concept ofinverse functionwas studied in

GeneralMathematics

course.

if

g(f(x)) = x for any x in the domain of f ,

andf(g(x)) = x for any x in the domain of g.

We denote the inverse function of a function f by f�1. The graphs of a functionand its inverse function are symmetric with respect to the line y = x.

208

In this lesson, we first restrict the domain of each trigonometric functionbecause each of them is not one-to-one. We then define each respective inversefunction and evaluate the values of each inverse trigonometric function.

3.7.1. Inverse Sine Function

All the trigonometric functions that we consider are periodic over their entiredomains. This means that all trigonometric functions are not one-to-one if weconsider their whole domains, which implies that they have no inverses over thosesets. But there is a way to make each of the trigonometric functions one-to-one.This is done by restricting their respective domains. The restrictions will give uswell-defined inverse trigonometric functions.

The domain of the sine function is the set R of real numbers, and its range isthe closed interval [�1, 1]. As observed in the previous lessons, the sine functionis not one-to-one, and the first step is to restrict its domain (by agreeing what theconvention is) with the following conditions: (1) the sine function is one-to-onein that restricted domain, and (2) the range remains the same.

The inverse of the (restricted) sine function f(x) = sin x, where thedomain is restricted to the closed interval

⇥�⇡

2 ,⇡

2

⇤, is called the inverse

sine function or arcsine function, denoted by f�1(x) = sin�1 x orf�1(x) = arcsin x. Here, the domain of f�1(x) = arcsin x is [�1, 1],and its range is

⇥�⇡

2 ,⇡

2

⇤. Thus,

y = sin�1 x or y = arcsin x

if and only ifsin y = x,

where �1 x 1 and �⇡

2 y ⇡

2 .

Throughout the lesson, we interchangeably use sin�1 x and arcsin x to meanthe inverse sine function.

Example 3.7.1. Find the exact value of each expression.(1) sin�1 1

2

(2) arcsin(�1)

(3) arcsin 0

(4) sin�1��1

2

�

Solution. (1) Let ✓ = sin�1 12 . This is equivalent to sin ✓ = 1

2 . This means thatwe are looking for the number ✓ in the closed interval

⇥�⇡

2 ,⇡

2

⇤whose sine is

12 . We get ✓ = ⇡

6 . Thus, we have sin�1 12 = ⇡

6 .

(2) arcsin(�1) = �⇡

2 because sin��⇡

2

�= �1 and �⇡

2 2⇥�⇡

2 ,⇡

2

⇤.

209

(3) arcsin 0 = 0

(4) sin�1��1

2

�= �⇡

6 2

As emphasized in the last example, as long as �1 x 1, sin�1 x is thatnumber y 2

⇥�⇡

2 ,⇡

2

⇤such that sin y = x. If |x| > 1, then sin�1 x is not defined in

R.We can sometimes find the exact value of sin�1 x (that is, we can find a value

in terms of ⇡), but if no such special value exists, then we leave it in the formsin�1 x. For example, as shown above, sin�1 1

2 is equal to ⇡

6 . However, as studiedin Lesson 3.2, no special number ✓ satisfies sin ✓ = 2

3 , so we leave sin�1 23 as is.

Example 3.7.2. Find the exact value of each expression.(1) sin

�sin�1 1

2

�

(2) arcsin�sin ⇡

3

�(3) arcsin(sin ⇡)

(4) sin�sin�1

��1

2

��

Solution. (1) sin�sin�1 1

2

�= sin ⇡

6 = 12

(2) arcsin�sin ⇡

3

�= arcsin

p32 = ⇡

3

(3) arcsin(sin ⇡) = arcsin 0 = 0

(4) sin�sin�1

��1

2

��= sin

��⇡

6

�= �1

2 2

From the last example, we have the following observations:

1. sin(arcsin x) = x for any x 2 [�1, 1]; and

2. arcsin(sin ✓) = ✓ if and only if ✓ 2⇥�⇡

2 ,⇡

2

⇤, and if ✓ 62

⇥�⇡

2 ,⇡

2

⇤, then

arcsin(sin ✓) = ', where ' 2⇥�⇡

2 ,⇡

2

⇤such that sin' = sin ✓.

To sketch the graph of y = sin�1 x, Table 3.28 presents the tables of valuesfor y = sin x and y = sin�1 x. Recall that the graphs of y = sin x and y = sin�1 xare symmetric with respect to the line y = x. This means that if a point (a, b) ison y = sin x, then (b, a) is on y = sin�1 x.

y = sin xx �⇡

2 �⇡

3 �⇡

4 �⇡

6 0 ⇡

6⇡

4⇡

3⇡

2

y �1 �p32 �

p22 �1

2 0 12

p22

p32 1

y = sin�1 xx �1 �

p32 �

p22 �1

2 0 12

p22

p32 1

y �⇡

2 �⇡

3 �⇡

4 �⇡

6 0 ⇡

6⇡

4⇡

3⇡

2

Table 3.28

210

The graph (solid thick curve) of the restricted sine function y = sin x is shownin Figure 3.29(a), while the graph of inverse sine function y = arcsin x is shownin Figure 3.29(b).

(a) y = sin x (b) y = sin�1 x

Figure 3.29

Example 3.7.3. Sketch the graph of y = sin�1(x+ 1).

Solution 1. In this solution, we use translation of graphs.

Because y = sin�1(x + 1) is equivalent to y = sin�1[x � (�1)], the graph ofy = sin�1(x + 1) is 1-unit to the left of y = sin�1 x. The graph below showsy = sin�1(x+ 1) (solid line) and y = sin�1 x (dashed line).

211

Solution 2. In this solution, we graph first the corresponding sine function, andthen use the symmetry with respect to y = x to graph the inverse function.

y = sin�1(x+ 1) () sin y = x+ 1 () x = sin y � 1

The graph below shows the process of graphing of y = sin�1(x + 1) from y =sin x� 1 with �⇡

2 x ⇡

2 , and then reflecting it with respect to y = x.Teaching Notes

Keep in mind that,because of the

restriction in thedomain, we have

the following:sin(sin�1

x) = x

for all x 2 [�1, 1].But sin�1(sinx) isnot always x. We

havesin�1(sinx) = x

only if�⇡

2 x ⇡2 .


1. Find the exact value of each expression.

(a) sin�1 1 Answer: ⇡

2

(b) arcsin⇣�

p22

⌘Answer: �⇡

2

(c) arcsin⇣p

32

⌘Answer: ⇡

3

(d) sin�1⇣�

p32

⌘Answer: �⇡

3

(e) sin⇣sin�1

p22

⌘Answer:

p22

(f) arcsin�sin ⇡

3

�Answer: ⇡

3

(g) arcsin�cos

��⇡

3

��Answer: ⇡

6

(h) sin�1�sin 4⇡

3

�Answer: �⇡

3

2. Sketch the graph of each equation.

(a) y = sin�1(x� 2)

212

(b) y = sin�1(2x)

3.7.2. Inverse Cosine Function

The development of the other inverse trigonometric functions follows similarlyfrom that of the inverse sine function.

213

y = cos�1 x or y = arccosx

meansTeaching Notes

Observe that thisdefinition of

cos�1x is

equivalent tocos�1

x =⇡2 � sin�1

x.

cos y = x,

where �1 x 1 and 0 y ⇡.

The graph (solid thick curve) of the restricted cosine function y = cosx isshown in Figure 3.30(a), while the graph of inverse cosine function y = arccosxis shown in Figure 3.30(b).

(a) y = cosx (b) y = cos�1 x

Figure 3.30

Example 3.7.4. Find the exact value of each expression.(1) cos�1 0

(2) arccos⇣�

p32

⌘

(3) cos⇣cos�1

⇣�

p32

⌘⌘

(4) cos�1�cos 3⇡

4

�

(5) arccos�cos 7⇡

6

�

(6) sin⇣cos�1

p22

⌘

Solution. (1) cos�1 0 = ⇡

2 because cos ⇡

2 = 0 and ⇡

2 2 [0, ⇡].

(2) arccos⇣�

p32

⌘= 5⇡

6

(3) cos⇣cos�1

⇣�

p32

⌘⌘= �

p32 because �

p32 2 [�1, 1]

(4) cos�1�cos 3⇡

4

�= 3⇡

4 because 3⇡4 2 [0, ⇡].

214

(5) arccos�cos 7⇡

6

�= arccos

⇣�

p32

⌘= 5⇡

6

(6) sin⇣cos�1

p22

⌘=

p22 2

Example 3.7.5. Simplify: sin�arcsin 2

3 + arccos 12

�.

Solution. We know that arccos 12 = ⇡

3 . Using the Sine Sum Identity, we have

sin�arcsin 2

3 + arccos 12

�

= sin�arcsin 2

3 +⇡

3

�

= sin�arcsin 2

3

�cos ⇡

3 + cos�arcsin 2

3

�sin ⇡

3

= 23 ·

12 + cos

�arcsin 2

3

�·p32

= 13 +

p32 cos

�arcsin 2

3

�.

We compute cos�arcsin 2

3

�. Let ✓ = arcsin 2

3 . By definition, sin ✓ = 23 , where

✓ lies in QI. Using the Pythagorean identity, we have

cos�arcsin 2

3

�= cos ✓ =

p1� sin2 ✓ =

p53 .

Going back to the original computations above, we have

sin�arcsin 2

3 + arccos 12

�= 1

3 +p32 cos

�arcsin 2

3

�

= 13 +

p32 ·

p53

= 2+p15

6 . 2

Example 3.7.6. Simplify: sin�2 cos�1

��4

5

��.

Solution. Let ✓ = cos�1��4

5

�. Then cos ✓ = �4

5 . Because cos ✓ < 0 and rangeof inverse cosine function is [0, ⇡], we know that ✓ must be within the interval�⇡

2 , ⇡⇤. Using the Pythagorean Identity, we get sin ✓ = 3

5 .

Using the Sine Double-Angle Identity, we have

sin�2 cos�1

��4

5

��= sin 2✓

= 2 sin ✓ cos ✓

= 2 · 35

��4

5

�

= �2425 . 2

Example 3.7.7. Sketch the graph of y = 14 cos

�1(2x).

Solution.

y =1

4cos�1(2x) () 4y = cos�1(2x) () x =

1

2cos(4y)

215

We graph first y = 12 cos(4x). The domain of this graph comes from the restriction

of cosine as follows:0 4x ⇡ =) 0 x ⇡

4.

Then reflect this graph with respect to y = x, and we finally obtain the graph ofy = 1

4 cos�1(2x) (solid line).

In the last example, we may also use the following technique. In graphingy = 1

4 cos�1(2x), the horizontal length of cos�1 x is reduced to half, while the

vertical height is reduced to quarter. This comparison technique is shown inthe graph below with the graph of y = cos�1 x in dashed line and the graph ofy = 1

4 cos�1(2x) in solid line.



