trigonometric functions problem solving approach

TRIGONOMETRICFUNCTIONS(Problem Solving Approach)

A. PanchishkinE. Shavgulidze

MirPublishersMoscow

TRIGONOMETRIC FUNCTIONS

A. A. Ilaasmmraa, E. T. Illasrvnanse

TPHI'OHO~1ETPl1qECRHE

A.PanchishkinE.Shavgulidze

TRIGONOMETRICFUNCTIONS(Problem-SolvingApproach)

MirPublishersMoscow

Translated from Russianby Leonid Levant

Pirst published 1988Revised from the 1986 Russian edition

Ha aH3AUUCnOM, R8bl1e

Printed in the Unton of Soviet Socialtst Republics

ISBN 5-03-000222-7 HSJJ;aTeJlhCTBO HayKa&, I'aaananpeAaK~HJI cPH8uRo-MaTeMaTBQeCROHaareparypra, 1986

Engl ish translation, Mir Publishers,1988

From the Authors

By tradition, trigonometry is an important componentof mathematics courses at high school, and trigonometryquestions are always set at oral and written examina-tions to those entering universi ties, engineering colleges,and teacher-training insti tutes.

The aim of this study aid is to help the student to mas-ter the basic techniques of solving difficult problems intrigonometry using appropriate definitions and theorem.sfrom the school course of mathematics. To present thematerial in a smooth way, we have enriched the textwith some theoretical material from the textbook Algebraand Fundamentals of Analysis edited by AcademicianA, N. Kolmogorov and an experimental textbook of thesame title by Professors N.Ya. Vilenkin, A.G. Mordko-vich, and V.K. Smyshlyaev, focussing our attention onthe application of theory to solution of problems. Thatis WIlY our book contains many worked competitionproblems and also some problems to be solved independ-ently (they are given at the end of each chapter, theanswers being at the end of the book).

Some of the general material is taken from ElementaryMathematics by Professors G.V. Dorofeev, M.I(. Potapov,and N.Kh. Rozov (Mir Publishers, Moscow, 1982), whichis one of the best study aids on mathematics for pre-college students.

We should like to note here that geometrical problemswhich can be solved trigonometrically and problemsinvolving integrals with trigonometric functions arenot considered.

At present, there are several problem books on mathe-matics (trigonometry included) for those preparing topass their entrance examinations (for instance, Problems

6 From the Authors

at Entrance Examinations in Mathematics by Yu.V. Nes-terenko, S.N. Olekhnik, and M.K. Potapov (Moscow,Nauka, 1983); A Collection of Competition Problems inMathematics with Hints and Solutions edited by A.I. Pri-lepko (Moscow, Nauka, 1986); A Collection of Problems inMathematics tor Pre-college Students edited by A.I. Pri-lepko (Moscow, Vysshaya Shkola, 1983); A Collection ofCompetition Problems in Mathematics for Those EnteringEngineering Institutes edited by M.I. Skanavi (Moscow,Vysshaya Shkola, 1980). Some problems have been bor-rowed from these for our study aid and we are gratefulto their authors for the permission to use them.

The beginning of a solution to a worked example ismarked by the symbol ~ and its end by the symbol ~.The symbol ~ indicates the end of the proof of a state-ment.

Our book is intended for high-school and pre-collegestudents. We also hope that it will be helpful for theschool children studying at the "smaller" mechanico-mathematical faculty of Moscow State University.

From the Authors

Contents

5Chapter 1. Definitions and Basic Properties of Trigono-

metric Functions 9

1.1. Radian Measure of an Arc. TrigonometricCircle . 9

1.2. Definitions of the Basic Trigonometric Func-tions . '18

1.3. Basic Properties of Trigonometric Functions 231.4. Solving the Simplest Trigonometric Equations.

Inverse Trigonometric Functions 31Problems 36

Chapter 2. Identical Transformations of TrigonometricEx pressions 4t

2.1. Addition Formulas 412.2. Trigonometric Identities for .Double, Triple,

and Half Arguments 552.3. Solution of Problems Involving Trigonometric

Transformat ions 63Problems 77

Chapter 3. Trigonometric Equations and Systems ofEquations 80

3.1. General 803.2. Principal Methods of Solving Trigonometric

Equations 873.3. Solving 'I'rigonometric Equations and Systems

of Equations in Several Un knowns 101Problems tog

Chapter 4. Investigating Tri.gonometric Functions 1Vl

4.1. Graphs of Basic Trigonometric Functions 1134.2. Computing Limits 126

8 Contents

4.3. Investigating Trigonometric Functions withthe Aid of a Derivative 132

Problems 146

Chapter 5. Trigonometric Inequalities 1495.1. Proving Inequalities Involving Trigonometric

Functions 14~5.2. Solving Trigonometric Inequalities 156Problems 162

Answers 163

Chapter I

Definitions and Basic Propertiesof Trigonometric Functions

1.1. Radian Measure of an Arc. Trigonometric Circle1. The first thing the student should have in mind whenstudying trigonometric functions consists in that thearguments of these functions are real numbers. The pre-college student is sometimes afraid of expressions suchas sin 1, cos 15 (but not sin 10, cos 15), cos (sin 1), and socannot answer simple questions whose answer becomesobvious if the sense of these expressions is understood.

When teaching a school course of geometry, trigonomet-ric functions are first introduced as functions of an angle(even only of an acute angle). In the subsequent study,the notion of trigonometric function is generalized whenfunctions of an arc are considered. Here the study is notconfined to the arcs enclosed within the limits of onecomplete revolution, that is, from 0 to 360; the studentis encountered with arcs whose measure is expressed byany number of degrees, both positive and negative. Thenext essential step consists in thatthe degree (or sexage-simal) measure is converted to a more natural radianmeasure. Indeed, the division of a complete revolutioninto 360 parts (degrees) is done by tradition (the divisioninto other number of parts, say into 100 parts, is alsoused). Radian measure of angles is based on measuringthe length of arcs of a circle. Here, the unit of measure-ment is one radian which is defined as a central anglesubtended in a circle by an arc whose length is equal tothe radius of the circle. Thus, the radian measure of anangle is the ratio of the arc it subtends to the radius ofthe circle i.n which it is the central angle; also calledcircular measure. Since the circumference of :1 circle ofa unit radius is equal to 2n, the length of the arc of 3600is equal to 2n radians. Consequently, to 180 there corre-spond 1t ra-dians. To change from degrees to radians and

10 1. Properties of Trigonometric Functions

vice versa, it suffices to remember that the relation be-tween the degree and radian measures of an arc is ofproportional nature.

Example 1.1.1. How IUUUy degrees are contained inthe arc of ono radian'?

~We wri te the proportion:If Jt radians ~ 180~\

and 1 radian == x,

then x=-=- 18HO ~ 57.29578 or 5717'44.S". ~

Jt

Example 1.1.2. How many degrees are contained inI f 35n di .,t ie arc O' 12 ra iansr

If n radians === 1800 ,

d 351t dian -- ra lans ~ x12 '

then x = ( ~~Jt_. 1800 ) 111:= 5250 ~Example 1.1.3. What is the radian measure of the arc

of 1U84?If Jt radians ~ 1800 ,

and y radians == 19840 ,1t 1984 496Jt 1

then y==: 180 ==~~1145Jl ~2. Trigonometric Circle. When considering ei ther the

degree or the radian measure of an arc, it is of importanceto know how to take into account the direction in whichthe arc is traced from the i ni tial pointAl to the terminalpoint A 2 The direction of tracing the arc anticlockwise isusually said to he positive (see Fig. 1a), while the direc-tion of tracing the are clockwise is said to be negative(Fig. 1h).

\Ve should. like to recall that a circle of unit radius witha given reference point and positive direction is called thetrigonometric (or coordinate) circle.

1.1. Radian Measure of an Arc 11

ba

Usually, the right-hand end point of the horizontaldiameter is chosen as the reference point. We arrangethe trigonometric circle on a coordinate plane wi th the

A2

Fig. 1

Fig. 2

IJ({l-/)

os = {(x, y): x2 -1- y2 = 1}.3. Winding the Real

Axis on the TrlgonometrleCircle. In the theory oftrigonometric functionsthe fundamental role isplayed by the mappingP: R ~ S of the set R of real numbers on the coordi-nate circle which is constructed as follows:

(1) The number t === on the real axis is associatedwith the point A: A == Po.

