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Module 2, Section 3

Trigonometric Identities

IntroductionCircular functions is a very lengthy topic. Therefore, it has been splitinto four sections. This section continues the topic of circularfunctions.

Section 3 — OutlineLesson 1 Elementary Identities

Lesson 2 Using Elementary Identities

Lesson 3 Sum and Difference Identities

Lesson 4 Double Angle Identities

Review

Principles of Mathematics 12 Section 3, Introduction 105

Module 2

Notes

106 Section 3, Introduction Principles of Mathematics 12

Module 2

Module 2

Lesson 1Elementary Identities

OutcomesUpon completing this lesson, you will be able to:• state the eight elementary identities• prove identities• find any of the six circular functions, when the value of one

circular function is given• use identities to simplify circular function expressions

OverviewAs circular functions become more complicated, it becomes moredifficult to sketch them and to solve equations involving these morecomplicated functions. However, there are relationships amongstthese functions which enable you to change them to simpler andmore manageable forms. For example, the graph of y = sin3 x + sin xcos2 x becomes the graph of y = sin x which is a much simpler form.These relationships are called identities.

Recall from Module 2, Section 1, the definition of the reciprocalcircular functions which translate into the reciprocal identities.

Reciprocal Identities

Next we have the quotient identities:

= =sin costan cot

cos sinx x

x xx x

( ) ( ) ( )cosecant secant cotangent

1 1 1csc sec cot

sin cos tanx x x

x x x= = =

↓ ↓ ↓

Principles of Mathematics 12 Section 3, Lesson 1 107

We will prove the first quotient identity, and as an activity, you canverify the second.

Once an identity is proven, any expression on one side of the “=” signcan be replaced by the expression on the other side of the sign.

The identities look like equations, but there is a subtle difference. Anequation usually has a fixed number of solutions. For example, theequation x2 – 4 = 0 has the numbers 2 and –2 as its solutions. Anidentity is true for every value in the domain of the expressions.

For example, x2 – 4 = (x – 2)(x + 2) is true for every real number x. Itis an identity.

If x = 5, then:Left Side: x2 – 4 = 52 – 4

= 25 – 4= 21

and if x = 5, then:Right Side: (x – 2)(x + 2) = (5 – 2)(5 + 2)

= 3(7)= 21

=

=

=

= ×

=

=

sinStatement: tancos

Proof: LHS =tan

(by defintion)

sinRHS =cos

LHS

yrxr

xx

xx

yx

xx

y rr xyx

108 Section 3, Lesson 1 Principles of Mathematics 12

Module 2

Note:

LHS = left-hand side

RHS = right-hand side

If x = 6, then:Left Side: x2 – 4 = 62 – 4

= 36 – 4= 32

and if x = 6, then:Right Side: (x – 2)(x + 2) = (6 – 2)(6 + 2)

= 4(8)= 32

No matter what the value of x, the left and right side will always bethe same.

In summary, an equation is true for some numbers, while anidentity is true for all numbers.

Using IdentitiesIf you are asked to sketch a function such as y = cot x sin x, you cansimply change this function to which

is just the graph of the basic y = cos x curve. The purpose of identitiesis to simplify your work!

Another example would be to solve an equation such as sin x tan x csc x – 1 = 0. You can change this equation into the

equation which is much

easier to solve.

There are three additional identities which you will find very useful.They are called the Pythagorean Identities because they resemblethe form of the Pythagorean Theorem.

Since the equation of the unit circle is x2 + y2 = 1 and the x-value on this circle is cos θ and the y-value is sin θ it follows that:

and, consequently, sin2 θ = 1 – cos2

θ and cos2θ = 1 – sin2

θ or byfactoring sin2

θ = (1 – cos θ)(1 + cos θ) and cos2θ =

(1 – sin θ)(1 + sin θ).

It is beneficial if you understand and remember all forms of thisidentity.

2 2cos sin 1θ θ+ =

sin tansin

tanx xx

x11 0 1 0− = − = or

y xx

x x= =cossin

sin cos ,


Module 2

Pythagorean Identity #1

Example 1

Find the value of tan θ if

SolutionMethod 1: Using IdentitiesSince cos2x + sin2x = 1, it follows that:

Since θ ∈ IV, it follows that sin θ < 0.

Therefore,

Method 2: Using Triangles

Using the Pythagorean Theorem the length of the third side is:

tanθ = =oppositesideadjacentside

43

aa

2 3 23 54

+ ==

3

5

θ

a

Furthermore, since tan , it follows that tan .