(a) cos�1(�1) Answer: ⇡

(b) arccos⇣�

p22

⌘Answer: 3⇡

4

216

(c) arccos�sin 5⇡

2

�Answer: 0

(d) cos�cos�1

��2

5

��Answer: �2

5

(e) cos�1�� tan 3⇡

4

�Answer: 0

(f) arccos�cos

��13⇡

3

��Answer: ⇡

3

2. Simplify each expression.

(a) cos�cos�1 2

3 � sin�1��1

3

��Answer:

4p2�

p5

9Solution. Let ↵ = cos�1 2

3 . This implies that cos↵ = 23 and ↵ in

QI. Using a Pythagorean identity, one gets sin↵ =p53 . Similarly, let

� = sin�1��1

3

�. This implies that sin � = �1

3 and � in QIV. Then,

we get cos � = 2p2

3 .

We now use Cosine Di↵erence Identity, and the given and computedvalues to simplify the expression.

cos

✓cos�1 2

3� sin�1

✓�1

3

◆◆= cos(↵� �)

= cos↵ cos � + sin↵ sin �

=

✓2

3

◆ 2p2

3

!+

p5

3

!✓�1

3

◆

=4p2�

p5

9.

(b) tan�arcsin

��1

2

�+ arccos 2

5

�Answer:

�17p3� 4

p21

9Solution. First, note that arcsin

��1

2

�= �⇡

6 . Second, we let ✓ =arccos 2

5 , which means that cos ✓ = 25 and ✓ is in QI. We get tan ✓ =

p212 .

We now simplify the expression as follows:

tan

✓arcsin

✓�1

2

◆+ arccos

2

5

◆= tan

⇣�⇡

6+ ✓

⌘

=tan(�⇡

6 ) + tan ✓

1� (tan(�⇡

6 ))(tan ✓)

=�p3

3 +p212

1� (�p3

3 )(p212 )

=�2

p3 + 3

p21

6� 3p7

217

=�17

p3� 4

p21

9

3. Sketch the graph of each equation.

(a) y = 2 cos�1(3x)

(b) y = 12 cos

�1(x+ 2)

3.7.3. Inverse Tangent Function and the Remaining InverseTrigonometric Functions

The inverse tangent function is similarly defined as inverse sine and inverse cosinefunctions.

218

y = tan�1 x or y = arctan x

meanstan y = x,

where x 2 R and �⇡

2 < y < ⇡

2 .

The graph (solid thick curve) of the restricted function y = tan x is shownin Figure 3.31(a), while the graph of inverse function y = arctan x is shown inFigure 3.31(b).

(a) y = tan x (b) y = tan�1 x

Figure 3.31

Example 3.7.8. Find the exact value of each expression.(1) tan�1 1

(2) arctan��p3�

(3) tan�tan�1

��5

2

��

(4) tan�1�tan

��⇡

6

��

(5) tan�1�tan 7⇡

6

�

(6) arctan�tan

��19⇡

6

��

Solution. Note the range of arctan is the open interval��⇡

2 ,⇡

2

�.

(1) tan�1 1 = ⇡

4

(2) arctan��p3�= �⇡

3

(3) tan�tan�1

��5

2

��= �5

2

(4) tan�1�tan

��⇡

6

��= �⇡

6 because �⇡

6 2��⇡

2 ,⇡

2

�.

219

(5) Here, note that 7⇡6 62

��⇡

2 ,⇡

2

�. Use the idea of reference angle, we know that

tan 7⇡6 = tan ⇡

6 .

tan�1

✓tan

7⇡

6

◆= tan�1

⇣tan

⇡

6

⌘=

⇡

6

(6) Here, we cannot use the idea of reference angle, but the idea can help in away. The number (or angle) �19⇡

6 is in QII, wherein tangent is negative, andits reference angle is ⇡

6 .

arctan

✓tan

✓�19⇡

6

◆◆= arctan

⇣tan

⇣�⇡

6

⌘⌘

= �⇡

62

Example 3.7.9. Find the exact value of each expression.(1) sin

�2 tan�1

��8

3

��(2) tan

�sin�1 3

5 � tan�1 14

�

Solution. (1) Let ✓ = tan�1��8

3

�. Then tan ✓ = �8

3 . Following the notations inLesson 3.2 and the definition of inverse tangent function, we know that ✓ liesin QIV, and x = 3 and y = �8. We get r =

p32 + (�8)2 =

p73.

Applying the Sine Double-Angle Identity (page 200) gives

sin

✓2 tan�1

✓�8

3

◆◆= sin 2✓

= 2 sin ✓ cos ✓

= 2 · yr· xr

= 2

✓� 8p

73

◆✓3p73

◆

= �48

73.

(2) Using the Tangent Di↵erence Identity, we obtain

tan

✓sin�1 3

5� tan�1 1

4

◆

=tan

�sin�1 3

5

�� tan

�tan�1 1

4

�

1 + tan�sin�1 3

5

�tan

�tan�1 1

4

�

=tan

�sin�1 3

5

�� 1

4

1 + tan�sin�1 3

5

�· 14

.

We are left to compute tan�sin�1 3

5

�. We proceed as in (1) above. Let

✓ = sin�1 35 . Then sin ✓ = 3

5 . From the definition of inverse sine function and

220

the notations used in Lesson 3.2, we know that ✓ lies in QI, and y = 3 andr = 5. We get x =

p52 � 32 = 4, so that tan ✓ = y

x

= 34 .

tan

✓sin�1 3

5� tan�1 1

4

◆=

tan�sin�1 3

5

�� 1

4

1 + tan�sin�1 3

5

�· 14

=34 �

14

1 + 34 ·

14

=8

192

?Example 3.7.10. A student isviewing a painting in a museum.Standing 6 ft from the painting,the eye level of the student is 5 ftabove the ground. If the paint-ing is 10 ft tall and its base is4 ft above the ground, find theviewing angle subtended by thepainting at the eyes of the stu-dent.

Solution. Let ✓ be the viewing angle, and let ✓ = ↵ + � as shown below.

We observe that

tan↵ =1

6and tan � =

9

6.

Using the Tangent Sum Identity, we have

tan ✓ = tan(↵ + �) =tan↵ + tan �

1� tan↵ tan �

=16 +

96

1� 16 ·

96

=20

9.

Using a calculator, the viewing angle is ✓ = tan�1 209 ⇡ 65.8�. 2

We now define the remaining inverse trigonometric functions.

221

Definecot�1 x =

⇡

2� tan�1 x.

It follows that the domain of y = cot�1 x is R and its range is (0, ⇡).

y = sec�1 x or y = arcsecx

meansTeaching Notes

Keep in mind thatthe domain

restrictions areconventions we set.

Other books andsources might have

di↵erent domainrestrictions. The

restrictions wemade aim to make

calculuscomputations

easier in the future.

sec y = x,

where |x| � 1 and y 2⇥0, ⇡2

�[⇥⇡, 3⇡2

�.

Definecsc�1 x =

⇡

2� sec�1 x.

This means that the domain of y = csc�1 x is (�1,�1] [ [1,1) andits range is

��⇡,�⇡

2

⇤[�0, ⇡2

⇤.

The graphs of these last three inverse trigonometric functions are shown inFigures 3.32, 3.33, and 3.34, respectively.

(a) y = cotx (b) y = cot�1 x

Figure 3.32

222

(a) y = secx (b) y = sec�1 x

Figure 3.33

(a) y = cscx (b) y = csc�1 x

Figure 3.34

Observe that the process in getting the value of an inverse function is thesame to all inverse functions. That is, y = f�1(x) is the same as f(y) = x. Weneed to remember the range of each inverse trigonometric function. Table 3.35summarizes all the information about the six inverse trigonometric functions.

223

Function Domain Range Graph

sin�1 x [�1, 1]⇥�⇡

2 ,⇡

2

⇤ Figure3.29(b)

cos�1 x [�1, 1] [0, ⇡]Figure3.30(b)

tan�1 x R��⇡

2 ,⇡

2

� Figure3.31(b)

cot�1 x R (0, ⇡)Figure3.32(b)

sec�1 x {x : |x| � 1}⇥0, ⇡2

�[⇥⇡, 3⇡2

� Figure3.33(b)

csc�1 x {x : |x| � 1}��⇡,�⇡

2

⇤[�0, ⇡2

⇤ Figure3.34(b)

Table 3.35

Example 3.7.11. Find the exact value of each expression.

(1) sec�1(�2)

(2) csc�1⇣�2

p3

3

⌘ (3) cot�1��p3�

(4) sin⇣sec�1

��3

2

�� csc�1

⇣�2

p3

3

⌘⌘

Solution. (1) sec�1(�2) = 4⇡3 because sec 4⇡

3 = �2 and 4⇡3 2

⇥⇡, 3⇡2

�

(2) csc�1⇣�2

p3

3

⌘= �2⇡

3

(3) cot�1��p3�= 5⇡

6

(4) From (2), we know that csc�1⇣�2

p3

3

⌘= �2⇡

3 . Let ✓ = sec�1��3

2

�. Then

sec ✓ = �32 . From defined range of inverse secant function and the notations

in Lesson 3.2, ✓ lies in QIII, and r = 3 and x = �2. Solving for y, we gety = �

p32 � (�2)2 = �

p5. It follows that sin ✓ = �

p53 and cos ✓ = �2

3 .

We now use the Sine Sum Identity.

sin

sec�1

✓�3

2

◆� csc�1

�2

p3

3

!!

= sin

✓✓ �

✓�2⇡

3

◆◆

= sin

✓✓ +

2⇡

3

◆

224

= sin ✓ cos2⇡

3+ cos ✓ sin

2⇡

3

=

�p5

3

!✓�1

2

◆+

✓�2

3

◆ p3

2

!

=

p5� 2

p3

62



(a) sec�1(�1) Answer: ⇡

(b) arctan(�1) Answer: �⇡

4

(c) arccsc�csc 5⇡

2

�Answer: ⇡

2

(d) cot (cot�1(�10)) Answer: �10

(e) sec�1�� tan 3⇡

4

�Answer: 0

(f) csc�csc�1

��3

8

��Answer: �3

8


(a) cos�arcsec 5

2 � arccot 3�

Answer:6p10 +

p210

50Solution. Let ↵ = arcsec 5

2 and � = arccot 3. These imply that sec↵ =52 and cot � = 3, where ↵ and � are both in QI. Using trigonometric

ratios, we obtain the following: cos↵ = 25 , sin↵ =

p215 , cos � = 3

p10

10 ,

and sin � =p1010 .

Using Cosine Di↵erence Identity and the above values, we simplify theexpression as follows:

cos(arcsec5

2� arccot 3) = cos(↵� �)

= cos↵ cos � + sin↵ sin �

=

✓2

5

◆ 3p10

10

!+

p21

5

! p10

10

!

=6p10 +

p210

50

(b) tan�arctan

��1

2

�+ arctan 5

3

�Answer:

7

11Solution

tan

✓arctan

✓�1

2

◆+ arctan

5

3

◆

225

=tan

�arctan

��1

2

��+ tan

�arctan 5

3

�

1� tan�arctan

��1

2

��tan

�arctan 5

3

�

=�1

2 +53

1��1

2

� �53

�

=7

11

Exercises 3.7


(a) sin�1��1

2

�Answer: �⇡

6

(b) cos�1 0 Answer: ⇡

2

(c) tan�1p3 Answer: ⇡

3

(d) csc�1 1 Answer: ⇡

2

(e) sec�1(�2) Answer: 4⇡3

(f) cot�1(�1) Answer: 3⇡4

(g) csc�1 12 Answer: Undefined

?2. Find the value of each expression using a calculator. Round your answer totwo decimal places.