(2) If t > 0, then, on the trigonometric circle, weconsider the arc AP b taking the point A == Po as theintitial poin t of the arc and tracing the path of length t

rectangular Cartesian coordinate system introduced(Fig. 2), placing the centre of the circle into the origin.1"hen the reference point has the coordinates (1, 0). Wedenote: A === A (1,0). Also, let B, C, D denote the pointsB (0, 1), C (- ~, 0), yD (0, - ~), respec~lvel.y. B(OII)

The trigonometrlc err-cle will be denoted by S.According to the afore-said,


round the circle in the positive direction. We denote theterminal point of this path by P t and associate the nUID-bar t with the point P t on the trigonometric circle. Orin other words: the point P t is the image of the pointA = Po when the coordinate plane is rotated about theorigin through an angle of t radians.

(3) If t < 0, then, starting from the point A round thecircle in the negative direction, we shall cover the path oflength It'. Let P t denote the terminal point of thispath which will just be the point corresponding to thenegative number t.

As is seen, the sense of the constructed mapping 1): R-+S consists in that the positive semiaxis is wound ontoS in the positive direction, while the negative semiaxisis wound onto S in the negative direction. This mappingis not one-to-one: if a point F ES corresponds to a num-ber t E R, that is, F == Ph then this point also corre-sponds to the numbers t + 2n, t - 2n: F = P t+21t ==P t -21t Indeed, adding to the path of length t thepath of length 2n (either in the positive or in the nega-tive direction) we shall again find ourselves at the point I?,since 2n is the circumference of the circle of unit radius.Hence it also follows that all the numbers going intothe point P t under the mapping 1) have the form t + Znk,where k is an arbitrary integer. Or in a briefer formula-t ion: the full inverse image 1)-1 (P t) of the point P tcoincides with the set

{t + 2nk: k EZ}.Remark. The number t is usually Identified with the

point P t corresponding to this number, however, whensolving problems, it is useful to lind out what object isunder considerati on.

Example 1.1.4. Find all the numbers {E R correspond-ing to the point F E S with coordinates (-1212,- 12/2) under the mapping P.

~ The point F actually lies on S, since

1.1. Radian Measure of an Arc i3

A

Fig. 3

Let X, Y denote the feetof the perpendiculars drop-ped from the [point F onthe coordinate axes Ox andOy (Fig. 3). Then I XO I ==I YO I = I XF I, and

~XFO is a right isoscelestriangle, LXOF = 45 ===n/4 radian. Thereforethe magnitude of the arc F(-qr~)AF is equalto n + ~ = ~,and to the point F therecorrespond the numbers5: + 2nk, k EZ, and only they. ~

Example 1.1.5. Find all the numbers corresponding tothe vertices of a regular N-gon inscribed in the trigono-

y

Fig. 4

metric circle so that one of the vertices coincides with thepoint PI (see Fig. 4 in which N = 5).


1 -I- 2;k ,where k EZ. The last assertion is verified in thefollowing way: any integer k EZ can be uniquely writtenin the form k == Nm + l, where O~ l~ N - 1 and m,1 EZ, l being the remainder of the division of the integerk hy N. It is now obvious that the equality 1 + 2;k =1 + 2;1 + 2nm is true since its right-hand side con-

c

sB=P'3n/z

E=~/sn/~

A=fl,a:

Fig. 5

tains the numbers which correspond to the points P 211:11+JV

on the trigonometric circle. ~Example 1.1.6. Find the points of the trigonometric

circle which correspond to the following numbers: (a) 3n/2,(b) 13n/2, (c) -15n/4, (d) -17n/6.

3n 3 3n~(a) 2 = T 2rr, therefore, to the number 2 therecorresponds the point D with coordinates (0, -1), sincethe arc AD traced in the positive direction has the mea-sure equal to i of a complete revolution (Fig. 5).

, 13n f n 13n(b) -2- == 32n +2' consequently, to the number 2there corresponds the point B (0, 1): starting from thepoint A we can reach the point B by tracing the trigono-metric circle in the positive direction three times and

1.1. Radian Measnre of an Arr: 15

then covering H quarter of revolution (Jt/2 radian) in thesame direction.

(c) Let us represent the number -t5n/4 ill the formZnk -i- to, where k is an integer, and to is a number suchthat O::s.; to < 2n. To do so, .it is necessary AJ1fl sufficientthat the Iollnwing inequalities he Iulfil led:

2nk ~ -151t/11~ 2n (k + 1).

Let liS write the number -15n/4 in the form -3 ~ n=-41[+ ~, whence it is clear that k= -2, to= '!t /4,and to the number t =!-15n/4 there corresponds apoint E = Pn/" such that the size of the angle EOA is1t/4 (or 45). Therefore, to construct the point P -lsn/4' wehave to trace the trigonometric circle twice in the negativedirection and then to cover the path of length n/4 corre-sponding to the arc of 45 in the positive direction. Thepoint E thus obtained has the coordinates CV2/2, V2/2).

(d) Similarly, - 1~n =-2 ~ n=-2n- ~n =-3n+~ , and in order to reach the point F == P -171t/6 (start-ing from A), we have to cover one and a half revolutions(3n radians) in the negative direction (as a result, wereach the point C (-1, 0) and then to return tracing anarc of length n/6 in the positive direction. The point Fhas the coordinates (-V3/2, -1/2). ~

Example 1.1.7. The points A = Po, B == Pn/2' C =P 31' D = P 331/2 divide the trigonometric circle intofour equal arcs, that is, into four quarters called quad-rants. Find in what quadrant each of the following pointslies: (a) PlO' (b) P 8' (c) P -8

~To answer this question, one must know the approxi-mate value of the number n which is determined as halfthe circumference of unit radius. This number has beencomputed to a large number of decimal places (here are thefirst 24 digits: n ~ 3.141 592 653 589 793 238 462 643).

To solve similar problems, it is sufficient to use far lessaccurate approximations, but they should be written in


the form of strict inequalities of type

3.1 < Jt < 3.23.14 < n < 3.15,

3.141 < jt < 3.112.

(1.1)(1.2)(1.3)

Inaccurate handling of approximate numbers is a flagranterror when solving problems of this kind. Such problemsare usually reduced to a rigorous proof of some inequali-ties. The proof of an inequality is, in turn, reduced toa certain obviously true estimate using equivalenttransformations, for instance, to one of the estimates(1.1)-(1.3) if from the hypothesis it is clear that suchan estimate is supposed to be known. In such cases, somestudents carry out computations with unnecessarilyhigh accuracy forgetting about the logic of the proof.Many difficulties also arise in the cases when we have toprove some estimate for a quantity which is usuallyregarded to be approximately known to some decimaldigits; for instance, to prove that n > 3 or that 1t < 4.The methods for estimating the number n are connectedwith approximation of the circumference of a circle withthe aid of the sum of the lengths of the sides of regularN-gons inscribed in, and circumscribed about, the trigo-nometric circle. This will be considered later on (inSec. 5.1); here we shall use inequality (1.1) to solvethe problem given in Example 1.1.7.

Let us find an integer k such that

(1.1)

Then the number of the quadrant in which the point P10 islocated will be equal to the remainder of the division ofthe number k -t- 1 by 4 since a complete revolution con-sists of four quadrants. Making use of the upper estimate

n6 91t < 3.~, we find that T < .6 for k = 6; at the same n(k+t) n7 0time, n > 3.1 and 2 = T> 3.1 3.f> == 1 .85.

Combining these inequalities with the obvious inequality

9.6 < 10 < 10.85,

1.1. Radian Measure of an Arc 17

\VO get a rigorous proof of the fact that inequality (1.4)is fulfilled for k = G, and the point P I O lies in the thirdquadrant since the remainder of the division of 7 by 4is equa1 to 3.

In similar fashion, we find that the inequalities~ < 8 < n(k+1)2 2

1' 1 f 1 5' n5 3.25 8 d rt-fare va l( or te=:: , Since -2- < -2- == an -2->

3.~.6 = 9.3. Consequently, the point e, lies in thesecond .quadrant, since the remainder of the division ofthe number k -t 1 == 6 by 4 is equal to 2. The point P -8,symmetric to the point ]J 8 with respect to the z-ax is, liesin the third quadrant. ..

Example 1.1.8. Find in which quadrant the pointp lr-; :1:-7 lies.- I ,,- J!


1.2. Definitions of the Basic Trigonometric Functions

1. The Sine and Cosine Defined. Here, recall that inschool textbooks the sine and cosine of a real number t EIt is defined wi th the aid of a trigonometric mapping

1-): R~ :3.