4sin 45

3cos 35

θ θθθ

= = = −−

sin .θ = −45

35

1

925

25251625

45

22

2

2

FHGIKJ + =

+ =

=

= ±

sin

sin

sin

sin

θ

θ

θ

θ

( )

θ θ ∈

∈

= IV

" IV" means "in Quadrant IV"

3cos and

5.


Module 2

Since θ ∈ IV, it follows that tan θ < 0. Hence, tan θ =

You may use either method. Use the one which is more comfortablefor you.

In order to help you remember the identities and become moreproficient in using them, you will be asked to prove certainrelationships. When proving identities, it is very important that yourealize you must show that the “=” sign belongs between the twosides. You cannot assume that it does and proceed to transposeterms from one side to the other.

Example 2Prove that 2 sin2x – 1 = 1 – 2 cos2x.

SolutionWorking on the left-hand side (LHS), you get:

This is one way to prove an identity: work only on one side of thegiven equality. Alter the expression on that side until it looks justlike the other side. Then you have proved that they are identical.You have your identity.

Example 3Prove that sec2 x csc2 x = sec2 x + csc2 x.

SolutionThis is the other way to prove identities: change both sides toexpressions involving sin x and cos x, simplify both sides until theyare exactly the same, and therefore LHS = RHS. But don’t move anyterms from one side of the “=” sign to the other side!

(Caution: The “=” sign does not exist until you have established thatboth sides are identical, so do not equate the two sides until the veryend.)

2

2

2

2(1 cos ) 1

2 2 cos 1

1 2 cos RHS

xxx

− −

− −− =

− 43

.


Module 2

Since LHS = RHS the identity has been proved.

Some common operations when proving identities are:1. Addition and/or subtraction of rational expressions, e.g.,

L.C.D.

Note: L.C.D. = lowest common denominator

2. Multiplication and/or division of rational expressions, e.g.,

3. Factoring, e.g.,sin3 x + sin x cos2 x = sin x(sin2 x + cos2 x).

Aside: Notice that

(1) the expression in number 1. simplifies to

(2) the expression in number 3. eventually simplifies to (sin x)(1) = sin x.

2 21

andcos sinx x

1 1 1cos sin cos sin

.x x x x

FHGIKJFHGIKJ =

++ = + =2 2 2 2

2 2 2 2 2 2 2 21 1 sin cos sin cos

.cos sin cos sin cos sin cos sin

x x x xx x x x x x x x

x x

x x

x xx x x x

x xx x

x x

⎛ ⎞ ⎛ ⎞+⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+

• •+

+

2 2

2 2

2 2

2 2 2 2

2 2

2 2

2 2

1 1RHS=

cos sin

1 1cos sin

1 sin cos 1cos sin cos sin

sin coscos sin

1cos sin

LHS= FHGIKJFHGIKJ

1 1

1

2 2

2 2

cos sin

cos sin

x x

x x


Module 2

The remaining two Pythagorean Identities are found by dividing bycos2x or by sin2x.

Example 4Find cos x if cot x = 2 and sin x < 0.

SolutionMethod 1: Using Triangles

Since the adjacent and opposite sides of the right

triangle are 2 and 1, respectively. You can use the PythagoreanTheorem to find the third side.

Since cot x > 0 (which means tan x > 0 → QI or QIII) and sin x < 0 (QIII or QIV), so x must terminate in QIII where cos x < 0.

= − = −adjacent 2Therefore, cos .hypotenuse 5

x

hh

h

2 2 1

2

2 1

4 1

5

= += +

=

2cot 2 ,

1x = =

= =2 adjacentGiven cot1 opposite

x1

2θ

h

cos2 x + sin2 x = 1

coscos

sincos cos

2

2

2

2 2

1xx

xx x

+ =cossin

sinsin sin

2

2

2

2 2

1xx

xx x

+ =

1 2 2+ =tan secx x cot csc2 21x x+ =

(dividing by cos2 x) (dividing by sin2 x)

Pythagorean Identity #2 Pythagorean Identity #3


Module 2

S A

T C

= cot x > 0= sin x < 0


Module 2

Method 2: Using Identities

Therefore

2 2

22 2 2

III, it follows that

Since sin 0, its reciprocal csc 0. it fol lows that1csc 5 and sin . Furthermore, cos sin 1,5

1 1 5 4so cos 1 or cos or cos .5 5 55

Since terminates in

x x

x x x x

x x x

x

< <

= − = − + =

−⎛ ⎞+ = + = =⎜ ⎟⎝ ⎠

cos .25

x = −

2 2

2 2

cot 1 csc

2 1 csc

5 csc

x xx

x

+ =+ =

± =

Guided Practice1. In which quadrant does θ terminate if:

2. Find the exact values of the other remaining circular functions if:

3. Prove that each of the following is an identity:

2 2 2

4 4 2

2

2 2 2 2

22

2

a) cos sec 1 b) csc sin 1c) tan cot 1 d) cot sin cos

sece) tan cos sin f) tan

csc

g) cos sin 1 2 sin

h) cos sin 1 2 sin1 1

i) 2 sec1 sin 1 sin

j) sec csc tan cot

csc 1k) cos

cscl) sec

x x x xx x x x x

xx x x xx

x x xx x x

θθ θ

θ θ θ θθ θθ

θ

= == =

= =

− = −

− = −

+ =− +

− = −

− =

2 3

22

2

tan sec sec1

m) sin costan cot

1 cosn) 2 csc 1

sin

x xx x

x xx

θ θ θ+ =

=+

+ = −

5a) sin and tan 013

3b) cos and csc 05

c) tan 5 and cos 0d) csc 2

13e) sec and tan 05

8f) cot 15

x x

x x

x

θ θ

θ θθ

= <

= − <

= <=

= >

= −

a) sin 0 and cos 0b) sec 0 and tan 0c) cos 0 and cot > 0d) csc 0 and tan < 0

θ θθ θθ θθ θ

> <> <<<


Module 2


Module 2

4. Simplify each expression as much as possible.

Check your answers in the Module 2 Answer Key.

a) b)

c) d)

e) f)

sin cos csccsc

cotcos csc

sin cos csccossin

sin

sin cos coscot

csc

x x xx

xx x

x x x xx

x

x x x xx

+

+

+ +

2

2

2 32

2

1

Lesson 2

Using Elementary Identities

OutcomesUpon completing this lesson, you will be able to:• solve equations involving circular functions• prove identities

OverviewThis lesson provides more practice with the elementary identitieswhich you learned in the previous lesson.

Example 1Solve tan x = 2 sin x where 0 ≤ x ≤ 2π.

Solution

Solution

π ππ 50, ,

3 3

x xx xx xx x xx x xx xx or xx x

x

x

π

π π

=

=≠

=− =− =

∴ = − == − = −

=

=

If tan 2sin(Using quotient identity #1, sin 2 sin

cos provided that cos 0)sin 2sin cossin 2 sin cos 0sin (1 2 cos ) 0sin 0 1 2cos 0

0, 2cos 1(Remember the quadrantals?) cos 0.5

5,3 3

(Special angles)


Module 2

Example 2Solve sin2 x + cos x = 1 where 0° ≤ x < 360°.

Solution

Solution

Note: This next equation is somewhat tricky. Since sin2 x = 1 – cos2 x,it follows that is a radical. Similarly,

is also a radical. Therefore, some equationswhich involve sums and differences of linear expressions of sine andcosine functions are frequently solved by squaring both sides untilthe “radicals” are eliminated. However, since both sides have beensquared, extraneous solutions may have been introduced. Therefore,the solutions must be checked.

Example 3

Solution

The terms cos x and act just like radicals. Therefore, begin by squaring both sides.

( ) ( )

( )

( )( )( )

x x

x x x

x x x

x x xx x

x x

x x

22

2 2

2 2

2 2

2

2

cos 1 3 sin

cos 2 cos 1 3 sin

cos 2 cos 1 3 1 cos

cos 2 cos 1 3 3cos

4 cos 2 cos 2 0

2 2 cos cos 1 0

2 2 cos 1 cos 1 0

+ =

+ + =

+ + = −

+ + = −+ − =

+ − =

− + =

3 sin x

Solve where cos sin , .x x x+ = ≤ <1 3 0 2π

cos sinx x= ± −1 2

2sin 1 cosx x= ± −

° ° °0 , 90 , 270

2

2

2

If sin cos 1

1 cos cos 1

0 cos cos0 cos (cos 1)

cos 0 or cos 190 , 270 , 0

+ =

− + =

= −= −

∴ = == ° ° = °

x xx xx x

x xx xx x


Module 2

Solution

Check:3

1 3LHS: cos 1 13 2 2

3 3RHS: 3 sin 33 2 2

LHS = RHS, therefore, is a true root.3

5Check:3

5 1 3LHS: cos 1 13 2 2

5 3 3RHS: 3 sin 33 2 2

5LHS RHS; therefore, is an extraneous roo t.3

Check:

π

π

π

π

π

π

π

π

+ = + =

⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠

+ = + =

⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎝ ⎠

≠

LHS: cos 1 1 1 0

3 sin 3(0) 0

LHS = RHS; therefore, is a root. The so lution is , .3

x

ππ

πππ π

+ = − + =

= =

=

π ππ 5, ,

3 3

π π π

∴ = = −

=

= =

2 cos 1 or cos 11cos2

5, or 3 3

x x

x

x x


Module 2

It’s only fair to tell you a (sometimes) simpler way to solve theseproblems: just graph them on your graphing calculator and find the zeros! You’ll get a decimal answer instead of an exact one like

most cases:

1. Divide the long decimal answer by π (use the 2nd π key).

2. If necessary, use the MATH 1:Frac function to turn theremaining decimal into the exact fraction.

Example: after graphing a trig equation, you find that oneroot is 4.71238898.

1. Divide by π. (To save time, use Ans ÷ 2nd π .) You get1.5. Therefore, the exact solution must have been 1.5π, or

2. If needed, you could have entered Ans MATH 1:Frac to

3see as the true fraction.

2

π3.

2

π3, but the following technique will yield the exact answer in

4


Module 2

Guided Practice1. Solve each equation for x where 0° ≤ x < 360°.

2. Solve each equation for x where 0 ≤ x < 2π.

3. Solve each equation for x where 0 ≤ x < 2π.

4. Test the following identity by substituting the stated values for αand β: (We will prove this identity in the next lesson.)Difference identity: cos (α – β) = cos α cos β + sin α sin β

5. Assume that the identity in Question 4,cos (α – β) = cos α cos β + sin α sin β, is valid. Use this identity toprove part (a) below. After part (a) has been proved, use it alongwith your knowledge of the other identities to prove the followingnew identities:

Incidentally, you demonstrated a) and b) graphically in the Section 2Review.

6. Why do you think the above identities are called co-functionidentities?

a) b)

c) d)

e) f)

cos sin sin cos

tan cot sec csc

csc sec cot tan

π π

π2 2

290

90 90

−FHGIKJ = −FHG

IKJ =

−FHGIKJ = °− =

°− = °− =

x x x x

x x x x

x x x x

b gb g b g

α β α βπα π β

= ° = ° = ° = °

= =

a) 90 , 30 b) 150 , 30

c) ,4

+ = + =− =

a) sin cos 1 b) sin cos 0c) sin 2 cos

x x x xx x

x x x

x π− = − =

⎛ ⎞− =⎜ ⎟⎝ ⎠

2 2

2

a) 2 cos 1 sin b) 4 cos 3

c) 4 sin 12

x x x xx x x

+ = ==

2 2a) tan sec 3 b) sin 2 cos 2c) tan sin tan


Module 2

7. Use the co-function and symmetry identities to simplify thefollowing:

8. Use the new co-function and/or symmetry identities to solve thefollowing equations over the set of real numbers:


a) cos 12

b) sin 12

c) sin 02

π

π

π

⎛ ⎞− =⎜ ⎟⎝ ⎠⎛ ⎞− =⎜ ⎟⎝ ⎠⎛ ⎞− =⎜ ⎟⎝ ⎠

x

x

x

2

a) sin b) tan2 2

c) cos d) sin ( )cot ( )2

1e) cos csc ( ) f)2 sec

2

x x

x x x

x xx

π π

π

ππ

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞+ − −⎜ ⎟⎝ ⎠⎛ ⎞− −⎜ ⎟ ⎛ ⎞⎝ ⎠ −⎜ ⎟⎝ ⎠


Module 2


Module 2

Lesson 3Sum and Difference Identities

OutcomesUpon completing this lesson, you will be able to:• derive the sum and difference identities• simplify expressions using identities• evaluate expressions using identities• prove identities

OverviewThe difference identity given to you in the last exercise will be calledthe Basic Difference Identity. You will prove it in this lesson, and useit to prove other sum and difference identities. You will then useyour enlarged set of identities to evaluate and simplify expressions.You will also prove some additional identities.

Basic Difference Identitycos (α – β) = cos α cos β + sin α sin β

Proof:

Transferring an arc to Standard Position

x

y

�

α β

=[cos (α−β), sin (α−β)]

P(0)=(1,0)

P(α−β)

–�

x

y

�

�

α

P(

β

)

P(α)

β

=(cos α, sin α)

=(cos β, sin β)


The coordinates of P(α) are (cos α, sin α) and the coordinates of P(β)are (cos β, sin β).

Transfer the arc joining the points P(β) and P(α) to a standardposition arc, to create the point P(α – β). The coordinates of P(α – β) are: (cos(α – β), sin(α – β)) since the length of the arc fromP(0) to P(α – β) is (α – β). See the figure on the previous page.