(a) sin�1(1/3) Answer: 0.34

(b) cos�1(�2/5) Answer: 1.98

(c) tan�1(100) Answer: 1.56

(d) csc�1(11/9) Answer: 0.96

(e) sec�1(�20/3) Answer: 1.72

(f) cot�1(5/7) Answer: 0.95


(a) cos�1�cos ⇡

3

�Answer: ⇡

3

(b) csc�1�tan ⇡

6

�Answer: undefined

(c) tan�1�tan 5⇡

4

�Answer: ⇡

4

(d) sin�1�cos

��⇡

4

��Answer: ⇡

4

(e) cos�1�csc ⇡

3

�Answer: ⇡

6


(a) sin�sin�1 1

2 + cos�1 12

�Answer: 1

226

(b) cos⇣tan�1

p3 + sin�1

⇣�

p32

⌘⌘Answer: 1

(c) tan (2 tan�1(�1)) Answer: undefined

(d) cos�tan�1 4

3 + cos�1 513

�Answer: �33

65

(e) sin⇣2 sin�1 1

2 � 3 tan�1p33

⌘Answer: �1

2

5. Solve for t in terms of x.

(a) x = sin 3t Answer: t = 13 sin

�1 x

(b) x = 2 tan(t+ 1) Answer: t = tan�1 x

2 � 1

(c) x = 12 cos(2t+ 1) Answer: t = cos�1(2x)�1

2

(d) x = 2� 32 sec(1� t) Answer: t = 1� sec�1

�23(2� x)

�

(e) x = 12 � cot(2� 3t) Answer: t = 2

3 �13 cot

�1�12 � x

�

6. Sketch the graph of each function.

(a) y = cos�1(x+ 1)

(b) y = sin�1(x� 2)

227

(c) y = sin�1 2x

(d) y = cos�1 x

2

228

(e) y = 2 cos�1(x� 1)

(f) y = �12 sin

�1(2x)

229

(g) y = 2 sin�1(2x+ 2)

(h) y = �2 cos�1(2x� 1)

230

7. Solve for x in the equation sin�1(x2 � 2x) = �⇡

2 . Answer: x = 1

8. Solve for x in the equation tan�1(4x2 + 5x� 7) = �⇡

4 . Answer: �2, 349. A woman is standing x ft from a wall with a billboard nailed on it. The

billboard is 15 ft tall, and its base is 6 ft above the eye level of the woman.Find the viewing angle subtended on the eyes of the woman from the base to

the top of the billboard. Answer: tan�1 15x

x2 + 126?10. During a leap year, the number of hours of daylight in a city can be modeled

by D(t) = 12 + 2.4 sin(0.017t� 1.377), where t is the day of the year (that is,t = 1 means January 1, t = 60 is February 29, and so on).

(a) Give one day of that year whose number of hours of daylight is about14.4. Answer: ⇡ 173 days, so the day would be June 21

(b) Find another day of that year whose number of hours of daylight is thesame as that of February 29.

Answer: ⇡ 287 days, so the day would be October 13

?11. After getting a job, a man started saving a percentage of his annual income,which can be modeled by

P (t) = 2.5 cos(0.157t) + 5.2,

where P (t) is the percentage of his annual income that he was able to save onyear t after he got a job.

231

(a) What percentage of his annual income did he save on the second year?

Answer: 7.58% of his annual income

(b) On what year right after getting a job did he save the least?

Answer: 20 years after getting a job

(c) On what year right after getting a job did he save the most? When wouldit happen again? Answer: 40 years after getting a job

(d) If he got his job at the age of 20, how much will he save on the year ofhis retirement (that is, when he is 60)? Answer: 7.7% of his annualincome


(a) cos (tan�1 x) =p1+x

2

1+x

2 , x 2 RSolution. Let ✓ = tan�1 x, where x is any real number. This implies thattan ✓ = x. One can think of a right triangle, with acute angle ✓ whoseopposite side is x and adjacent side as 1. Solving for the hypotenuse,

we getp1 + x2. Thus cos ✓ = adjacent side

hypotenuse = 1p1+x

2 =p1+x

2

1+x

2 . But

✓ = tan�1 x, therefore, cos(tan�1 x) =p1+x

2

1+x

2 .

(b) sin (tan�1 2x) = 2xp1+4x2 , x 2 R

Solution. Let ✓ = tan�1 2x, where x is any real number. This implies thattan ✓ = 2x. One can think of a right triangle, with acute angle ✓ whoseopposite side is 2x and adjacent side as 1. Solving for the hypotenuse, we

getp1 + 4x2. Thus sin ✓ = opposite side

hypotenuse = 2xp1+4x2 . But ✓ = tan�1 2x,

therefore, sin(tan�1 2x) = 2xp1+4x2 .

(c) tan�1 x+ tan�1 1x

= ⇡

2 , x > 0

Solution. To prove the identity, one can prove an equivalent identity;that is, to show that sin(tan�1 x+ tan�1 1

x

) = 1, x > 0.

Let ↵ = tan�1 x and � = tan�1 1x

. The same techniques as above areapplied to get the following: sin↵ = xp

1+x

2 , cos↵ = 1p1+x

2 , sin � = 1p1+x

2 ,and cos � = xp

1+x

2 .

We now do the following manipulations:

sin(tan�1 x+ tan�1 1

x)

= sin(↵ + �) = sin↵ cos � + cos↵ sin �

=

✓xp

1 + x2

◆✓xp

1 + x2

◆+

✓1p

1 + x2

◆✓1p

1 + x2

◆

=x2 + 1

1 + x2= 1.

It follows that tan�1 x+ tan�1 1x

= ⇡

2 .

232

(d) sin�1 x+ cos�1 x = ⇡

2 , x 2 [�1, 1]

Solution. Same as in (c), we show that sin(sin�1 x+ cos�1 x) = 1, wherex 2 [�1, 1].

Let ↵ = sin�1 x and � = cos�1 x. It follows that sin↵ = x and cos � = x.Using fundamental identities, we obtain the following: cos↵ =

p1� x2 =

sin �.

Then, we do the following manipulations:

sin(sin�1 x+ cos�1 x) = sin(↵ + �)

= sin↵ cos � + cos↵ sin �

= x(x) + (p1� x2)(

p1� x2)

= x2 + 1� x2

= 1.

Therefore, sin�1 x+ cos�1 x = ⇡

2 .

4

Lesson 3.8. Trigonometric Equations




(1) solve trigonometric equations; and

(2) solve situational problems involving trigonometric equations.

Lesson Outline

(1) Definition of a trigonometric equation

(2) Solution to a trigonometric equation

(3) Techniques of solving a trigonometric equation

Introduction

We have studied equations in Lesson 3.4. We di↵erentiated an identity froma conditional equation. Recall that an identity is an equation that is true for allvalues of the variable in the domain of the equation, while a conditional equationis an equation that is not an identity.

In this lesson, we mostly study conditional trigonometric equations. Wehave started it unnoticeably in the preceding lesson. For example, the equa-tion sin x = 1

2 has the unique solution x = sin�1 12 = ⇡

6 in the closed interval

233

⇥�⇡

2 ,⇡

2

⇤. However, if we consider the entire domain (not the restricted domain)

of the sine function, which is the set R of real numbers, there are solutions (otherthan ⇡

6 ) of the equation sin x = 12 . This current lesson explores the techniques of

solving (conditional) trigonometric equations.

We divide the lesson into two groups of equations: the ones with a basic wayof solving and those that use more advanced techniques.

3.8.1. Solutions of a Trigonometric Equation

Any equation that involves trigonometric expressions is called a trigonometricequation. Recall that a solution or a root

Teaching Notes

The word“solution” has two

meanings in ourdiscussion. One is

a process of solvinga problem, and theother is a number

that makes anequation true. Theintended meaning

depends on thecontext of its

usage.

of an equation is a number in the domainof the equation that, when substituted to the variable, makes the equation true.The set of all solutions of an equation is called the solution set of the equation.

Technically, the basic method to show that a particular number is a solutionof an equation is to substitute the number to the variable and see if the equationbecomes true. However, we may use our knowledge gained from the previouslessons to do a quicker verification process by not doing the manual substitutionand checking. We use this technique in the example.

Example 3.8.1. Which numbers in the set�0, ⇡6 ,

⇡

4 ,⇡

3 ,⇡

2 ,2⇡3 ,

3⇡4 ,

5⇡6 , ⇡, 2⇡

are

solutions to the following equations?Teaching Notes

In the process ofshowing that a

number is asolution of an

equation, note thatwe cannot assume

yet that it is asolution. Thismeans that wecannot use the

equality sign yet inthe process.

(1) sin x = 12

(2) tan x = 1

(3) 3 secx = �2p3

(4)p3| cot x| = 1

(5) sec2 x� tan2 x = 1

(6) sin x+ cosx = 0

(7) cos2 x = cos 2x+ sin2 x

(8) sin x+ cos 2x = 0

(9) 2 sin x+ tan x� 2 cosx = 2

(10) sin2 x+ cos2 x = 2

(11) sin 2x = sin x

(12) 2 tan x+ 4 sin x = 2 + secx

Solution. Note that the choices (except 2⇡) are numbers within the interval [0, ⇡].To quickly determine which numbers among the choices are solutions to a par-ticular equation, we use some distinctive properties of the possible solutions.Teaching Notes

In the actualclassroom

discussion, youmay only choose

some of theseitems.

(1) The sine function is positive on (0, ⇡). From Lesson 3.2, we recall that ⇡

6 isan obvious solution. We may imagine the graph of y = sin x. We may alsouse the idea of reference angle. Thus, among the choices, only ⇡

6 and 5⇡6 are

the only solutions of sin x = 12 .

(2) Since tan x = 1 > 0, any solution of the equation among the choices mustbe in the interval

�0, ⇡2

�(that is, in QI). Again, among the choices, the only

solution to tan x = 1 is ⇡

4 .

234

(3) Here, the given equation is equivalent to secx = �2p3

3 .Teaching Notes

For convenience inshowing or findinga solution of anequation, we mayuse an equivalentequation. Bydefinition, thesolutions of theequivalent equationare exactly thesame as thesolutions of theoriginal equation.

Among the choices,

the only solution of the equation 3 sec x = �2p3 is 5⇡

6 .

(4) Eliminating the absolute value sign, the given equation is equivalent to cot x =p33 or cot x = �

p33 . Among the choices, the only solution of cot x =

p33 is ⇡

3 ,

while the other equation has 2⇡3 . Thus, the only solutions of

p3| cot x| = 1

from the given set are ⇡

3 and 2⇡3 .

(5) The given equation is one of the Pythagorean Identities (page 181). It meansthat all numbers in the domain of the equation are solutions. The domainof the equation is R \ {x : cosx = 0}. Thus, all except ⇡

2 are solutions ofsec2 x� tan2 x = 1.