_____--+-----.-.l.-_+__~A

y

o

Definition. Let the mapping P associate a number t E Rwi th the poi lit. J)/ on tl: 0 trigonometric circle. Then the

ordinate y of P I is calledthe sine of the number t and

Jt - I{(COSf, sin t) is symbolized sin t ; andthe abscissa x of P t is calledthe cosine of the num-ber t and is denoted bycos t.

Let us drop perpendicu-lars from the point P t onthe coordinate axes Ox andOy. Let X t and Y t denotethe feet of these perpendic-ulars. Then the coordinateof the point Y t on the

y-axis is equal to Sill t, and the coordinate of thepoint X t on the z-axis is equal to cos t (Fig. 6).

The lengths of the line segments OY t and OX t do notexceed 1, therefore sin t and cos t are functions definedthroughout the number line whose values lie in the closedin terval [- 1, 1]:

D (sin t) == D (cos t) = R,E (sin t) ~ E' (cos t) = [ -1, 1].

The important property of the sine and cosine (thefundamental trigonometric identity): for any t E R

sin ~ t -~- cos'' t === 1.Indeed.tthe coordinates(x, y) of the point 1) t on the trig-

onometric circle satisfy the relationship .y2 + y2 === 1,and consequently cos" t + sin'' t = 1.

.Example 1.2.1. Find sin t and cos t if: (a) t === 31[/2,(b) t = 13n12, (c) t = -15n/4, (d) t = -171[/6.

1.2. Definitions 19

A

Fig. 7

IIIk 8 E~Pn--2

P,

~ In Example 1.1.6, it was shown thatP3n/2=I!(0, -1), P131t/2==B(O, 1),

P_t 511 / 4 = ECV2/2, V2/2), P_171t/6==F(-V3i2, -1/2).Consequently, sin (3n/2) ::::--: -1, cos (3n/2) == 0;

sin (13n/2) = 1, cos (13n/2) == 0; sin (-15n/4) == V2/2,cos (-15n/4) == V2/2; sin (-17rt/6) == -1/2,cos (-17n/6) = - Y3/2. ~

Example 1.2.2. Compare the numbers sin 1 and sin 2.~Consider the points PI and P 2 on the trigonometriccircle: P1 lies in the first quadrant and P2 in the secondquadrant since n/2 < 2< rt.Through the point P 2' wepass a line parallel to thez-axis to intersect the cir-cle at a point E. Then thepoints E and P 2 have equalordinates. Since LAOE = cLP20C, E = P 1t -2 (Fig. 7), ~-~---f,----I--J---+-----:~consequently, sin 2 ==sin (n - 2) (this is a partic-ular case of the reductionformulas considered below).The inequality rt - 2 > 1is valid, therefore sin (n -2) > sin 1, since bothpoints PI and P n-2 lie in the first quadr~nt, and whena movable point traces the arc of the first quadrantfrom A to B the ordinate of this point increases from 0to 1 (while its abscissa decreases from 1 to 0). Conse-quently, sin 2 > sin 1. ~

Example 1.2.3. Compare the numbers cos 1 and cos 2.~ The point P 2 lying in the second quadrant has a nega-tive abscissa, whereas the abscissa of the point PI ispositive; consequently, cos 1 > 0 > cos 2. ~

Example 1.2.4. Determine the signs of the numberssin 10, cos 10, sin 8, cos 8.

~ It was shown in Example 1.1.7 that the point P 1 0 Iiesin the third quadrant, while the point P 8 is in the secondquadrant. The signs of the coordinates of a point on the'trigonometric circle are completely determined by the-2*


position of a given point, that is, by the quadrant inwhich the point is Iouud. For instance, both coordinatesof any point lying in the third quadrant are negative,while a point lying in the second quadrant has a negativeabscissa and a positive ordinate. Consequently, sin 10 < 0,cos 10 < 0, sin 8 > 0, cos 8 < O. ~

Example 1.2.5. Determine the signs of the numberssin (Ys + V"7) and cos (YS+ Vi]).

~ From what was proved in Example 1..1.8, it follows that

rr< V5 + =:;7 < 3n/2.Consequently, the point P JI"5+~/7 lies in the thirdquadrant; therefore

sin (V:5 +V7) < 0, cos (YS+V7) < O. ~Note for further considerations that sin t = 0 if and

only if the point P t has a zero ordinate, that is, P t = Aor C, and cos t = 0 is equivalent to that P t = B or D(see Fig. 2). Therefore all the solutions of the equationsin t == 0 are given by the formula

t = nn, nEZ,and all the solutions of the equation cos t = 0 have theform

3tt=T+1'['n, nEZ.

2. The Tangent and Cotangent Defined.Definition. The ratio of the sine of a number t E R to

the cosine of this number is called the tangent of thenumber t and is symbolized tan t. The ratio of the cosineof the number t to the sine of this number is termed thecotangent of t and is denoted by cot t.

By definition,

t t sin t t t cos tan == cos t' co ==-= sin t

. sin t II IThe expression -- has sense for a rea valuescos t

'Of t, except those for which cos t ==0, that is, except

1.2. Dejtntuon 21

Fig. M

y

o

11:for the values t ="2 + stk;k EZ, and the expressioncot t has sense for all val-ues of t, except those forwhich sin t = 0, that is,except for t = stk, k E Z. C'Thus, the function tan t is --+---,defined on the set of allreal numbers except the

1tnumbers t = 2 +nk, k EZ.The function cot t is definedon the set of all real num-bers except the numberst == ttk, k EZ.

Graphical representation of the numbers tan t and cot twi th the aid of the trigonometric circle is very useful.Draw a tangent AB' to the trigonometric circle throughthe point A = Po, where B' = (1, 1). Draw a straightline through the origin 0 and the point P t and denote thepoint of its intersection with the tangent AB' by Zt(Fig. 8). The tangent AB' can be regarded as a coordi-nate axis with the origin A so that the point B' has thecoordinate 1 on this axis. Then the ordinate of the pointZ t Oil this axis is equal to tan t. This follows from thesinrilari ty of the triangles OX tP t and OAZ t and the defini-tion of tho function tan t. Note that the point of inter-section is absent exactly for those values of t for whichP t = B or D, that is, for t == ~ + nn, nEZ, when thefunction tan t is not defined.

Now, draw a tangent BB' to the trigonometric circlethrough the point B and consider the point of intersectionW t of the 1ine 01) t and the tangent. TIle a bscissa of W tis equal to cot t. The point of intersection W t isabsent exactly for those t for which P t = A or C, that is,when t = nn, nEZ, and the function cot t is not defined(Fig. 9).

In this graphical representation or tangent and cotan-gent, the tangent linesAR' and BB' to the trigonometriccircle are called the line (or axis) of tangents and the line(axis) of cotangents, respectively.


Example 1.2.6. Determine the signs of the numbers:tan 10, tan 8, cot 10, cot 8.

.f)

Fig. 9

...- In Example 1.2.4, it was shown that sin 10 < andcos 10 < 0, sin 8 > and cos 8 < 0, consequently,tan 10 > 0, cot 10 :> 0, tan 8 < 0, cot 8 < 0.

(-VJ, /)

o

h

Fig. 10

Example 1.2.7. Determine the sign of the numbercot (l!5 + V7) .....- In Example 1.2.5, it was shown that sin (V 5 + 1/7) 0. ~

1.3. Basic Properties 23

3n 17nExample 1.2.8. Find tan t and cot t if t~ 4 ~,1731 i1:)".

--0-' -0-

~ As in Item 3 of See. 1.1, we locate the points P31t/4.P171t/4 (Fig. iDa), P-171t/6' P11n/6 (Fig. 10b) on the trig-onometric circle and compute their coord ina tes:

P~JT/4(-Y2/2, Y2/2), P171t/l.(V2/2, V2/2), ~p -17:t/6 ( - V3/2~ -1/2), [J U 1t/ 6 ("V' 3/2, - 1/2),

therefore tan (3n/4) = cot (3n/4) === -1., tan (t7n/4) ~cot (17n/4) ~ 1, tan (-17Jt/6) == 1/Y3" z :': V3/:~,cot (-17rr./6)==V3, tan (11n/6) = -V3/3, cot(11n/6)=-=-V3. ~

1.3. Basic Properties of Trigonometric Functions

1. Periodicity. A function / with domain of deJinitionX === D (/) is said to be periodic if there is a nonzero num-bel' T such that for any x E X

.r + T E X and ,x - 11 EX,and the following equality is true:

f (z - .T) ~ f (.x) == f (z + T).The number T is then called the period of the functionf (x). A periodic function has infinitely many periodssince, along with T, any number of the form n'I', where nis an integer, is also a period of this function. The sm al-lest positive period of the function (if such period exists)is called the fundamental period.