Since both arcs are the same length it follows that the chords d1 andd2 are the same length.

d1 and d2

Now express d1 and d2 using the Distance Formula for the distancebetween the end points:

Example 1Use the above identity and the fact that:

SolutionUsing the basic difference identify, cos (α – β) = cos α cos β + sin α sinβ, you get:

cos cos cos cos sin sin12 3 4 3 4 3 4

1 1 3 1 1 32 22 2 2 2

π π π π π π π⎛ ⎞ ⎛ ⎞= − = +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠+⎛ ⎞ ⎛ ⎞= + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

π π π π⎛ ⎞= − ⎜ ⎟⎝ ⎠to find cos exactly.

12 3 4 12

x

�[cos (α−β), sin (α−β)]

(1,0)�

d2

x

�

�(cos α, sin α)

(cos β, sin β)d1

2 2 2 2

2 2 2 2

2 2

(cos cos ) (sin sin ) [cos( ) 1] [sin( ) 0]

cos 2cos cos cos sin 2sin sin sin

cos ( ) 2cos( ) 1 sin ( )2 2cos cos 2sin sin 2 2cos( )

2cos cos 2sin sin 2cos( )cos cos

α β α β α β α β

α α β β α α β βα β α β α β

α β α β α βα β α β α βα

− + − = − − + − −

− + + − + =− − − + + −− − = − −− − = − −

sin sin cos( )β α β α β+ = −

(square both sides)

(since sin2A + cos2A = 1)

You can use the Basic Difference Identity, along with the co-functionand symmetry identities, to prove the following set of sum anddifference identities:

1.

Proof: Expanding the left side of the equation,

2.

Proof:

3.

Proof:

Now you are capable of finding the coordinates of any point on theunit circle that is a sum or a difference of the special arc lengths,without using a calculator.

sin( ) sin( ( ))sin cos( ) cos sin( )sin cos cos [ sin ]sin cos cos sin

α β α βα β α βα β α βα β α β

+ = − −= − − −= − −= +

sin(α + β) = sin α cos β + cos α sin β

Because sin and cos aresin ( ) cos ( )

2 co-functions. (Lesson 2)

cos2

cos cos sin sin2 2

sin cos cos sin

πα β α β

π α β

π πα β α β

α β α β

⎡ ⎤− = − −⎢ ⎥⎣ ⎦⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

sin(α – β) = sin α cos β – cos α sin β

( ) ( )( ) ( ) (

( ) ( )( )

cos cos cos cos sin sin

The last expansion used the Basic Differ ence Identity.Since sine is odd and cosine is even, it follows that

cos cos sin sin

=cos cos sin sin

cos cos si

α β α β α β α β

α β α β

α β α βα β

+ = − − = − + −

− + −

+ −= − n sin .α β

cos(α + β) = cos α cos β – sin α sin β


Module 2

Example 2

Find the coordinates of

Solution

The x-coordinate is as shown in Example 1.

To find the y-coordinate, you use the identity sin (α – β) = sin α cos β – cos α sin β.

Example 3

a)

Note: QII = means second quadrant

b) In what quadrant does the angle (a + β) terminate?

Solutiona) To find the coordinates of P(a + β), you must find the values of

cos(a + β) and sin(a + β).2

2

22

3 3Since sin , it follows that cos =1. 5 5

4But lies in the second quadrant, so cos . 5

5 5Also, cos and tan 0; therefore, sin 1 13 13

12yielding sin . Because sin is positi13

α α

α α

β β β

β β

⎛ ⎞= + ⎜ ⎟⎝ ⎠

= −

⎛ ⎞= > + =⎜ ⎟⎝ ⎠

= ve and so is tan

(given), is in the first quadrant.

β

β

3 5If sin , with QII, and cos , with tan 0,

5 13find the coordinates of P( ).

α α β β

α β

= ∈ = >

+

1 3 3 1 2 6 6 2The coordinates of P , or , .

12 4 42 2 2 2π ⎛ ⎞ ⎛ ⎞+ − + −⎛ ⎞ = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

sin sin cos cos sinπ π π π π π3 4 3 4 3 4

32

12

12

12

3 12 2

−FHGIKJ = − = F

HGIKJ −FHGIKJ =

−

1 32 2+

Pπ12FHGIKJ.


Module 2

b) Since cos (α + β) < 0 and sin(α + β) < 0, it follows that (α + β)terminates in the third quadrant.