(6) For the sum of sin x and cos x to be 0, they must have equal absolute valuesbut di↵erent signs. Among the choices, only 3⇡

4 satisfies these properties, andit is the only solution of sinx+ cosx = 0.

(7) This equations is one of the Double-Angle Identities for Cosine. This meansthat all numbers in the domain of the equation are its solutions. Because thedomain of the given equation is R, all numbers in the given set are solutionsof cos2 x = cos 2x+ sin2 x.

(8) We substitute each number in the choices to the expression on the left-sideof the equation, and select those numbers that give resulting values equal to1.

Teaching Notes

If one side of anequation isconstant and if thenon-constantexpression is a bitcomplicated, thebasic method ofshowing that anumber is asolution of theequation is moreappropriate; thatis, to start on thenon-constant side,then substitute thenumber to thevariable, simplifythe expression, andend on the value ofthe constant on theother side.

x = 0: sin 0 + cos 2(0) = 0 + 1 = 1

x = ⇡

6 : sin⇡

6 + cos 2(⇡6 ) =12 +

12 = 1

x = ⇡

4 : sin⇡

4 + cos 2(⇡4 ) =p22 + 0 =

p22

x = ⇡

3 : sin⇡

3 + cos 2(⇡3 ) =p32 � 1

2 =p3�12

x = ⇡

2 : sin⇡

2 + cos 2(⇡2 ) = 1� 1 = 0

x = 2⇡3 : sin

2⇡3 + cos 2(2⇡3 ) =

p32 � 1

2 =p3�12

x = 3⇡4 : sin

3⇡4 + cos 2(3⇡4 ) =

p22 + 0 =

p22

x = 5⇡6 : sin

5⇡6 + cos 2(5⇡6 ) =

12 +

12 = 1

x = ⇡: sin ⇡ + cos 2⇡ = 0 + 1 = 1

x = 2⇡: sin 2⇡ + cos 2(2⇡) = 0 + 1 = 1

From these values, the only solution of sinx+ cos 2x = 0 among the choicesis ⇡

2 .

(9) We again substitute the numbers in the given set one by one, and see whichresulting values are equal to 1.

235

x = 0: 2 sin 0 + tan 0� 2 cos 0 = �2

x = ⇡

6 : 2 sin⇡

6 + tan ⇡

6 � 2 cos ⇡

6 = 3�2p3

3

x = ⇡

4 : 2 sin⇡

4 + tan ⇡

4 � 2 cos ⇡

4 = 1

x = ⇡

3 : 2 sin⇡

3 + tan ⇡

3 � 2 cos ⇡

3 = 2p3� 1

x = ⇡

2 : Since tan ⇡

2 is undefined, this value of x cannot be a solution of theequation.

x = 2⇡3 : 2 sin

2⇡3 + tan 2⇡

3 � 2 cos 2⇡3 = 1

x = 3⇡4 : 2 sin

3⇡4 + tan 3⇡

4 � 2 cos 3⇡4 = 2

p2� 1

x = 5⇡6 : 2 sin

5⇡6 + tan 5⇡

6 � 2 cos 5⇡6 = 3+2

p3

3

x = ⇡: 2 sin ⇡ + tan ⇡ � 2 cos⇡ = 2

x = 2⇡: 2 sin 2⇡ + tan 2⇡ � 2 cos 2⇡ = �2

Thus, the only solution of 2 sinx+tan x� 2 cosx = 2 from the given set is ⇡.

(10) This equation has no solution because one of the Pythagorean Identities sayssin2 x+ cos2 x = 1.

(11) We substitute each number in the given set to the expression of each side ofthe equation, and see which resulting values are equal.

Teaching Notes

If both sides of anequation are both

non-constantexpressions, one

method of showingthat a number is a

solution of theequation is tosubstitute the

number to bothexpressions (butnever join them

with equality signbecause they are

not yet equallogically), and

check if theresulting values are

equal.

x = 0: sin 2(0) = 0; sin 0 = 0

x = ⇡

6 : sin 2(⇡

6 ) =p32 ; sin ⇡

6 = 12

x = ⇡

4 : sin 2(⇡

4 ) = 1; sin ⇡

4 =p22

x = ⇡

3 : sin 2(⇡

3 ) =p32 ; sin ⇡

3 =p32

x = ⇡

2 : sin 2(⇡

2 ) = 0; sin ⇡

2 = 1

x = 3⇡4 : sin 2(

3⇡4 ) = �1; sin 3⇡

4 =p22

x = 5⇡6 : sin 2(

5⇡6 ) = �

p32 ; sin ⇡

3 = 12

x = ⇡: sin 2⇡ = 0; sin ⇡ = 0

x = 2⇡: sin 2(2⇡) = 0; sin 2⇡ = 0

Thus, among the numbers in the given set, the solutions of sin 2x = sin x are0, ⇡

3 , ⇡, and 2⇡.

(12) We employ the same technique used in the previous item.

236

x = 0 : 2 tan 0 + 4 sin 0 = 0

2 + sec 0 = 3

x = ⇡

6 : 2 tan ⇡

6 + 4 sin ⇡

6 = 2p3+63

2 + sec ⇡

6 = 2p3+63

x = ⇡

4 : 2 tan ⇡

4 + 4 sin ⇡

4 = 2p2 + 2

2 + sec ⇡

4 =p2 + 2

x = ⇡

3 : 2 tan ⇡

3 + 4 sin ⇡

3 = 4p3

2 + sec ⇡

3 = 4

x = ⇡

2 : Both tan ⇡

2 and sec ⇡

2 are undefined.

x = 2⇡3 : 2 tan 2⇡

3 + 4 sin 2⇡3 = 0

2 + sec 2⇡3 = 0

x = 3⇡4 : 2 tan 3⇡

4 + 4 sin 3⇡4 = 2

p2� 2

2 + sec 3⇡4 = 2�

p2

x = 5⇡6 : 2 tan 5⇡

6 + 4 sin 5⇡6 = 6�2

p3

3

2 + sec 5⇡6 = 6�2

p3

3

x = ⇡ : 2 tan ⇡ + 4 sin ⇡ = 0

2 + sec⇡ = 1

x = 2⇡ : 2 tan 2⇡ + 4 sin 2⇡ = 0

2 + sec 2⇡ = 3

After checking the equal values, the solutions of 2 tan x + 4 sin x = 2 + secxamong the given choices are ⇡

6 ,2⇡3 , and

5⇡6 . 2


In each equation, list down its solutions from the set��⇡

3 ,�⇡

4 ,⇡

6 ,⇡

4 ,2⇡3 , ⇡,

3⇡2

.

(1)p3 sec ✓ = 2 Answer: ⇡

6

(2) (sin x)(tan x+ 1) = 0 Answer: �⇡

4 , ⇡

(3) 2 + cos ✓ = 1 + 2 sin2 ✓ Answer: �⇡

3 , ⇡

(4) cos ✓ tan2 ✓ = 3 cos ✓ Answer: 2⇡3 ,

3⇡2

237

3.8.2. Equations with One Term

From the preceding discussion, you may observe that there may be more solutionsof a given equations outside the given set. We now find all solutions of a givenequation.

We will start with a group of equations having straightforward techniquesof finding their solutions. These simple techniques involve at least one of thefollowing ideas:

(1) equivalent equations (that is, equations that have the same solutions as theoriginal equation);

(2) periodicity of the trigonometric function involved;

(3) inverse trigonometric function;

(4) values of the trigonometric function involved on the interval [0, ⇡] or [0, 2⇡](depending on the periodicity of the function); and

(5) Zero-Factor Law: ab = 0 if and only if a = 0 or b = 0.

To “solve an equation” means to find all solutions of the equation. Here,unless stated as angles measured in degrees, we mean solutions of the equationthat are real numbers (or equivalently, angles measured in radians).

Example 3.8.2. Solve the equation 2 cosx� 1 = 0.

Solution. The given equation is equivalent to

cos x =1

2.

On the interval [0, 2⇡], there are only two solutions of the last equation, and theseare x = ⇡

3 (this is in QI) and x = 5⇡3 (in QIV).

Because the period of cosine function is 2⇡, the complete solutions of theequation are x = ⇡

3 + k(2⇡) and x = 5⇡3 + k(2⇡) for all integers k. 2

In the preceding example, by saying that the “complete solutions are x =⇡

3 + k(2⇡) and x = 5⇡3 + k(2⇡) for all integers k,” we mean that any integral

value of k will produce a solution to the given equation.Teaching Notes

Any particularsolution in a familyof solutions can be

used as a “seedsolution” to

produce the othersolutions in the

family.

For example, whenk = 3, x = ⇡

3 + 3(2⇡) = 19⇡3 is a solution of the equation. When k = �2,

x = 5⇡3 + (�2)(2⇡) = �7⇡

3 is another solution of 2 cosx � 1 = 0. The family ofsolutions x = ⇡

3 + k(2⇡) can be equivalently enumerated as x = 19⇡3 + 2k⇡, while

the family x = 5⇡3 + k(2⇡) can also be stated as x = �7⇡

3 + 2k⇡.

Example 3.8.3. Solve: (1 + cos ✓)(tan ✓ � 1) = 0.

238

Solution. By the Zero-Factor Law, the given equation is equivalent to

1 + cos ✓ = 0 or tan ✓ � 1 = 0

cos ✓ = �1 tan ✓ = 1

✓ = ⇡ + 2k⇡, k 2 Z ✓ = ⇡

4 + k⇡, k 2 Z.

Therefore, the solutions of the equation are ✓ = ⇡ + 2k⇡ and ✓ = ⇡

4 + k⇡ for allk 2 Z. 2

Example 3.8.4. Find all values of x in the interval [�2⇡, 2⇡] that satisfy theequation (sinx� 1)(sin x+ 1) = 0.

Solution.

sin x� 1 = 0 or sin x+ 1 = 0

sin x = 1 sin x = �1

x = ⇡

2 or � 3⇡2 x = 3⇡

2 or � ⇡

2

Solutions: ⇡

2 , �3⇡2 ,

3⇡2 , �⇡

2 2

Example 3.8.5. Solve: cosx = 0.1.

Solution. There is no special number whose cosine is 0.1. However, because0.1 2 [�1, 1], there is a number whose cosine is 0.1. In fact, in any one-periodinterval, with cosx = 0.1 > 0, we expect two solutions: one in QI and another inQIV. We use the inverse cosine function.

From Lesson 3.7, one particular solution of cosx = 0.1 in QI is x = cos�1 0.1.We can use this solution to get a particular solution in QIV, and this is x =2⇡ � cos�1 0.1, which is equivalent to x = � cos�1 0.1.

From the above particular solutions, we can produce all solutions of cosx =0.1, and these are x = cos�1 0.1+2k⇡ and x = � cos�1 0.1+2k⇡ for all k 2 Z. 2

Example 3.8.6. Solve: 3 tan ✓ + 5 = 0.