Theorem 1.1. The junctions f (z) ::.= sin x and f (x) ==cos x are periodic ioiih. [undamenial period 231.

Theorem 1.2. The functions f (z') == tan :1: and f (.1') ::=cot x are periodic ioith. fundamental period JT.

It is natural to carry 011t the proof of Theorems 1.1and 1.2 using tho graphical representation of sine, eosine.tangent, and cot nngen t wi th the aid of the tr-igonornetriccircle. To the real numbers x, :1; + 2n, and x - 231,there corresponds one and the same point P x on the


trigonometric circle, consequently, these numbers havethe same sine and cosine. At the same time, no positivenumber less than 2n can be the period of the functionssin x and cos x. Indeed, if T is the period of cos x, thencos T == cos (0 + T) = cos 0 ::= 1. Hence, to the num-ber T, there corresponds the point PT with coordinates(1, 0), therefore the number T has the form T === 2nn(n EZ); and since it is positive, we have T~ 2n. Similar-ly, if T is the period of the function sin x, then sin (~ +T) = sin ~ = i, and to the number ~ + T there

corresponds the point P n with coordinates (0, 1).T+T

Hence, ~ + T === ~ + 21tn (n E Z) or T = 2nn, that is,T~ 21t. ~

To prove Theorem 1.2, let us note that the points P tand P t +n are symmetric with respect to the origin forany t (the number 1t specifies a half-revol ution of thetrigonometric circle), therefore the coordinates of thopoints P t and Pt+1t are equal in absolute valueand have unlike signs, that is, sin t == -sin (t + n),

sin tcos t = -cos (t -t a). Consequently, tan t =-= --:.==

cos t-sin (t+n) t (t +) t t cos t -- cos (t +n)---~ == an 1t co - --= -- cos (t+ n) ,- sin t - sin (t+ rr)

cot (t + rr). Therefore rr is the period of the functionstan t and cot t. To make sure that n is the fundamentalperiod, note that tan 0 == 0, and the least positive valueof t for which tan t == 0 is equal to rr. The same rea-soning is applicable to the function cot t. ~

Example 1.3.1. Find the fundamental period of thefunction f (t) == cos! t -+- sin t.

~ The function f is periodic sincef (t + 2n) == cos4 (t + 2n) -f- sin (t + 2n)

= cos" t + sin t.No positive number T, smaller than 2rf., is the periodof ~he function f (t) since f ( - ~ ) =F f ( - ~ + T ) ::00f ( T ). Indeed, the numbers sin ( - ~ ) and sin ~


are dist.iuct from zero and have unlike signs, thenumbers cos ( - ~) and cos + coincide, therefore

( T) T) l' Tcos- -""2 + sin ( -""2 =1= cos- "2 +sin "2' ..Example 1.3.2. Find the fundamental period of the

function f eVSx) if it is known that T is the fundamentalperiod of the function f (x).

~ First of all, let 118 note that the points x - t, x,x +. t belong to the domain of definition of the func-tion g(x)=fCV5x) if and only if the points xV5-t V5, x V5, x V5 -t- t ~r5 belong to the domain of def-ini tion of the function f (x). The defini tion of the Iunc-tion g (x) implies that the equalities g (x - t) =g (x) = g (x + t) and f (xV5 - tVS) == f (xV5) ==f (xV 5 + tV' 5) are equivalent. Therefore, since T' is thefundamental period of the function f (.:r), the numberT/1!5 is a period of the function g (x); it is the funda-mental period of g (x), since otherwise the function g (x)would have a period t < I'IV s: and, hence, the functionf (x) would have a period tV 5, strictly less than T. ~

Note that a more general statement is valid: if a Iunc-tion f (x) has the fundamental period 1', then the Iunc-tion g (x) = f (ax -+- b) (a =1= 0) has the Iundamentalperiod 1'/1 a I.

Example 1.3.3. Prove that the Iunction / (x) =sin VTXT is not periodic. VX-, therefore the Iol lowiuginequality holds true:

Vx+ l'~2n+ Vx.Both sides of this inequality contain positive numbers,consequently, when squared, this inequality will be re-placed by an equivalent one: .T -1-- T~ 4n2 -I" 4n'V'x-+-x or

T~4n2+4n Vx.


Fig. t t

y

o

The obtained inequality leads to a contradiction sincethe number Vxcan be chosen arbitrarily large and, inparticular, so that the inequality is not valid for thegiven fixed number 1'. ~

2. Evenness and Oddness. Recall that a function f issaid to be even if for any x from its domain of defini tion-x also belongs to this domain, and the equalityj (-x) === f (x) is valid. A function f is said to be odd if,

under the same conditions,the equality f (-x) ==-/ (x) holds true.

A couple of examples ofeven functions: / (x) == x2 ,f (x) === x 4 +V 5x2 + rr.

A couple of examples ofA a odd functions: f(.7;) == -II X3,

f (x) = 2x5 + nx3 Note that many functions

are nei ther even nor odd.For instance, the functionf (x) = x3 + x2 + 1 is noteven since f( -x) ==( - X)3 + (- X)2 + 1 ==

- x3 + x2 -t- 1 =1= f (x) for x * O. Similarly, the functionj (x) is not odd since f (-x) =1= - j (x).

Theorem 1.3. The junctions sin x, tan x, cot x are odd,and the function cos x is even.

Proof. Consider the arcs AP t and AI) -t of the trigono-metric circle having the same length It' but oppositedirections (Fig. 11). These arcs are symmetric withrespect to the axis of abscissas, therefore their end points

P t (cos t, sin t), P -t (cos (-t), sin (-t)

have equal abscissas but opposite ordinates, that is,cos (-t) == cos t, sin (-t) ~ -sin t. Consequently,the function sin t is odd, and the function cos t is even.

Further, by the definition of the tangent and cotangent,

tan ( _ t) == sin ( - t) ==. cos (- t)

-sin tcos t = -tan t


if cos t =1= 0 (here also cos (-t) = cos t =1= 0), and

a:A

Fig. 12

o

t ( t) - cos (- t) - cos t - t tco - - sin(-t) - -sint - -co

if sin t =1= 0 (here also sin (-t) == -sin t =F 0). Thus,the functions tan t and cot t are odd. ~

Example 1.3.4. Prove that the function f (t) ==sin" 2t cos 4t + tan 5t is odd.

~Note that f (-t) = (sin (-2t))3 cos (-2t) +tan (-5t) == (-sin 2t)3 cos 2t - tan 5t ~ - f (t) forany t from the domain ofdefinition of the function(that is, such thatcos 5t =1= 0). ~

3. Monotonicity. Recallthat a function f definedin an interval X is said tobe increasing in this inter-val if for any numbers Xl'X 2 EX such that 'Xl < X 2the inequality f (Xl) < f (x2)holds true; and if a weakinequality is valid, that is,t (Xl)~ f (x2 ) , then the func-tion f is said to be nonde-creasing on the interval X. The notion of a decreasingfunction and a nonincreasing function is introduced ina similar way. The properties of increasing or decreasingof a function are also called monotonicity of the function.The interval over which the function increases or decreasesis called the interval of monotonicity of the function.

Let us test the functions sin t and cos t for monotonici-ty. As the point P l moves along the trigonometric circleanticlockwise (that is, in the positive direction) from thepoint A === Po to the point B (0, 1), it keeps rising anddisplacing to the left (Fig. 12), that is, with an increasein t the ordinate of the point increases, while the abscissadecreases. But the ordinate of P t is equal to sin t , itsabscissa being equal to cos t. Therefore, on the closedinterval [0, n/21, that is, in the first quadrant, the func-tion sin t increases from 0 to 1, and cos t decreasesfrom 1 to O.

28 1. Properties 0/ Trigonometric Functions

o

y

----+----r-~---.---+---___+.~X

1f3 0Fig. 13 Fig. 14

y

o

]}

Fig. 15

Similarly, we can investigate the behaviour of these'functions as the point P t moves in the second, third, andfourth quadrants. Hence, we may formulate the followingtheorem.

Theorem 1.4. On the interval [0, n/2l the junction,sin t increases from 0 to 1, ich!le cos t decreases [rom. 1 to o.On the interual [n/2,. n] the function sin t decreases from 1to 0, while cos t decreases from 0 to -1. On the interval[1[, 31[/2] the junction sin t decreases from 0 to -1,. tohilecos t increases [rom. -1 to O. On the interval [331/2, 2rt]the function sin t increases from -1 to 0, tohile cos tincreases from 0 to 1.