α β α β α β

α β α β α β

α β

+ = −− −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − − = = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

+ = +−⎛ ⎞ ⎛ ⎞⎛ ⎞= + − = = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞+ − −⎜⎝ ⎠

cos( ) cos cos sin sin

4 5 3 12 20 36 56and

5 13 5 13 65 65sin( ) sin cos cos sin

3 5 4 12 15 48 335 13 5 13 65 65

56 33The coordinates of P( ) are ,

65 65 ⎟.


Module 2

Guided Practice1. Without using a calculator, evaluate:

2. Determine the exactvalue of

in the relevant identity.

3. Find the coordinates of

4. Use identities to find the exact value of:a) cos 105°b) sin 105°c) tan 105°(Hint: Express 105° as a sum or difference of two special angles.)

5. Given that and neither P(α) nor

P(β) are in the first quadrant, find:

9. Express as a function, or functions, of only θ.cos

3π θ⎛ ⎞+⎜ ⎟⎝ ⎠

( ) ( )

( ) ( )

( )

a) sin b) cos

c) tan d) sec

e) the quadrant in which terminates

+ +

+ +

+

α β α β

α β α β

α β

7 9sin and cos

25 41α β= =

P712πFHGIKJ.

3πβ =

7sin bysetting and

12 4π πα⎛ ⎞ =⎜ ⎟⎝ ⎠

5 5a) sin cos cos sin9 36 9 36

2 3 2 3b) cos sin sin cos5 5 5 5

7 7c) cos cos sin sin12 4 12 4

5 5d) cos cos sin sin9 9 9 9

π π π π

π π π π

π π π π

π π π π

+

+

+

−


Module 2

6. Express in terms of only θ.

7. Prove each of the following identities. (Note: (b) is Extra for Experts.)

8. Use identities to simplify to a single circular function.


2 2

a) sin 2 cos 3 cos 2 sin 3

b) sin cos 6 cos sin 6

c) cos 3 sin 3

d) 2 sin 5 cos 5

e) cos 2 cos 3 sin 2 sin 3

α α α α

α α

+

+

−

+

x x x x

x x

a a a a

( )

( ) ( ) 2 2

a) sin sin ( ) 2 sin cos

b) cos cos cos sinx y x y x y

+ + − =

+ − = −

α β α β α β

πθ⎛ ⎞−⎜ ⎟⎝ ⎠tan

6


Module 2

Notes


Module 2

Lesson 4Double Angle Identities

OutcomesUpon completing this lesson, you will be able to:• derive the double angle identities• simplify expressions using identities• evaluate expressions using identities• prove identities

OverviewWe will now prove identities for sin 2α, cos 2α, and tan 2α. Theseare called double angle identities for obvious reasons.

1.

There are three different expressions for cos 2α . . . .

2. a) 2 2

2 2

cos 2 cos – sin

Proof: LHS cos( )

cos cos sin sin

(using the identity for cos( )

cos – sin

RHS

α α α

α α

α α α α

α β

α α

=

= +

= −

+

=

=

α α α

α α

α α α α

α α

=

= +

= +

=

=

sin 2 2 sin cos

Proof: LHS sin( )

sin cos cos sin

2sin cos

RHS


Module 2

b)

c)

And, finally, the Double Angle Identify for tan.

Record these five identities and have them handy for future use.

Example 1Use the double angle identities to change the following into functionsof angles which are half the given angles:a) sin 2A b) sin 4Ac) sin 10x d) cos 4α

2 2

2

2

sin 2 2 sin cos

cos sin

cos 2 2 cos 1

1 2 sin

α α α

α αα α

α

=

⎧ −⎪

= −⎨⎪ −⎩

α α

α

α α

α α

α

=

=

= −

= − −

=

=

2

2 2

2 2

2

cos2 1 – 2sin

Proof: LHS cos2

cos sin [proven in (a)]

1 sin sin [Pythagorean Identity #1]

1 – 2sin

RHS

α α

α

α α

α α

α α

α

=

=

= −

= − −

= +

=

=

2

2 2

2 2

2 2

2

cos2 2cos – 1

Proof: LHS cos2

cos sin [proven in (a)]

cos (1 cos ) [Pythagorean Identity #1]

cos – 1 cos

2cos – 1

RHS


Module 2

Solutiona) sin 2A = 2 sin A cos A

b) sin 4A = 2 sin 2A cos 2ANotice that the coefficient of 2 is not required. To use thisidentity, all you need is one angle to be twice as large as the other— that’s all!

c) sin 10x = 2 sin 5x cos 5xAgain, notice that 10x = 2(5x).

d) Any one of the three following choices:

Example 2On the unit circle, the coordinates of the circular point P(θ) are(0.588, 0.809). Find the coordinates of P(2θ).