Solution.3 tan ✓ + 5 = 0 =) tan ✓ = �5

3

We expect only one solution in any one-period interval.

tan ✓ = �53 =) ✓ = tan�1

��5

3

�+ k⇡, k 2 Z 2

?Example 3.8.7. The voltage V (in volts) coming from an electricity distribut-ing company is fluctuating according to the function V (t) = 200+170 sin(120⇡t)at time t in seconds.

239

(1) Determine the first time it takes to reach 300 volts.

(2) For what values of t does the voltage reach its maximum value?

Solution. (1) We solve for the least positive value of t such that V (t) = 300.

200 + 170 sin(120⇡t) = 300

sin(120⇡t) =100

170

120⇡t = sin�1 100

170

t =sin�1 100

170

120⇡⇡ 0.00167 seconds

(2) The maximum value of V (t) happens when and only when the maximumvalue of sin(120⇡t) is reached. We know that the maximum value of sin(120⇡t)is 1, and it follows that the maximum value of V (t) is 370 volts. Thus, weneed to solve for all values of t such that sin(120⇡t) = 1.

sin(120⇡t) = 1

120⇡t =⇡

2+ 2k⇡, k nonnegative integer

t =⇡

2 + 2k⇡

120⇡

t =12 + 2k

120⇡ 0.00417 + 0.017k

This means that the voltage is maximum when t ⇡ 0.00417+0.017k for eachnonnegative integer k. 2


1. Solve each equation.

(a) tan x = �1 Answer: �⇡

4 + k⇡, k 2 Z

(b) sin x = 12 Answer: ⇡

6 + 2k⇡, 5⇡6 + 2k⇡, k 2 Z

(c) (cos x� 1)(tan x+ 1) = 0 Answer: 2k⇡, �⇡

4 + k⇡, k 2 Z

2. Find all values of the variable in the interval [�2⇡, 2⇡] that satisfy the givenequation.

(a) (sin ✓ + 1)(tan ✓) = 0 Answer: �⇡

2 ,3⇡2 , 0, ⇡, 2⇡,�⇡,�2⇡

(b) sec ✓ + 2 = 0 Answer: 2⇡3 ,�

4⇡3 ,

4⇡3 ,�

2⇡3

240

3.8.3. Equations with Two or More Terms

We will now consider a group of equations having multi-step techniques of find-ing their solutions. Coupled with the straightforward techniques discussed in thepreceding discussion, these more advanced techniques involve factoring of expres-sions and trigonometric identities. The primary goal is to reduce a given equationinto equivalent one-term equations.

Example 3.8.8. Solve: 2 cosx tan x = 2 cos x.

Solution.

2 cosx tan x = 2 cos x

2 cosx tan x� 2 cosx = 0

(2 cosx)(tan x� 1) = 0 Teaching Notes

The method forsolving

trigonometric

equations follows

the usual way of

solving nonlinear

equations; that is,

transform theequation so that

one side is 0, and

then factor.

2 cosx = 0 or tan x� 1 = 0

cos x = 0 tan x = 1

x = ⇡

2 + 2k⇡ orx = 3⇡

2 + 2k⇡,k 2 Z

x = ⇡

4 + k⇡,k 2 Z

Solutions: ⇡

2 + 2k⇡, 3⇡2 + 2k⇡, ⇡

4 + k⇡, k 2 Z 2

Example 3.8.9. Solve for x 2 [0, 2⇡): sin 2x = sin x.

Solution.

sin 2x = sin x

sin 2x� sin x = 0

2 sin x cos x� sin x = 0 Sine Double-Angle Identity

(sin x)(2 cosx� 1) = 0

sin x = 0 or 2 cosx� 1 = 0

x = 0 or x = ⇡ cos x = 12

x = ⇡

3 or x = 5⇡3

Solutions: 0, ⇡, ⇡

3 ,5⇡3 2

241

Tips in Solving Trigonometric Equations

(1) If the equation contains only one trigonometric term, isolate thatterm, and solve for the variable.

(2) If the equation is quadratic in form, we may use factoring, findingsquare roots, or the quadratic formula.

(3) Rewrite the equation to have 0 on one side, and then factor (ifappropriate) the expression on the other side.

(4) If the equation contains more than one trigonometric function,try to express everything in terms of one trigonometric function.Here, identities are useful.

(5) If half or multiple angles are present, express them in terms of atrigonometric expression of a single angle, except when all anglesinvolved have the same multiplicity wherein, in this case, retainthe angle. Half-angle and double-angle identities are useful insimplification.

Example 3.8.10. Solve for x 2 [0, 2⇡): 2 cos2 x = 1 + sin x.

Solution.

2 cos2 x = 1 + sin x

2(1� sin2 x) = 1 + sin x Pythagorean Identity

2 sin2 x+ sin x� 1 = 0

(2 sin x� 1)(sin x+ 1) = 0 Factoring

2 sin x� 1 = 0 or sin x+ 1 = 0

sin x = 12 sin x = �1

x = ⇡

6 or x = 5⇡6 x = 3⇡

2

Solutions: ⇡

6 ,5⇡6 ,

3⇡2 2

Example 3.8.11. Solve for x 2 [0, 2⇡) in the equation 3 cos2 x+ 2 sin x = 2.

Solution.

3 cos2 x+ 2 sin x = 2

3(1� sin2 x) + 2 sin x = 2 Pythagorean Identity

(3 sin x+ 1)(sin x� 1) = 0 Factoring

242

3 sin x+ 1 = 0 or sin x� 1 = 0

sin x = �13 sin x = 1

x = sin�1(�13)+ 2⇡

orx = ⇡ � sin�1(�1

3)

x = ⇡

2

Solutions: Teaching Notes

Using theOdd-Even Identity,sin�1(� 1

3 ) =

� sin�1( 13 ).

2⇡ � sin�1(13)+, ⇡ + sin�1(13),⇡

2 2

One part of the last solution needs further explanation. In the equationsin x = �1

3 , we expect two solutions in the interval [0, 2⇡): one in (⇡, 3⇡2 ) (whichis QIII), and another in (3⇡2 , 2⇡) (which is QIV). Since no special number satisfiessin x = �1

3 , we use inverse sine function. Teaching Notes

Using the referenceangle of sin�1( 13 ),we get twosolutions (QIII andQIV), and theseare ⇡ + sin�1( 13 )

and 2⇡ � sin�1( 13 ).

Because the range of sin�1 is [�⇡

2 ,⇡

2 ], weknow that �⇡

2 < sin�1(�13) < 0. From this value, to get the solution in (3⇡2 , 2⇡),

we simply add 2⇡ to this value, resulting to x = sin�1(�13) + 2⇡. On the other

hand, to get the solution in (⇡, 3⇡2 ), we simply add � sin�1(�13) to ⇡, resulting to

x = ⇡ � sin�1(�13).

Example 3.8.12. Solve: sin2 x+ 5 cos2 x

2 = 2.

Solution.

sin2 x+ 5 cos2 x

2 = 2

sin2 x+ 5�1+cosx

2

�= 2 Cosine Half-Angle Identity

2 sin2 x+ 5 cosx+ 1 = 0

2(1� cos2 x) + 5 cosx+ 1 = 0 Pythagorean Identity

2 cos2 x� 5 cosx� 3 = 0

(2 cosx+ 1)(cos x� 3) = 0

2 cosx+ 1 = 0 or cos x� 3 = 0

cos x = �12 cos x = 3

x = 2⇡3 + 2k⇡ or

x = 4⇡3 + 2k⇡,k 2 Z

no solution

Solutions: 2⇡3 + 2k⇡, 4⇡

3 + 2k⇡, k 2 Z 2

Example 3.8.13. Solve for x 2 [0, 2⇡) in the equation tan 2x� 2 cosx = 0.

243

Solution.

tan 2x� 2 cosx = 0

sin 2x

cos 2x� 2 cosx = 0

sin 2x� 2 cosx cos 2x = 0

Apply the Double-Angle Identities for Sine and Cosine, and then factor.

2 sin x cos x� 2(cosx)(1� 2 sin2 x) = 0

(2 cosx)(2 sin2 x+ sin x� 1) = 0

(2 cosx)(2 sin x� 1)(sin x+ 1) = 0

2 cosx = 0 or 2 sin x� 1 = 0 or sin x+ 1 = 0

cos x = 0 sin x = 12 sin x = �1

x = ⇡

2 orx = 3⇡

2

x = ⇡

6 orx = 5⇡

6

x = 3⇡2

These values of x should be checked in the original equation because tan 2x maynot be defined. Upon checking, this is not the case for each value of x obtained.The solutions are ⇡

2 ,3⇡2 ,

⇡

6 ,5⇡6 , and

3⇡2 . 2

?Example 3.8.14. A weight is suspended from a spring and vibrating verticallyaccording to the equation

f(t) = 20 cos�45⇡

�t� 5

6

��,

where f(t) centimeters is the directed distance of the weight from its centralposition at t seconds, and the positive distance means above its central position.

(1) At what time is the displacement of the weight 5 cm below its centralposition for the first time?

(2) For what values of t does the weight reach its farthest point below its centralposition?

Solution. (1) We find the least positive value of t such that f(t) = �5.

20 cos�45⇡

�t� 5

6

��= �5

cos�45⇡

�t� 5

6

��= �1

4

There are two families of solutions for this equation.

244

• 45⇡

�t� 5

6

�= cos�1

��1

4

�+ 2k⇡, k 2 Z

t = 56 +

cos�1(� 14)+2k⇡

45⇡

In this family of solutions, the least positive value of t happens whenk = 0, and this is

t =5

6+

cos�1��1

4

�+ 2(0)⇡

45⇡

⇡ 1.5589.

• 45⇡

�t� 5

6

�= 2⇡ � cos�1

��1

4

�+ 2k⇡, k 2 Z

t = 56 +

2⇡�cos�1(� 14)+2k⇡

45⇡

Here, the least positive value of t happens when k = �1, and this is

t =5

6+

2⇡ � cos�1��1

4

�+ 2(�1)⇡

45⇡

⇡ 0.1078.

Therefore, the first time that the displacement of the weight is 5 cm belowits central position is at about 0.1078 seconds.