The ])1"00/ of the theorem is graphically illustrated inFigs. 12-15, where points P tl' P iI' P 13 are such thatt l < t 2 < t 3 ~

Theorem 1.5. The function tan t increases on the interval(-n/2, n/2), and the junction cot t decreases on the inter-val (0, n).

Proof. Consider the function tan t. We have to showthat for any numbers t1 , t 2 such that -n/2 < t1 < t 2 cos t 2 > 0,whence

sin tl < sin t2cos t1 co~ t2

Consequently, tan t1 < tan t 2 (2) -n/2 < t} < 0 < t 2 < n/2. In this case,tan t 1 < 0, and tan t 2 > 0, therefore tan tt < tan t 2 (3) -'!t/2 < t l < t 2~ O. By virtue of Theorem 1.4,

sin t l < sin t2~ 0, < cos t1 < cos t 2,therefore

sin it ~ sin t2cos t 1

2 _ 1cos t -- 1+ tan2 t

~o 1. Properties of Trigonometric Functions

quadrant since these numbers lie on the closed interval[0, 1], and 1 < n/2. Therefore, we may once again applyTheorem 1.4 which implies that for any tt, t2 E[0, n/2lsuch that tt < t 2 , the following inequalities are valid:

sin (cos tt) > sin (cos t 2) , cos (sin tt) > cos (sin t 2) ,that is, sin (cos t) and cos (sin t) are decreasing functionson the interval [0, n/2]. ~

4. Relation Between Trigonometric Functions of Oneand the Same Argument. If for a fixed value of the argu-ment the value of a trigonometric function is known,then, under certain conditions, we can find the values ofother trigonometric functions. Here, the most importantrelationship is the principal trigonometric identity (seeSec. 1.2, Item 1):

sin2t + cos'' t = 1. (1.9)Dividing this identity by cos'' t termwise (providedcos t =F 0), we get

1+ tan2 t = _1- (1.10)cos2 t '

where t=l= ; +:rtk, k EZ. Using this identity, it ispossible to compute tan t if the value of cos t and the signof tan t are known, and, vice versa, to compute cos tgiven the value of tan t and the sign of cos t. In turn,the signs of the numbers tan t and cos t are completelydetermined by the quadrant in which the point P t icorresponding to the real number t, lies.

Example 1.3.6. Compute cos t if it is known thattan t =cc 152 and t E(31, 3;).

~ From formula (1.10) we find:1 14425 169

1+ 144

Consequently, I cos t I = 12/13, and therefore eithercos t = 12/13 or cos t == -12/13. By hypothesis,t E (rt, 3n/2), that is, the point P t lies in the third quad-rant. In the third quadrant, cos t is negative: consequent-ly, cos t = -12/13. ~

1.4. Solving the Simplest Equations 31

Dividing both sides of equality (1.9) by sin2 t (forsin t =F 0) termwise, we get

1 + 2 _ 1cot t .. _. -.-, Sln2 t '

(1.it)

where t =1= nk, k EZ. Using this identity, we can corn-pute cot t if the value of sin t and the sign of cot tareknown and compute sin t knowing the value of cot t andthe sign of sin t.

Equalities (1.9)-(1.11) relate different trigonometricfunctions of one and the same argument.

1.4. Solving the Simplest Trigonometric Equations.Inverse Trigonometric Functions

a:A

Fig. 16

y

o

n

___--i----y-=-m.....;."lml ~ I

1. Solving Equations of the Form sin t == m, Arc, Sine.To solve the equation of the form sin t = m; it is neces-sary to find all real numbers t such that the ordinateof the corresponding pointP t is eq ual to m. To thisend, we draw a straight liney == m and find the pointsof its intersection with thetrigonometric circle. Thereare two such points if cI m I < 1 (points E and' F I------+-----+-------:~in Fig. 16), one point ifI m I = 1, and no pointsof intersection for Im I >1.Let I m I~ 1. One of thepoints of intersection liesnecessarily in the. right-hand half-plane, wherex> O. This point can be written in the form E == P to'where to is some number from the closed interval [-n/2,n/2]. Indeed, when the real axis is being wound on thetrigonometric circle, the numbers from the interval[-n/2, n/2] go into the points of the first and fourthquadrants on the trigonometric circle, the points Band Dincluded. Note that the ordinate sin to of the pointE = P t is equal to m: sin to = m,

32 1. Properties oj Trigonometric Functions

Definition. Tho arc sine of a numher m is a number to,-n/2~ to~ n/2, such that sin to :== In. 'I'he followingnotation is used: to :== arcsin In (or sin'? m).

Obviously, the expression arcsin ni has senRe only forI m I~ 1. By definition, \ve have:

si n (arcsin In) === m, -n/2~ arcsin m~ n/2.The Iollowing equality hollis true:

arcsin (-nt) :== -arcsin m.Note that the left-hand point F of intersection of the

line y = In with the trigonometric circle can be written inthe form F ==: P 1( -to' therefore all the solutions of the equa-

tion sin t== m,lml~1,U are given by the formulas

t == arcsin m. -f- 2nk, I: EZ,t :== Jt - arcsin 111. 1- 2nk,

k EZ,a which are usually united

!/=-1/2 into one formula:--tt-----+--~-----

t === (_1)n arcsin m -1- nil,n EZ.

l"ig. 17 Example 1.4.1. Solve theequation si n t := -1/2.~ Consider the points of intersection of the line y == -1/2and the trigonometric circle S. Let X and Y be the feet ofthe perpendiculars dropped from the right-hand point Eof intersection on the coordinate axes (Fig. 17). In theright triangle XOE, we have: I EX I == 1/2, I OE J ~ 1,that is, LXOE == 30. Consequently, LAOE is measuredby an arc of -n/G radian, and E == P -1[/6" Thereforearcsin (-1/2) == -n/6, and the general solution of theequation sin t = -1/2 has the form t:= (_1)n+l ~ +nn, n EZ. ~

Example 1.4.2. What is the value of arcsin (sin iO)?~ In Example 1.1.7 it was shown that the point P I O liesin the third quadrant. Let t == arcsin (sin 10), thensin r == sin 10 < Oand -'JrJ2~ t~ n/2.Consequently, thepoint P t lies in the fourth quadrant and has the same or-


A

x=m x=mfml~1 1m 1>1,

F

E

o

diuate as the point PI O; therefore r, == Pn,-lO (Fig. 18),and the equality t ~ n - 10 + 2nk holds for some inte-ger k. For the condition -n/2 < t < 0 to be fulfilled,

y

Fig. 18 Fig. 19

it is necessary to set k = 2. Indeed,

- n/2 < 3n - 10 < 0(these inequalities follow from the estimate 3.1 < n 1 ata point Zt with ordinate equal to m (Fig. 21). The equa-tion of the st.raight line passing through the origin and P tis given by the formula y == mx. For an arbitrary real


y

Fig. 21

number m there are exactly two points of intersection ofthe l ino y == mx wi th the trigonometric circle. One ofthese points lies in tho right-hand half-plano and can berepresented in the form E == P to' where to E (-rt/2, n/2).

Definition. The arc tangent of a number m is a num-ber to, lying on the open interval (-n/2, n/2), suchthat tan to == m. It is de-noted by arctan m (ortan -1 m).

The general solution ofthe equation tan t == m(see Fig. 21) is:t = arctan m + nk, k EZ.For all real values of m thefoIl owing equal i ties hoI d:

tan (arctan m) = m,arctan (- m) == -arctan m,

Arc cotangent is intro-duced in a similar way: forany mER the numbert = arccot m is uniquelydefined by two conditions:o< t < rr, cot t ==m.

The general solution of the equation cot t = m ist == arccot m + stk; k EZ.

The following identities occur:

cot (arccot m) = m, arccot (-m) = rt - arccot In.Example 1.4.5. Prove that arctan (-2/5) =

arccot (-5/2) - n.~ Let t == arccot (-5/2), then 0 < t < n, cot t =-5/2, and tan t = -2/5. The point P t lies in thesecond quadrant, consequently, the point P t _j(, lies inthe fourth quadrant, tan (t - n) == tan t == -2/5,and the condition -n/2 < t - n < n/2 is satisfied.Consequently, the number t - n satisfies both condi-tions defining arc tangent, i.e.

t -1t = arctan (-2/5) == arccot (-5/2) - n.~

30 1. Properties 0 f Trigonometric Functions

Example 1.4.6. Prove the identitysin (arctan x) == xIV-t-+-X-"'2.~ Let t == arctan x. Then tan t == x, and -n/2 < t 0 since -11/2 < t < n/2.By virtue of (1.10), 1/cos2 t === 1 -t- tan" t == 1 -t- x2 orcos t == 1tV1 -t- x2 , whence sin t ~ tan tcost::=:xVi + x2 ~

It is clear from the above examples that when solvingproblems, it is convenient to use more formal definition ofinverse trigonometric functions, for instance: t == are-cos Inif (1) cos t == m, and (2) O~ t~ J1. To solve problemson computation involving inverse trigonometric Iunc-tions, it suffices to rememher the above defini tions andbasic trigonometric formulas.