SolutionThe coordinates of P(2θ) will be (cos 2θ, sin 2θ).cos 2θ = 2 cos2

θ – 1 = 2(0.588)2 – 1 = – 0.309 sin 2θ = 2 sin θ cos θ = 2(0.809)(0.588) = 0.951Therefore, the coordinates of P(2θ) = (– 0.309, 0.951).

P(2θ)

P(θ) = (0.588, 0.809)

x

y

2 2

2

2

cos (2 ) sin (2 )

cos 4 2 cos (2 ) 1

1 2 sin (2 )

α αα α

α

⎧ −⎪

= −⎨⎪ −⎩


Module 2

Guided Practice1. Write sin 4A in terms of functions of A.

2. In Example 2, you found the coordinates of P(2θ). What would bethe coordinates of P(4θ)?

3. You will have to think about this one, but, using the coordinatesof P(θ) in Example 2, try to find the coordinates

of

4. Write each of the following in terms of only one circular function:

5. Use a double angle identityto find

6.

7. If cos θ = p, find the value of cos 4θ in terms of p.

8. Given that and neither P(α) nor

P(β) are in the first quadrant, find:

9. Write each of the following in terms of only one circular functionor a constant.

2a) (sin cos )

sin 2b)

cos

+x x

xx

a) sin 2 b) cos ( )

c) cos 2 d) sin ( )

α α β

β α β

−

+

sin cos ,α β= =45

513

and

If and find the exact values of

and

02

35

2

2

< < =θ π θ θ

θ

sin , sin

cos .

cos if cos 0.95.5 10π π⎛ ⎞ ⎛ ⎞ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2

2

2

a) cos 4 sin 4 b) 2 sin 2 cos 2

c) 4 sin cos d) 1 2 sin 5

e) 2 cos 10 1 f) sin cos

x x x x

x x α

α α α

−

−

−

Pθ2FHGIKJ.


Module 2

10. Prove the following identities:


Review Section 3 before attempting the review questions followingthe summary on the next page. These questions should help youconsolidate your knowledge as you prepare for Section Assignment2.3.

4 4

2

2

3

3

a) tan cot 2 csc 2

b) cos sin cos 22 sin(c) secsin 2

secd) sec 22 sec

e) sin 3 3 sin 4 sin

f) cos3 4cos 3cos

x x x

x x xx xxx x

xx x x

x x x

+ =

− =

=

=−

= −

= −


Module 2

Notes:


Module 2

Summary of Section 3

You have been introduced to many trigonometric identities in thissection. You do not need to remember all of them, but if you do, yourtask of using them quickly and effectively will become much easier.At least you should know and quickly recognize the Pythagorean,Reciprocal, and Double Angle Identities.

Pythagoreansin2

θ + cos2 θ = 1

1 + tan2θ = sec2

θ1 + cot2

θ = csc2 θ

Remember that the identity sin2 θ + cos2

θ = 1 is often written as sin2

θ = 1 – cos2 θ, or cos2

θ = 1 – sin2 θ.

Reciprocal and Rational IdentitiesSin2

θ and cos2 θ can be used to define other trigonometric functions.

You use the reciprocal and rational identities to prove identitiesinvolving tan, cot, sec, and csc by changing everything to sin and cos.

Double Angle Identities

2 2

2

2

sin2 2sin cos

cos2 cos sin

2cos 1

1 2sin

θ θ θ

θ θ θ

θ

θ

=

= −

= −

= −

sintancos

1 coscottan sin

1seccos

1cscsin

θθθ

θθθ θ

θθ

θθ

=

= =

=

=

Principles of Mathematics 12 Section 3, Summary 137

Module 2

These identities are more important than the Sum and DifferenceIdentities they are derived from. You also need to be able torecognize composition of functions with double angle identities.

e.g., sin 4θ = 2 sin 2θ cos 2θ

(4θ = 2(2θ))

cos 6θ = 2 cos2 3θ – 1

Sum and Difference Identitiessin(α + β) = sin α cos β + cos α sin β

sin(α – β) = sin α cos β – cos α sin β

cos (α + β) = cos α cos β – sin α sin β

cos (α – β) = cos α cos β + sin α sin β

Simplifying Expressions and Proving IdentitiesYou need to have at your fingertips the identities AND thetechniques of algebra, such as dividing fractions, factoring(especially difference of squares), and simplifying complex fractions.