(2) The minimum value of f(t) happens when and only when the minimumvalue of cos 4

5⇡�t� 5

6

�is reached. The minimum value of cos 4

5⇡�t� 5

6

�is

�1, which implies that the farthest point the weight can reach below itscentral position is 20 cm. Thus, we need to solve for all values of t such thatcos 4

5⇡�t� 5

6

�= �1.

cos 45⇡

�t� 5

6

�= �1

45⇡

�t� 5

6

�= cos�1(�1) + 2k⇡, k � 0

45⇡

�t� 5

6

�= ⇡ + 2k⇡

t = 56 +

⇡+2k⇡45⇡

= 2512 +

52k

Therefore, the weight reaches its farthest point (which is 20 cm) below itscentral position at t = 25

12 +52k for every integer k � 0. 2



(a) 2 sin2 ✓ = sin ✓ + 1 Answer: 7⇡6 + 2k⇡, 11⇡

6 + 2k⇡, ⇡

2 + 2k⇡, k 2 Z

(b) tan2 x+ tan x = 6 Answer: tan�1(�3) + k⇡, tan�1 2 + k⇡, k 2 Z

(c) sin x = 1 + cosx Answer: ⇡

2 + 2k⇡, ⇡ + 2k⇡, k 2 Z

245

2. Find the solutions in the interval [0, 2⇡).

(a) sin 2✓ = cos ✓ Answer: ⇡

2 ,3⇡2 ,

⇡

6 ,5⇡6

(b) 3 cos2 x+ sin x = 3 Answer: 0, ⇡, sin�1 13 , ⇡ � sin�1 1

3

Exercises 3.8


(a) 2 sin x+ 1 = 0 Answer: �⇡

6 + 2k⇡, 7⇡6 + 2k⇡, k 2 Z

(b) sin x tan x = 0 Answer: k⇡, k 2 Z(c) tan x+ 1 = 0 Answer: 3⇡

4 + k⇡, k 2 Z(d) cos 3x = 0 Answer: 1

6⇡(2k � 1), k 2 Z(e) tan 4x� 1 = 0 Answer: 1

16⇡(4k + 1), k 2 Z(f) sec2 x� 1 = 0 Answer: k⇡, k 2 Z(g) sec2 x+ 6 tan x+ 4 = 0 Answer: �⇡

4 + k⇡, tan�1(�5) + k⇡, k 2 Z(h) cos 2x+ 3 = 5 cosx Answer: �⇡

3 + 2k⇡, ⇡

3 + 2k⇡, k 2 Z(i) cos2 x+ sin2 x

2 = 1 Answer: 2⇡3 + 2k⇡, 4⇡

3 + 2k⇡, 2k⇡, k 2 Z(j) 6 sec2 x

2 + 3 = 7 tan x

2

Answer: 2 tan�1��1

3

�+ 2k⇡, 2 tan�1 3

2 + 2k⇡, k 2 Z

2. Find the solutions in the interval [0, 2⇡).

(a) 4 sin2 x� 1 = 0 Answer: ⇡

6 ,5⇡6 ,

7⇡6 ,

11⇡6

(b) 2 cos2 x+ 3 cosx� 2 = 0 Answer: ⇡

3 ,5⇡3

(c) tan x� cot x = 0 Answer: ⇡

4 ,5⇡4

(d) 2 sin 2x =p3 Answer: ⇡

6 ,⇡

3 ,7⇡6 ,

4⇡3

(e) sec2 x� 4 = 0 Answer: ⇡

3 ,2⇡3 ,

4⇡3 ,

5⇡3

(f) 2 sin2 x� 5 sin x = 3 Answer: 7⇡6 ,

11⇡6

(g) tan x+ secx = 0 Answer: 3⇡2

(h) 2 sin 2x = 3 sin x Answer: 0, ⇡, cos�1 34 , 2⇡ � cos�1 3

4

(i) tan2 x = 1 + secx Answer: 0, ⇡

3 ,5⇡3

(j) tan x+p3 = secx Answer: 11⇡

6

3. Find the solutions in the interval [0�, 180�).

(a) sin x� cos x = 0 Answer: 45�

(b) cot 4x� 1 = 0 Answer: 11.25�, 56.25�, 101.25�, 146.25�

246

(c) 3 cos 2x� 3 cosx = 0 Answer: 0�, 120�

?(d) 6 sec2 x+ 3 = 7 tan x Answer: 161.6�, 56.3�

?(e) tan2 x+ tan x = 6 Answer: 63.4�, 108.4�

?4. A weight is suspended from a spring and vibrating vertically according to theequation

f(t) = 25.2 sin(3.8t� 2.1),

where f(t) centimeters is the directed distance of the weight from its cen-tral position at t seconds, and the positive distance means above its centralposition.

(a) Find the times when the weight is at its central position.

Solution. We solve the equation f(t) = 0.

25.2 sin(3.8t� 2.1) = 0

sin(3.8t� 2.1) = 0

3.8t� 2.1 = k⇡ k nonnegative integer

t =2.1 + k⇡

3.8t ⇡ 0.55 + 0.83k

Therefore, the weight is at its central position at t ⇡ 0.55+0.83k seconds(where k is a nonnegative integer). In other words, it is at central positionwhen t = 0.55 s, t = 1.38 s, t = 2.21 s, etc.

(b) For what values of t does the weight reach its farthest point below itscentral position?

Solution. The weight reaches it farthest point below the central positionwhen sin(3.8t� 2.1) = �1. Solving for t, we get

sin(3.8t� 2.1) = �1

3.8t� 2.1 =3⇡

2+ 2k⇡ where k is a whole number

t =3⇡+4k⇡

2 + 2.1

3.8where k is a whole number

t =(3 + 4k)⇡ + 4.2

7.6where k is a whole number

t ⇡ 1.79 + 1.65k where k is a whole number

Therefore, the weight reaches it farthest point below the central positionat t ⇡ 1.79 + 1.65k seconds (where k is a whole number). For instance,at t = 1.79 s, t = 3.44 s, t = 5.09 s, etc.

247

?5. The finance department of a car company conducted a study of their weeklysales in the past years, and came out with the following approximating func-tion:

s(t) = 12.18 cos(0.88t� 7.25) + 20.40, t � 0,

where s(t) represents weekly car sales in million pesos at week t (t = 0 repre-sents the start of the study).

(a) Find the weekly sales at the start of the study.

Solution.

s(0) = 12.18 cos(0.88(0)� 7.25) + 20.40

= 12.18 cos(�7.25) + 20.40

= (12.18)(0.5679) + 20.40

= 27.32

The weekly sales of the car company at the start of the study is approx-imately 27.32 million pesos.

(b) Find the projected maximum and minimum weekly sales of the company.

Solution. The projected maximum and minimum weekly sales of thecompany are attained when the cosine values are 1 and �1, respectively.Thus, the maximum weekly sales is 12.18 + 20.40 = 32.58 million pesos,and the minimum weekly sales is �12.18 + 20.40 = 8.22 million pesos.

(c) If the company were able to reach its maximum sales this week, whenwill the next projected maximum weekly sales and upcoming projectedminimum weekly sales be?

Solution. The next projected maximum weekly sales will be attained afterone period. That is, P = 2⇡

0.88 ⇡ 7.14. Hence, if the company were ableto reach its maximum sales this week, then the next projected maximumweekly sales will be after about 7 weeks.

On the other hand, the upcoming minimum weekly sales is projected afterhalf the period. That is, 1

2P = 12

2⇡0.88 ⇡ 3.57. Hence, if the company were

able to reach its maximum sales this week, then the upcoming projectedminimum weekly sales will be after about 3.5 weeks.

(d) After the start of the study, when did the company experience a weeklysales of only 10 million for the first time?

Solution. Here, we want to solve s(t) = 10 for the least nonnegative valueof t.

12.18 cos(0.88t� 7.25) + 20.40 = 10

cos(0.88t� 7.25) = �520

609

248

We get

0.88t� 7.25 = cos�1

✓�520

609

◆+ 2k⇡, k 2 Z =) t ⇡ 11.19 + 7.14k

or

0.88t� 7.25 = � cos�1

✓�520

609

◆+ 2k⇡ =) t ⇡ 5.29 + 7.14k.

Among these solutions, the least nonnegative value of t is t = 11.19 +7.14(�1) ⇡ 4.05. Thus, about 4 weeks after the start of the study, thecompany experienced a weekly sales of only 10 million for the first time.

?6. After many years in business, the financial analyst of a shoe company projectedthat the monthly costs of producing their products and monthly revenues fromthe sales of their products are fluctuating according to the following formulas:

C(t) = 2.6 + 0.58 sin(0.52t� 7.25)

andR(t) = 2.6 + 1.82 cos(0.52t� 7.25),

where C(t) and R(t) are the costs and revenues in million of pesos at month t(t = 0 represents January 2010). About how many months after January 2010did the company experience a zero profit for the first time?

Solution. The profit is zero when the revenue is the same as the cost.

2.6 + 0.58 sin(0.52t� 7.25) = 2.6 + 1.82 cos(0.52t� 7.25)

tan(0.52t� 7.25) =91

29

0.52t� 7.25 = tan�1

✓91

29

◆+ k⇡, k 2 Z

t =7.25 + tan�1

�9129

�+ k⇡

0.52⇡ 16.37 + 6.04k

The company experienced zero profit for the first time about 16.37+6.04(�2) ⇡4.29 or 4 months after January 2010.

7. If x be a real number such that

cos 2x =2

3and cos x� sin x =

1

2,

what is cosx+ sin x?

249

Solution.

cos 2x =2

3

cos2 x� sin2 x =2

3

(cos x+ sin x)(cos x� sin x) =2

3

(cos x+ sin x)

✓1

2

◆=

2

3

cos x+ sin x =4

3

8. Solve for x 2 [�⇡, ⇡): 16 sin4 x+ 1 = 8 sin2 x.

Solution.

16 sin4 x+ 1 = 8 sin2 x

16 sin4 x� 8 sin2 x+ 1 = 0

(4 sin2 x� 1)2 = 0

4 sin2 x� 1 = 0

sin x = ±1

2

x = ±5⇡

6,±⇡

6

9. Find the smallest positive value of ✓ that satisfies the equation

sin⇣⇡3+ ✓

⌘+ sin

⇣⇡3� ✓

⌘=

3

8.

Solution.

sin⇣⇡3+ ✓

⌘+ sin

⇣⇡3� ✓

⌘=

3

8

sin⇡

3cos ✓ + cos

⇡

3sin ✓ + sin

⇡

3cos ✓ � cos

⇡

3sin ✓ =

3

8

2 sin⇡

3cos ✓ =

3

8p3 cos ✓ =

3

8

✓ = cos�1

p3

8

!

250

10. In 4ABC, angles A and B (in degrees) satisfy

3 sinA+ cosB = 3 and sinB + 3 cosA = 2.

Prove that C = 30�.

Solution. (3 sinA+ cosB = 3

3 cosA+ sinB = 2

Square both equations:(9 sin2 A+ 6 sinA cosB + cos2 B = 9

9 cos2 A+ 6 cosA sinB + sin2 B = 4

Add and solve:

9 + 6 sin(A+B) + 1 = 13

6 sin(A+B) = 3

sin(A+B) =1

2A+B = 30�, 150�

If A + B = 150�, then C = 30�. If A + B = 30�, then A < 30�, and3 sinA+cosB < 3 · 12 +1 = 5

2 < 3, contradicting the first equation. Therefore,we only have C = 30�.

4

Lesson 3.9. Polar Coordinate System




(1) locate points in polar coordinate system;

(2) convert the coordinates of a point from rectangular to polar system and viceversa; and

(3) solve situational problems involving polar coordinate system.