PROBLEMS1. t. Locate the points by indicating the quadrants:(a) t'. (b) P10-~/2' (c) PV26+V21.2. Given a regular pentagon inscribed in the trigono-

metric circle with vertices A h == P2nh/r), k == 0,1,2,3, 4.Find on which of the arcs joining two neighbonring ver-tices the following points lie: (a) PIO' (b) P -11' (c) ])12.

1.3. Given a regular heptagon inscribed in the trigono-metric circle wi th vertices B h = P 2+2nh./7' k === 0, 1, 2, 3,4, 5, 6. Find on which of the arcs joining the neighbouringvertices the point Pl!29 lies.

1.4. Prove th at there is a regul ar N -gon inscribed inthe trigonometric circle such that its vertices incl nde thepoints PY2+41t/1 t' PY2+71t/13 Find the least possiblenumber N.

1.5. Prove that any two points of the form P 3tX' P ny'where x, yare rational numbers, are vertices of a regularN-gon inscribed in the trigonometric circle.

1.6. Find a necessary and sufficient condition for thepoints Pa and Pf'J to be vertices of one and the same regu-lar N -gOIl inscribed in the trigonometric circle.

t.7. Compare the following pairs of numbers:(a) sin 1 and sin ( 1.+ ~Jt ) ,

Problems 37

(b) cos ( 1 --I- 2; ) and cos ( 1 j-_ 4; ).1.8. Determine the sign of the indicated number:(a) cos ( 1 + 6; ),(b) cos ( 1 --I- 2; )cos ( 1 + 4; )cos ( 1 + 6; )X

cos (1+ 8; ).1.9. Can the sine of an angle be equal to:(a) log a+-1-

1- (a>O, a+ 1),

oga

(b) ( 113-1 _ )-1?V5-!fg

1.10. Determine the sign of the product sin 2 -sin 3 -sin 5.1.11. Evaluate:

. tnoire f23n .(a) SIll -6-' (b) cos-4 - , (c) SIn (-117n/4),(d) cos ( - 205n/6).1.12. Determine the sign of the number tan 11.t.13. Evaluate:

1011n 1(101n(a) tan -4-' (b) cot -6-.-.1.14. Prove that for an arbitrary real number a E It and

an integer N > 1 the following equalities are valid:N-1

~ sin (a + 2;k )= 0,k=O

N-i

~ cos ( a + Z;k )= O.h=O

. ttt f3tt .15. Prove that the function f (t) == tan 34+cot 54is periodic and find its fundamental period.

1.16. Is the function f (t) === sin (2x + cos eV2 x))periodic?

1.17. Prove that the function f (x) == cos (x2 ) is not. periodic.

1.18. Prove that the function tan (V2x) + cot (1/3x) isnot periodic.


1.19. Prove tha t the function f (t) == si n 3t -1- cos 5tis periodic and find its fundamental period.

1.20. Prove that the function !(x)=-=-cos(VlxI2) isnot periodic.

1.21. Find the fundameu tal period of the fUIlC tion:

(a) y==-cosnx+sin ~x ,

(b) y == sin x + cos ~ + tan ~) .1.22. Find the fundamental period of the function y ===

15 sin" 12x + 12 sin" 15x.1.23. Prove that the function of the form f (x) ==

cos (ax + sin (bx, where a and b are real nonzero num-bers, is periodic if and only if the number alb is rational.

1.24. Prove that the function of the form f (x) =cos (ax) + tan (bx), where a and b are real nonzero num-bers, is periodic if and only if the ratio alb is a rationalnumber.

1.25. Prove that the function y == tan 5x -t- cot 3x -l-4 sin x cos 2x is odd.

1.26. Prove that the function y = cos 4x + sin" ~ Xtan x +6x 2 is even.

1.27. Represent the function y = sin (x + 1) Sill3 (2x-3) as a sum of an even and an odd function.

1.28. Represent the function y = cos (x + ~ )+sin (2x - ;2 ) as a sum of an even and an odd function.

1.29. Find all the values of the parameters a and b forwhich (a) the function f (t) == a sin t + b cos t is even,

(b) the function f (t) == a cos t -1- b sin t is odd.In Problems 1.30 to 1.32, without carrying out COIll-

putations, determine the sign of the given difference.1.30. (a) sin 2; - sin l~rt , (h) cos 3.13 - sin 3.13.1.31. (a) sin 1 - sin 1.1, (b) sin 2 - sin 2.1,

(c) sin 131 - sin 130, (d) sill 200 - sin 201.1.32. (a) cos 71 - cos 72, (b) cos 1 - cos 0.9,

(c) cos 100 - cos 99, (d) cos 3.4 - cos 3.5.

Problems 39

1.33. Is the function cos (sin t) increasing 0[' decreasingon the closed interval [-Jt/2, Ol?

1.34. Is the function sin (cos t) increasing or decreasingon the closed interval In, 3n/21?

1.35. Prove that the function tan (cos t) is decreasingon the closed interval [0, n/2J.

1.36. Is the function cos (sin (cos t)) increasing or de-creasing on' the closed interval [Jt/2, rt]?

In Problems 1.37 to 1.40, given the value of one Iunc-tion, find the val ues of other trigonometric functions.

1.37. (a) sin t == 4/5, n/2 < t < n ,(b) sin t = -5/13, n < t < 3n/2,(c) sin t = -0.6, -n/2 < t < O.1.38. (a) cos t = 7/25, 0 < t < n/2,(b) cos t = -24/25, 11: < t < 311:12,(c) cos t = 15/17, 3n/2 < t < 2TC.1.39. (a) tan t . 3/4, 0 < t < n/2,(b) tan t = -3/4, n/2 < t < rr.1.40. (a) cot t = 12/5) rt < t < 3n/2,(b) cot t == -5/12, 3n/2 < t < 21t.1.41. Solve the given equation:(a) 2 cos" t - 5 cos t + 2 = 0, (b) 6 cos" t + cos t -

1 == O.1.42. Find the roots of the equation cos t = - 1/2 be-

longing to the closed interval [-211:, 6nl.1.43. Solve the equations:(a) tan t = 0, (b) tan t = 1, (c) tan 2t == V3,(d) tan 2t = - Va, (e) tan ( t - ~ ) - '1 = 0,(f) V3 tan ( t + ~ ) == 1.1.44. Compute:(a) arcsin 0 -~ arccos 0 + arctan 0,.

. f va V3(b) arcsm 2"" +arccos -2- + arctan -3- ,.V3 (It;"3 ) (V3 )(c) arCSln-2- + arccos --2- -arctan -~

In Problems 1.45 to 1.47, prove the identities.1.45. (a) tan I arctan x I = I x I, (b) cos (arctan x) =

1lV 1 + .1,2.


1.46. (a) cot I arccot x I = x,(b) tan (arccot x) = 11x if x =1= 0,(c) sin (arccot x) = 1/V 1+ x2 ,(d) cos (arccot x) zzz: x/V 1-t- x2

and arccos x ==x1.47. (a) arcsin x ~ arctan ---V1-x 2

xarceot y __ for O~x< 1,

1-x2

V1-x2(b) arcsin x = arceot for 0 < x~ 1,x

(c) arctan ~ = arceot xz

. 1 x= arCSIn == arccos---Y1+x 2 Y1+x 2 '

arccot ~ = arctan xx

. x 1= arCSIn ~/__ = arccos

y 1+x2 Y1+x2for x> o.

. 3 12 51.48. Express: (a) arcsin "5' (b) arccos 13' (c) arctan 12'(d) arctan: in terms of values of each of the threeother inverse trigonometric functions.

1.49. Express: (a) arccos ( - ~ ), (b) arctan ( - 2: )I(c) arccot ( - ;4) in terms of values of each of thethree other inverse trigonometric functions.