138 Section 3, Summary Principles of Mathematics 12

Module 2

Module 2, Section 3

Review

1. Simplify each expression as much as possible.

2. Prove each identity.

3. Find the exact values of the remaining circular functions if:

5a) sin and tan 013

3b) cos and csc 05

3c) tan and cos 04

8d) cot15

x x

x

θ θ

θ θ

= − <

= <

= − <

=

2 2 2

4 4 2

2

2 2 2 2

2 2

2 2

a) cos 2 sin 1 sin

b) cos sin 2 cos 1

1 1c) 2 tan 2

1 sin 1 sin

d) sec cot tan csc

e) sin 2 sin cos sin cos 2 sin

f) cos 2 sin 2 cos 4

x x x

x x x

x

x x x x x x

x x x

θ θ

θ θ θ θ

+ = +

− = −

+ = +− +

+ = +

− =

= +

3 2

a) sin cos csc tan

b) sin cos sin

c) sin 2 sec

d) sin cos( ) sin( ) cos2 2

e) 4 sin 4 cos 4

f) sin 5 cos 2 cos 5 sin 2

x x x x

x x x

x x

x x x x

x x

x x x x

π π

+

⎛ ⎞ ⎛ ⎞− − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−

Principles of Mathematics 12 Section 3, Review 139

Module 2

4.

5. Solve each equation for x, where 0° ≤ x < 360°.

6. Without the use of a calculator, find the exact value of:

Extra for Experts

7. Sketch y = cos x + sin x. State the domain, range, amplitude, y-intercept, and period.


Now do the section assignment which follows this section. When it iscomplete, send it in for marking.

a) sin12

b) cos8

π

π

a) sin 2 sin

b) sin 2 cos 0

=

+ =

x x

x x

a) sin( )b) cos( )

c) sin 2

d) the quadrant in which ( ) terminates

α βα β

α

α β

++

+

3 5Given that sin , and cos and neither P( ) nor5 13

P( ) are in the first quadrant, find:

α β α

β

= =

140 Section 3, Review Principles of Mathematics 12

Module 2

Student's StudentName ____________________________ No. _______________________

Address ____________________________ School _______________________

____________________________Student's

____________________________ FAX No. _______________________

Instructor'sName ____________________________Date Sent _______________________

PRINCIPLES OF MATHEMATICS 12

Section Assignment 2.3

Your instructor will fill in this box with your percentage mark and lettergrade on the Section Assignment:

Comments/Questions:

Principles of Mathematics 12 Section Assignment 2.3 141

Version 05 Module 2


Address ____________________________ School _______________________

____________________________Student's

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Version 05 Module 2

142 Section Assignment 2.3 Principles of Mathematics 12

Module 2


Trigonometric Identities

1. If find the exact value of sin θ.

2. State the amplitude of the curve with equationy = 4 sin(2x – π) + 5.

3. Write the expression in terms of only one

circular function.( )2 2

2 2

cos sintan 1 sin

x xx x−

tan cos ,θ θ= <713

0 and


Module 2

Total Value: 53 marks(Mark values in margins)

(3)

(2)

(4)

4. Solve each equation algebraically for x where 0 ≤ x < 2π.

b) sin cos 1x x+ =

a) sin2 cosx x=


Module 2

(5)

(6)

2d) sin cos2

x xπ − =

c) sin 3 cosx x=


Module 2

(4)

(6)

5.

b) cos 2βb g

a) sin α β−b g

Given: and < <32

and and

Find the exact value of:

sin sin

.

α π α π β

π β π

= − =

< <

45

1213

2


Module 2

(6)

6. Prove the identities:

7. Find the exact value of cos 75°.

b) sin sin cos sinx y x y x y+ − − =b g b g 2

a)1

12

+= −

sinsec

tancosx

x xx


Module 2

(5)

(4)

(3)

8.

A math teacher, I. M. Wright, told the class that theequation for the above graph was:

Brent, the whiz kid, piped up and said the equation shouldbe written as:

Kathy quickly replied that the equation should be written as:

4cos 2 14

y xπ = − +

4cos 2 12

y xπ = − +

4cos 2 12

y xπ = − + +

x

y

�

�

�

�

� (π, 1)(0, 1)

( , 3)3π 4

( , 5)π4

–


Module 2

a) Who of the above is/are correct? Justify your answer.

b) Wayne said, “I can write an equation for this graph involvingthe sine function.” Write such a possible equation.

Send in this work as soon as you complete this section.


Module 2

(3)

(2)

Total: 53 marks

Notes


Module 2

trigonometric identitiesmedia.openschool.bc.ca/ocrmedia/pmath12/pdf/mod2_section3.pdf · the...

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