Lesson Outline

(1) Polar coordinate system: pole and polar axis

251

(2) Polar coordinates of a point and its location

(3) Conversion from polar to rectangular coordinates, and vice versa

(4) Simple graphs and applications

Introduction

Two-dimensional coordinate systems are used to describe a point in a plane.We previously used the Cartesian or rectangular coordinate system to locate apoint in the plane. That point is denoted by (x, y), where x is the signed dis-tance of the point from the y-axis, and y is the signed distance of the pointfrom the x-axis. We sketched the graphs of equations (lines, circles, parabolas,ellipses, and hyperbolas) and functions (polynomial, rational, exponential, log-arithmic, trigonometric, and inverse trigonometric) in the Cartesian coordinateplane. However, it is often convenient to locate a point based on its distancefrom a fixed point and its angle with respect to a fixed ray. Not all equationscan be graphed easily using Cartesian coordinates. In this lesson, we also useanother coordinate system, which can be presented in dartboard-like plane asshown below.

3.9.1. Polar Coordinates of a Point

We now introduce the polar coordinate system. It is composed of a fixed pointcalled the pole (which is the origin in the Cartesian coordinate system) and afixed ray called the polar axis (which is the nonnegative x-axis).

252

In the polar coordinate system, a point is described by the ordered pair (r, ✓),where the radial coordinate r refers to the directed distance of the point from thepole and the angular coordinate ✓ refers to a directed angle (usually in radians)from the polar axis to the segment joining the point and the pole.

Because a point in polar coordinate system is described by an order pair ofradial coordinate and angular coordinate, it will be more convenient to geomet-rically present the system in a polar plane, which serves just like the Cartesianplane. In the polar plane shown below, instead of rectangular grids in the Carte-sian plane, we have concentric circles with common center at the pole to identifyeasily the distance from the pole (radial coordinate) and angular rays emanatingfrom the pole to show the angles from the polar axis (angular coordinate).

253

Example 3.9.1. Plot the following points in one polar plane: A(3, ⇡3 ), B(1, 5⇡6 ),C(2, 7⇡6 ), D(4, 19⇡12 ), E(3,�⇡), F (4,�7⇡

6 ), G(2.5, 17⇡4 ), H(4, 17⇡6 ), and I(3,�5⇡3 ).

Solution.

As seen in the last example, unlike in Cartesian plane where a point has aunique Cartesian coordinate representation, a point in polar plane have infinitelymany polar coordinate representations. For example, the coordinates (3, 4) inthe Cartesian plane refer to exactly one point in the plane, and this particularpoint has no rectangular coordinate representations other than (3, 4). However,the coordinates (3, ⇡3 ) in the polar plane also refer to exactly one point, butthis point has other polar coordinate representations. For example, the polarcoordinates (3,�5⇡

3 ), (3,7⇡3 ), (3,

13⇡3 ), and (3, 19⇡3 ) all refer to the same point as

that of (3, ⇡3 ).

The polar coordinates (r, ✓ + 2k⇡), where k 2 Z, represent the samepoint as that of (r, ✓).

In polar coordinate system, it is possible for the coordinates (r, ✓) to havea negative value of r. In this case, the point is |r| units from the pole in theopposite direction of the terminal side of ✓, as shown in Figure 3.36.

254

Figure 3.36

Example 3.9.2. Plot the following points in one polar plane: A(�3, 4⇡3 ), B(�4, 11⇡6 ),C(�2,�⇡), and D(�3.5,�7⇡

4 ).

Solution. As described above, a polar point with negative radial coordinate lieson the opposite ray of the terminal side of ✓.

255

Points in Polar Coordinates

1. For any ✓, the polar coordinates (0, ✓) represent the pole.

2. A point with polar coordinates (r, ✓) can also be represented by(r, ✓ + 2k⇡) or (�r, ✓ + ⇡ + 2k⇡) for any integer k.


1. Plot the following points in one polar plane:

A(2,�⇡

2 ) B(3, 7⇡4 ) C(4,�⇡

6 )

D(2,�4⇡3 ) E(�4, ⇡4 ) F (1, 41⇡12 )

G(�3,�2⇡3 ) H(�4,� ⇡

12) I(�2,�11⇡2 )

Answer:

2. Give the polar coordinates (r, ✓) with indicated properties that represent thesame point as the given polar coordinates.

256

(a) (2, ⇡); r > 0, �2⇡ < ✓ 0 Answer: (2,�⇡)

(b) (5, 3⇡4 ); r < 0, 0 < ✓ 2⇡ Answer: (�5,�7⇡4 )

(c) (�5, 4⇡3 ); r < 0, �2⇡ < ✓ 0 Answer: (�5,�2⇡3 )

(d) (1, 0); r < 0, 0 < ✓ 2⇡ Answer: (�1, ⇡)

(e) (2, sin�1(0.6)); r < 0, �2⇡ < ✓ 0 Answer: (�2, sin�1(0.6)� ⇡)

(f) (�3, cos�1(�0.4)); r > 0, 0 < ✓ 2⇡ Answer: (3, cos�1(�0.4) + ⇡)

3.9.2. From Polar to Rectangular, and Vice Versa

We now have two ways to describe points on a plane – whether to use the Carte-sian coordinates (x, y) or the polar coordinates (r, ✓). We now derive the conver-sion from one of these coordinate systems to the other.

We superimpose the Cartesian and polar planes, as shown in the followingdiagram.

Figure 3.37

Suppose a point P is represented by the polar coordinates (r, ✓). From Lesson3.2 (in particular, the boxed definition on page 138), we know that

x = r cos ✓ and y = r sin ✓.

Conversion from Polar to Rectangular Coordinates

(r, ✓) �!

8<

:x = r cos ✓

y = r sin ✓�! (x, y)

Given one polar coordinate representation (r, ✓), there is only onerectangular coordinate representation (x, y) corresponding to it.

257

Example 3.9.3. Convert the polar coordinates (5, ⇡) and (�3, ⇡6 ) to Cartesiancoordinates.

Solution.

(5, ⇡) �!

8<

:x = 5 cos ⇡ = �5

y = 5 sin ⇡ = 0�! (�5, 0)

Teaching Notes

One can also easilyconvert the polarcoordinates (5,⇡)

to itscorresponding

rectangularcoordinates (�5, 0)by simply plotting

the point.

(�3, ⇡6 ) �!

8<

:x = �3 cos ⇡

6 = �3p3

2

y = �3 sin ⇡

6 = �32

�! (�3p3

2 ,�32) 2

As explained on page 254 (right after Example 3.9.1), we expect that thereare infinitely many polar coordinate representations that correspond to just onegiven rectangular coordinate representation. Although we can actually determineall of them, we only need to know one of them and we can choose r � 0.

Suppose a point P is represented by the rectangular coordinates (x, y). Re-ferring back to Figure 3.37, the equation of the circle is

x2 + y2 = r2 =) r =p

x2 + y2.

We now determine ✓. If x = y = 0, then r = 0 and the point is the pole. Thepole has coordinates (0, ✓), where ✓ is any real number.

If x = 0 and y 6= 0, then we may choose ✓ to be either ⇡

2 or 3⇡2 (or their

equivalents) depending on whether y > 0 or y < 0, respectively.

Now, suppose x 6= 0. From the boxed definition again on page 138, we knowthat

tan ✓ =y

x,

where ✓ is an angle in standard position whose terminal side passes through thepoint (x, y).

Conversion from Rectangular to Polar Coordinates

(x, y) = (0, 0) �! (r, ✓) = (0, ✓), ✓ 2 R

(0, y)y 6=0

�! (r, ✓) =

8<

:(y, ⇡2 ) if y > 0

(|y|, 3⇡2 ) if y < 0

(x, 0)x 6=0

�! (r, ✓) =

8<

:(x, 0) if x > 0

(|x|, ⇡) if x < 0

258

(x, y)x6=0, y 6=0

�! (r, ✓)

r =p

x2 + y2

tan ✓ = y

x

✓ same quadrant as (x, y)

Given one rectangular coordinate representation (x, y), there aremany polar coordinate representations (r, ✓) corresponding to it. Theabove computations just give one of them.

Example 3.9.4. Convert each Cartesian coordinates to polar coordinates (r, ✓),where r � 0.(1) (�4, 0)

(2) (4, 4)

(3) (�3,�p3)

(4) (6,�2)

(5) (�3, 6)

(6) (�12,�8)

Solution. (1) (�4, 0) �! (4, ⇡)Teaching Notes

Plotting the pointson thesuperimposedCartesian andpolar planes is aquicker approachin convertingrectangularcoordinates topolar.

(2) The point (4, 4) is in QI.

r =p

x2 + y2 =p42 + 42 = 4

p2

tan ✓ = y

x

= 44 = 1 =) ✓ = ⇡

4

(4, 4) �!�4p2, ⇡4

�

(3) (�3,�p3) in QIII

r =q

(�3)2 + (�p3)2 = 2

p3

tan ✓ = �p3

�3 =p33 =) ✓ = 7⇡

6

(�3,�p3) �!

�2p3, 7⇡6

�

(4) (6,�2) in QIV

r =p

62 + (�2)2 = 2p10

tan ✓ = �26 = �1

3 =) ✓ = tan�1��1

3

�

(6,�2) �!�2p10, tan�1

��1

3

��

(5) (�3, 6) in QII

r =p

(�3)2 + 62 = 3p5

tan ✓ = 6�3 = �2 =) ✓ = ⇡ + tan�1(�2)

Teaching Notes

Recall thattan�1(�2) is inQIV.

(�3, 6) �!�3p5, ⇡ + tan�1(�2)

�

259

(6) (�12,�8) in QIII

r =p

(�12)2 + (�8)2 = 4p13

tan ✓ = �8�12 = 2

3 =) ✓ = ⇡ + tan�1 23Teaching Notes

We may also use✓ = tan�1 2

3 � ⇡. (�12,�8) �!�4p13, ⇡ + tan�1 2

3

�2


1. Convert each polar coordinates to Cartesian coordinates.

(a) (2,�⇡) Answer: (�2, 0)

(b) (4, 3⇡4 ) Answer: (�2p2, 2

p2)

(c) (6, 3⇡2 ) Answer: (0,�6)

(d) (�2, 2⇡3 ) Answer: (2,�p3)

(e) (�4, 7⇡6 ) Answer: (2p3, 2)

(f) (�3,�2⇡3 ) Answer: (32 ,

3p3

2 )

(g) (1, sin�1(�13)) Answer: (2

p2

3 ,�13)

(h) (�2, tan�1 43) Answer: (�6

5 ,�85)

2. Convert each Cartesian coordinates to polar coordinates (r, ✓), where r � 0.

(a) (0, 6) Answer: (6, ⇡2 )

(b) (3,�3) Answer: (3p2,�⇡

4 )

(c) (�3p3, 3) Answer: (6, 5⇡6 )

(d) (�1,�p3) Answer: (2, 4⇡3 )

(e) (1, 4) Answer: (p17, tan�1 4)

(f) (�2, 4) Answer: (2p5, ⇡ + tan�1(�2))

(g) (�6,�2) Answer: (2p10, ⇡ + tan�1 1

3)

(h) (�1, 1) Answer: (p2, 3⇡4 )

3.9.3. Basic Polar Graphs and Applications

From the preceding session, we learned how to convert polar coordinates of apoint to rectangular and vice versa using the following conversion formulas:

r2 = x2 + y2, tan ✓ =y

x, x = r cos ✓, and y = r sin ✓.