1.50. Find sin a if tan ex = 2 and 1t < a < 331/2.

Chapter 2

Identical Transformationsof Trigonometric Expressions

2.1. Addition FormulasThere are many trigonometric formulas. Most causedifficulties to school-pupils and those entering college.Note that the two of them, most important formulas,are derived geometrically. These are the fundamentaltrigonometric identity sin" t + cos" t = 1 and the for-mula for the cosine of the sum (difference) of two numberswhich is considered in this section. Note that the basicproperties of trigonometric functions from Sec. 1.3(periodicity, evenness and oddness, monotonicity) arealso obtained from geometric considerations. Any ofthe remaining trigonometric formulas can be easily ob-tained if the student well knows the relevant definitionsand the properties of the fundamental trigonometricfunctions, as well as the two fundamental trigonometricformulas. For instance, formulas (1.10) and (1.11) fromItem 4 of Sec. 1.3 relating cos t and tan t, and also sin tand cot t are not fundamental; they are derived Irom thefundamental trigonometric identity and the definitionsof the functions tan t and cot t. The possibility of derivinga variety of trigonometric formulas from a few funda-mental formulas is a certain convenience, but requires anattentive approach to the logic of proofs. At the sametime, such subdivision of trigonometric Iormulas intofundamental and nonfundamental (derived from thefundamental ones) is conventional. The student shouldalso remember that among the derived formulas there arecertain formulas which are used most frequently, e.g.double-argument and half-argument formulas, and also theformulas for transforming a product into a slim. To solveproblems successfully, these formulas should be kept inmind so that they can be put to use straight. away. A goodtechnique to memorize such formulas consists in follow-

42 2. Identical Transformations

ing attentively the way they are derived and solvinga certain number of problems pertaining' to identical trans-Iormations of trigonometric ex pressions.

1. The Cosine of the SUIll and Difference of Two RealNumbers, One should not think that there arc severalbasic addition formulas. We are going to derive the for-mula for the cosine of the sum of two real numbers and

o I

Fig. 22

then show that other addition formulas are derived fromit provided that the properties of evenness and oddness ofthe basic trigonometric functions are taken into consid-eration. .

1'0 prove this formula, we shall need the following note.Under the trigonometric mapping

P: R-+Sof the real axis R onto the trigonometric circle S (seeItem 3, Sec. 1.1), line segments of equal length go intoarcs of equal size. More precisely, this means the follow-ing. Let on the number line be taken four points: t1 , t 2 ,' t 4 such that the distance from t1 to t 2 is equal to thedistance from t: J to t 4 , that is, such that I tt - t2 I ===I t 3 --: t 4 I, ~nd let P i 1 ' P i 2 ' P 13 , P i 4 be points on thecoordinate CIrcle corresponding to those points. Then thearcs PtlP t , and Pt~Pt~ are congruent (Fig. 22). Hence it

2.1. Addition Formulas 43

follows that the chords P til) t. and P t3P if, are also COIl-grueu t: I e,, I == I PtaP t" I

Theorem 2.1: For any real numbers t and s the followingidentity is valid:

cos (t + s) == cos t cos s - Sill t sin s. (2.1)

!I

Fig. 23

o

P l +s = (cos (t+8),sin (t + s,

p -8 == (cos (-s), sin (-s)= (cos 8, -sin s).

Proof. Consider the point Po of intersection of the unitcircle and the positive direction of tIle z-axis, Po === (1, 0).Then the corresponding points P b P t +$' and P -s are im-ages of the point Po whenthe plane is rotated aboutthe origin through anglesof t, t + s, and -8 radians

(~-"ig. 23).By the definition of the

sine and cosine, the coordi-nates of the points P 17 P t +s'and P -8 will be:

P t = (cos t, sin t),

Here we have also used the basic properties of the evennessof cosine and oddness of sine. We noted before the proofof the theorem that the lengths of the line segmentsPoP t +s and P -sP t are equal, hence we can equate thesquares of their lengths. Thus, we get identity (2.1).Knowing the coordinates of the points Po and P t +s, wecan find the square of the length of the line segmentPoP t +8: I PoP t +8 12 == (1- cos (t + S))2 + sin" (t + s) =1 - 2 cos (t + s) + GOS2 (t + s) -~- sin" (t -f- S) Of, hyvirtue of the Iundnmental trigonometric identity,

r PoP t +s 12 == 2 - 2 cos (t + s].


On the other hand,I P -sP t p~ = (cos s - cos tr~ -t- (-sin s - sin t):!

== cos" S - 2 cos s cos t -f- cos" t+ sin" 8 + 2 sin s -sin t + sin? t= 2 - 2 cos t cos s + 2 sin t sin s.

Here we have used the Iundurucul.al trigonometric identi-ty (1.9) once again. Consequently, from the equalityI PoP t +s 12 = I r ;, 12 we get

2 - 2 cos (t + s) == 2 - 2 cos t cos s + 2 sin t sin s,whence (2.1) follows. ~

Corollary 1. For any real numbers t and s we havecos (t - s) == cos t cos s -~- sin t sin s. (2.2)

Proof. Let us represent the number t - s as (t + (-8)and apply Theorem 2.1:

cos (t - s) == cos (t +- (-s== cos t cos (-s) - sin t sin (-s).

Taking advan tage of the properties of the evenness of co-sine and oddness of sine, we get:

cos (t - s) = cos t cos s -t- sin t sin s. ~In connection with the proof of identities (2.1) and (2.2),

we should like to note that it is necessary to make surethat in deriving a certain identity we do not rely on anoth-er identity which, in turn, is obtained from the identityunder consideration. For instance, the property of theevenness of cosine is sometimes proved as follows:cos (-t) = cos (0 - t) ==cos 0 cos t -t- sin 0 Sill t == cos t,that is, relying Oil the addition formula (2.2). \Vhen deriv-ing formula (2.2), we use the property of the evennessof the function cos t. Therefore the mentioned proof ofthe evenness of cos t may be recognized as correctonly provided that the student can- justify the additionformula cos (t - s) without using this property of cosine.

Example 2.1.1. Compute cos 8= cos 15.


...... Si rr j[. n t f f 1

...... ince 12 ==- 3" -- 4"' we ge l'OlnorlllU at=n./3 and s~'!t/4:

(2.2) for

rr ( nIt) rr 1t I _ 1t 1tcos 12 = cos 3 - 7; == cos 3 COg 4 _t- sm "3 SIn T

==:~ . V2 + V"'3 . \/2 == ~./2+ {/fi22 22 4~

2. Reduction Formulas,Corollary 2. For any real number t tne hare

(2.3)

(2.4)

(2.7)

(2.5)(2.6)

(2.8)(2.9)

(2.10)

cos ( ~ - t ) = sin t,cos ( ~ + t ) =0 - sin t,

cos (rr- t) ~~ - cos t,cos (n + t) =-- - cos i :

Proof. Let ns use formula (2.2):( 11: t) j[ n: tcos 2- == cos 2 cos t+ sm 2 SIn

== (l- cos t -~ 1 gin t ~ sin t,cos (n - t) == cos n cos t + sin n sin t

:::::: (-1).eos t - Osin t = -cos t,which yields identities (2.3) and (2.5). We now use (2.1):

(n) 1t . 11: cos 2 -1- t == cos 2: cos t - Sin 2"" SIn t:-:::: Ovcos t-1. sin t=:= - sin t,

cos (n 1- t) == cos n cos t _. sin rt sin t= (--1)cos t-O-Rin t==-cos t.

These equalities prove identities (2.4) and (2.6). ~Corollary 3. For any real number t uie have

sin ( ~ - t) ~. cos t,sin ( ~ + t ) = cos t ;

sin (rr - t) :::-: sin t,sin (n -~ t):=: - sin t,

46 2. Identical Trans formations

tan (~ +t) --", -cott, t=l=nk,cot ( ~ -+ t ) ~.7 - tan t,

t =1= ~ -t- rt If ,

Proof. Let us use identity (2.3):cos t = cos ( ~ - ( ~ - t ) ) ==siu ( ~ -+ t) ,

which yields identity (2.7). From identity (2.3) and theproperty of evenness of cosine there follows:

cos t = cos ( - t) =- cos ( ~ - ( ~ --I t ) ) = sin ( ~ --I t) ,which proves identity (2.8).

Further, by virtue of (2.8) and (2.3), we have:sin (31 - t) = sin ( ~ -+ ( ~ -- t ) ) = cos ( ; - t ) = sin t,

and from identities (2.8) and (2.4) we get:sin (31 -+ t) = sin ( ~ -+ ( ~ + t ))

= cos ( ~ -+ t) =- -sin t. ~Remark. Using formulas (2.3)-(2.10), we can easi ly

obtain reduction formulas for cos ( 3; + t) , cos (231- t),sin ( ~1t + t) , sin (231 - t). The reader is invited to dothis as an exercise.