Because a graph is composed of points, we can identify the graphs of some equa-tions in terms of r and ✓.

260

Graph of a Polar Equation

The polar graph of an equation involving r and ✓ is the set of allpoints with polar coordinates (r, ✓) that satisfy the equation.

As a quick illustration, the polar graph of the equation r = 2�2 sin ✓ consistsof all points (r, ✓) that satisfy the equation. Some of these points are (2, 0), (1, ⇡6 ),(0, ⇡2 ), (2, ⇡), and (4, 3⇡2 ).

Example 3.9.5. Identify the polar graph of r = 2, and sketch its graph in thepolar plane.

Solution. Squaring the equation, we get r2 = 4. Because r2 = x2 + y2, we havex2+ y2 = 4, which is a circle of radius 2 and with center at the origin. Therefore,the graph of r = 2 is a circle of radius 2 with center at the pole, as shown below.

In the previous example, instead of using the conversion formula r2 = x2+y2,we may also identify the graph of r = 2 by observing that its graph consists ofpoints (2, ✓) for all ✓. In other words, the graph consists of all points with radialdistance 2 from the pole as ✓ rotates around the polar plane. Therefore, thegraph of r = 2 is indeed a circle of radius 2 as shown.

Example 3.9.6. Identify and sketch the polar graph of ✓ = �5⇡4 .

Solution. The graph of ✓ = �5⇡4 consists of all points (r,�5⇡

4 ) for r 2 R. Ifr > 0, then points (r,�5⇡

4 ) determine a ray from the pole with angle �5⇡4 from

the polar axis. If r = 0, then (0,�5⇡4 ) is the pole. If r < 0, then the points

(r,�5⇡4 ) determine a ray in opposite direction to that of r > 0. Therefore, the

261

graph of ✓ = �5⇡4 is a line passing through the pole and with angle �5⇡

4 withrespect to the polar axis, as shown below.

Example 3.9.7. Identify (and describe) the graph of the equation r = 4 sin ✓.

Solution.

r = 4 sin ✓

r2 = 4r sin ✓

x2 + y2 = 4y

x2 + y2 � 4y = 0

x2 + (y � 2)2 = 4

Therefore, the graph of r = 4 sin ✓ is a circle of radius 2 and with center at (2, ⇡2 ).

262

?Example 3.9.8. Sketch the graph of r = 2� 2 sin ✓.

Solution. We construct a table of values.

x 0 ⇡

6⇡

4⇡

3⇡

22⇡3

3⇡4

5⇡6 ⇡

r 2 1 0.59 0.27 0 0.27 0.59 1 2

x 7⇡6

5⇡4

4⇡3

3⇡2

5⇡3

7⇡4

11⇡6 2⇡

r 3 3.41 3.73 4 3.73 3.41 3 2

This heart-shaped curve is called a cardioid. 2

?Example 3.9.9. The sound-pickup capability of a certain brand of microphoneis described by the polar equation r = �4 cos ✓, where |r| gives the sensitivity ofthe microphone to a sound coming from an angle ✓ (in radians).

(1) Identify and sketch the graph of the polar equation.

(2) Sound coming from what angle ✓ 2 [0, ⇡] is the microphone most sensitiveto? Least sensitive?

Solution. (1) r = �4 cos ✓

r2 = �4r cos ✓

x2 + y2 = �4x

x2 + 4x+ y2 = 0

(x+ 2)2 + y2 = 4

This is a circle of radius 2 and with center at (2, ⇡).

263

(2) We construct a table of values.

x 0 ⇡

6⇡

4⇡

3⇡

22⇡3

3⇡4

5⇡6 ⇡

r �4 �3.46 �2.83 �2 0 2 2.83 3.46 4

From the table, the microphone is most sensitive to sounds coming fromangles ✓ = 0 and ✓ = ⇡, and least sensitive to sound coming from an angle✓ = ⇡

2 . 2


1. Identify (and describe) the graph of each polar equation.

(a) ✓ = 2⇡3

Answer: Line passing through the pole with angle 2⇡3 with respect to the

polar axis

(b) r = �3

Answer: Circle with center at the pole and of radius 3

(c) r = 2 sin ✓

Answer: Circle of radius 1 and with center at (1, ⇡2 )

(d) r = 3 cos ✓

Answer: Circle of radius 1.5 and with center at (1.5, 0)

(e) r = 2 + 2 cos ✓ Answer: A cardioid

264

2. Sketch the graph of each polar equation.

(a) r = �3

(b) r = �2 sin ✓

(c) r = 2 + 2 sin ✓

265

(d) r = 4 cos ✓

3. The sound-pickup capability of a certain brand of microphone is described bythe polar equation

r = 1.5(1 + cos ✓),

where |r| gives the sensitivity of the microphone of a sound coming from anangle ✓ (in radians).

(a) Identify and sketch the graph of the polar equation.

Answer: A cardioid

266

(b) Sound coming from what angle ✓ 2 [0, 2⇡) is the microphone most sensi-tive to? Least sensitive?

Answer: Most sensitive at ✓ = 0; least sensitive at ✓ = ⇡

Exercises 3.9

1. Plot the following points in one polar plane:

A(�2, ⇡2 ) B(1,�7⇡3 ) C(�2, ⇡4 )

D(�3,�2⇡3 ) E(4,�⇡

4 ) F (�3, 7⇡12 )

G(4,�8⇡3 ) H(�2,�11⇡

12 ) I(1, 15⇡2 )

Answer:

267

2. Give the polar coordinates (r, ✓) with indicated properties that represent thesame point as the given polar coordinates.

(a) (�3, 2⇡); r > 0, 0 < ✓ 2⇡ Answer: (3, ⇡)

(b) (10,�4⇡3 ); r < 0, 0 < ✓ 2⇡ Answer: (�10, 5⇡3 )

(c) (�4, 3⇡2 ); r < 0, �2⇡ < ✓ 0 Answer: (�4,�⇡

2 )

(d) (�1,�⇡); r < 0, 0 < ✓ 2⇡ Answer: (�1, ⇡)

(e) (�2, cos�1 23); r > 0, �2⇡ < ✓ 0 Answer: (2,�⇡ + cos�1 2

3)

3. Convert each polar coordinates to Cartesian coordinates.

(a) (4, ⇡) Answer: (�4, 0)

(b) (�4, 7⇡4 ) Answer: (2p2,�2

p2)

(c) (2,�2⇡3 ) Answer: (�1,�

p3)

(d) (�5,�3⇡) Answer: (5, 0)

(e) (8,�11⇡6 ) Answer: (4

p3, 4)

4. Convert each Cartesian coordinates to polar coordinates (r, ✓), where r � 0and 0 ✓2⇡.

(a) (0,�6) Answer: (6, 3⇡2 )

(b) (�5,�5) Answer: (5p2, 5⇡4 )

(c) (�2, 6) Answer: (2p10, ⇡ + tan�1(�3))

(d) (1,�4) Answer: (p17, 2⇡ + tan�1(�4))

(e) (1,�p3) Answer: (2, 5⇡3 )

5. Identify and sketch the graph of each polar equation.

(a) ✓ = �⇡

3

Answer: A line passing through the pole and with angle �⇡

3 with respectto the polar axis

268

(b) r = �3 sin ✓

Answer: A circle tangent to the x-axis with center at (0,�1.5)

(c) r = cos ✓

Answer: A circle tangent to the y-axis with center at (0.5, 0)

(d) r = 2� 2 cos ✓

Answer: A cardioid

269

(e) r = 1 + sin ✓

Answer: A cardioid

6. The graph of the polar equation r = 2 cos 2✓ is a four-petaled rose. Sketch itsgraph.

Answer:

?7. A comet travels on an elliptical orbit that can be described by the polarequation

r =1.164

1 + 0.967 sin ✓

with respect to the sun at the pole. Find the closest distance between the sunand the comet.

Answer: Closest distance occurs when sin ✓ = 1, so r = 1.1641.967 ⇡ 0.59 units.

270

?8. Polar equations are also used by scientists and engineers to model motion ofsatellites orbiting the Earth. One satellite follows the path

r =36210

6� cos ✓,

where r is the distance in kilometers between the center of the Earth and thesatellite, and ✓ is the angular measurement in radians with respect to a fixedpredetermined axis.

(a) At what value of ✓ 2 [0, 2⇡) is the satellite closest to Earth, and what isthe closest distance?

Answer: The satellite is closest to Earth when cos ✓ = �1, and this occurswhen ✓ = ⇡. The closest distance is, therefore, r = 36210

6�(�1) ⇡ 5182.86kilometers.

(b) How far away from Earth can the satellite reach?

Answer: The satellite can reach as far as r = 362106�1 ⇡ 7242 km away from

the Earth.

9. The graph of the polar equation

r =15

3� 2 cos ✓

is a conic section. Identify and find its equation in rectangular coordinatesystem. Answer: Ellipse, 5x2 + 9y2 � 60x� 225 = 0

Solution

r =15

3� 2 cos ✓r(3� 2 cos ✓) = 15

3r � 2r cos ✓ = 15

3p

x2 + y2 � 2r · xr= 15

3p

x2 + y2 = 2x+ 15⇣3p

x2 + y2⌘2

= (2x+ 15)2

9x2 + 9y2 = 4x2 + 60x+ 225

5x2 + 9y2 � 60x� 225 = 0

(x� 6)2

81+

y2

45= 1 an ellipse

10. The graph of the polar equation

r =6

3 + 3 sin ✓

271

is a parabola. Find its equation in rectangular coordinate system.

Answer: y = �14x

2 + 1

11. For what values of ✓ 2 [0, 2⇡) will the graphs of r = 4 cos ✓ and r cos ✓ = 1intersect? Answer: ⇡

3 ,2⇡3 ,

4⇡3 ,

5⇡3

12. Convert the polar equation

r =2 sin 2✓

cos3 ✓ � sin3 ✓

into Cartesian equation. Answer: x3 = y3 + 4xy

4

272

References

[1] R.N. Aufmann, V.C. Barker, and R.D. Nation, College Trigonometry, HoughtonMi✏in Company, 2008.

[2] E.A. Cabral, M.L.A.N. De Las Penas, E.P. De Lara-Tuprio, F.F. Francisco,I.J.L. Garces, R.M. Marcelo, and J.F. Sarmiento, Precalculus, Ateneo deManila University Press, 2010.

[3] R. Larson, Precalculus with Limits, Brooks/Cole, Cengage Learning, 2014.

[4] L. Leithold, College Algebra and Trigonometry, Addison Wesley LongmanInc., 1989, reprinted by Pearson Education Asia Pte. Ltd., 2002.

[5] M.L. Lial, J. Hornsby, and D.I. Schneider, College Algebra and Trigonometryand Precalculus, Addison-Wesley Educational Publisher, Inc., 2001.

[6] J. Stewart, L. Redlin, and S. Watson, Precalculus: Mathematics for Calculus,Brooks/Cole, Cengage Learning, 2012.

[7] M. Sullivan, Algebra & Trigonometry, Pearson Education, Inc., 2012.

[8] C. Young, Algebra and Trigonometry, John Wiley & Sons, Inc., 2013.

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