Example 2.1.2. Derive the reduction formula forcos ( ~1t - t ) .

~ Using formulas (2.6) and (2.3), we havecos ( ~1t - t ) = cos (31 + ( ~ - t ) ) = - cos ( ~ - t )

==-sint. ~.Corollary 4. The reduction [ormulas for tangent and

cotangent are

tan (; -t) =cott,cot ( ~ - t ) = tan t,

() n. ktan n - t == - tan t , t =1= "2 -1,- Jt ,cot (n- t) =--: --cot t ; t =f= sck (k E Z).


Proof. All these formulas can he obtained from thedefinition of the functions tun t and cot t applying theappropriate reduction formulas. ~

Note that the reduction formulas together with theperiod ici ty properties of the basic trigonometric functionsmake it possible to reduce the computation of the value ofa trigonometric function at any point to its computationat a point in the closed interval [0, n/4].

To facilitate the memorizing of the reduction formulas,the Iol lowi ng mnemonic rule is recommended:

(1) Assuming that 0 < t < ~ , find in which quad-rant the point ].J 1tk ,kEZ, lies, dotermine the sign

2tthe given expression has in this qnadrant, and put thissign before the obtained result.

(2) When replacing the argument 3t + t or 2n - t hy t,the name of the function is retained.

(3) When replacing the argument ~ t or ;Jt + tby t, the name of the function should be changed: sine forcosine and cosine for sine, tangent for cotangent andcotangen t for tangent.

3. Sine of a Sum (a Difference).Corollary 5. For any real numbers t and s the [olloioing

identities are valid:sin (t + s) == sin t cos s -1- cos t sins, (2.11)sin (t - s) == sin t cos s - cos t sin s. (2.12)

Proof. Using the reduction formula (2.3), we reduce sineto cosine:

sin (t--j- s) = cos ( ; - (t +s) ) = cos ( ( ~ - t ) - s) .We now app1.y the formula for the cosine of a differ-

ence (2.2):sin (t+ s) =-0 cos ( ~ - t ) cos s +sin ( ~ - t ) sin s

== sin t cos S +COR t sin S(in the last equality, we have used the reduction formu-las (2.3) and (2.7)). To prove formula (2.12), it suffices


to represen t the difference t - s as t -1- (-s) and usc theproved identity (2.11) and the properties of the evennessof cosine and oddness of sine:

sin (t - s) == sin (t + (-s) == Rill t cos (-s)+ cos t sin (-s) === sin t cos s - cos t sin s. ~

4. Tangent of a Sum (a Difference).Corollary li. For any real numbers t and s, except for

n n nI===T-t- nk, S=--=T1-nm, t-r-s:-~=2-~-nn u, m, nEZ),the [ollouiing identity holds:

tan (t + s) =::: tan t+tan s i-tan t tan s (2.13)

Remark. The domains of permissible values of thearguments t and s are different for the right-hand and left-hand sides of (2.13). Indeed, the left-hand side is definedfor all the values of t and s except t + s === ~ -+- sin,nEZ, while the right-hand side only for the values of tand .': mentioned in Corollary G. Thus, for t = ~ andS == : the left-hand side is defined, while the right-handside is not.

Prooj. Let us apply the definition of tangent and for-mulas (2.11) and (2.1):

tan (t _~_ s) ==:-: sin (tt s) == sin t COg s+ cos t sin scos (t +s) cos t cos s - sin t sin s

Since cos t =t= 0 and cos s =t= 0 (by hypothesis), both thenumerator and denominator of the fraction can be dividedby cos t cos s, then

tan (t -t- s) -1sin t -t sin scos t cos s tan t + tan s

1 __ sin t sin s - 1 -- tan t tan s' ~cos t cos s

Corollary 7. For any real t and s , except for t ==n k n nT+ n , s==Ti- nm, t-s~2+nn (k, m, nEZ), the


following identity holds:tan t-tan $

tan (t-s) == 1+tan t tans (2.14)

Remark. As in (2.13), the domains of permissible valuesof the arguments of the right-hand and left-hand sidesof the identity are different. . .

Proof. It suffices to replace s by -s in (2.13) and makeuse of the oddness property of tangent. ~

Example 2.1.3. Find tan ( ~ + t) if tan t= : .~ Use formula (2.13), bearing in mind that tan ~ = 1.We have

1+tan ti-tan ttan ( ~ + t) =

ntanT+tan t

ni-tauT tan t

i+~43 =7. ~

1-4Example 2.1.4. Evaluate tan (arcsin ~ +arccos 153 ) ~ Let t = arcsin ~ , s = arccos ;3' Then, by the def-inition of inverse trigonometric functions (Sec. 1.4),we have:

sin t = 3/5, 0 < t < n/2,cos s = 5/'13, 0 < s < n/2.

Let us now find tan t and tan s, noting that tan t > 0 andtan s ; 0:

2 _ sin 2 t _ 9 _ ( 3 ) 2tan t- 1-sin2t -16- T '

1 ( 13 ) 2 ( 12 ) 2tan2s=---1 == - -1 = -cos2 S 5 5

(we have used formulas (1.9) and (1.10). Thereforetan t = : ' s = ~2 ; and we may use formula (2.13) for-01644

50 2.. Identical Transiormattons

the tangen t of a sum:3 12--t--4 5

3 121----4 fi

tan t + tan stun (t + s) == i-tan t tan s ------

245+12724. 12 -- fl. 7-tan (t-I- s) ~ -

15+48 G320-3fl == -we ~

Example 2.1.5. Evaluate

tan ( arccos ( - ;5 )+ arcsin ( - ~~ ) ) .~ \Ve use the properties of inverse trigonometric functions(see Items 1 and 2 of Sec. 1.4)

arccos ( --: ;5 )= rr- arccos ;5 '. ( 12 ) . 12

arcsin -13 == - arcsin 13 '

rl 7 12an set t ~ arccos 2fi' s =:-~ arcsin 13 to get

tan ( arccos ( - ;5 )+ arcsin ( ~ ~~ ) )== tan (n - t-s) ==- -tan (t + s),

cos t == 7/25, 0 < t < n/2,sin s == 12/13, 0 < s < n/2.

Proceeding as in the preeeding example we have: tan t ==:24 127 and tan s==-5- ,

24 1;l7+5

24 121---7-. 55. Cotangent of a SUID (8 Difference).Corollary 8. For any real numbers t and s, except t = sik,

s = ttm, t --1- s == sin (k, m, n EZ), the following identityholds true:

t (t + )_., cot t cot s-1co s -- .cot t+cot.,: (2.15 )


The proof Iollows from the definition of cotangent andIorrnulus (2.1) and (2.11):

eot(t+s):::.:::: cot (t+s) == costcoss-sintsinssin (t-+- s) sin t cos s+ cos t sin s

Since the product sin t sin s is not equal to zero since t =t=nk, S "* scm, (k, m EZ), both the numerator and denomi-nator of the fraction may be divided by sin t sin e. Then

cot t cot s-1cot t+cot s

cos t . cos s -1sint sins

C?S t + C?S Ssin t sin s

cot (t+s) == ------

Corollary 9. For any real values of t and s, except t = nk,s == scm, and t - s = rtn (k, m, n EZ), the jolloioingidentity is valid:

t(t-)- cottcots+1co s -- cot i-cot s (2.16)

(2.17)

(2.18)

The proof follows from identity (2.15) and oddness prop-erty of cotangent. ~

6. Formulas of the Sum and Difference of Like Trigo-nometric Functions. The formulas to be considered hereinvolve the transformation of the sum and difference oflike trigonometric functions (of different arguments) intoa product of trigonometric functions. These formulas arewidely used when solving trigonometric equations totransform the left-hand side of an equation, whose right-hand side is zero, into a product. This done, the solutionof such equations is usually reduced to solving elementarytrigonometric equations considered in Chapter 1. Allthese formulas are corollaries of Theorem 2.1 and arefrequently used.

Corollary 10. For any real numbers t and s the [olloioingidentities are valid:

. 2' t+s t-sSIn t +SIn s;:::: SIn -2- cos -2- ,. t 2 t-s t+ssin -Slns== 81n-2 - cos - 2 - .

The proof is Lased on the formulas of the si ne of asum and a difference. Let us write the number t in the

52 2. Identical Trnnsjormntions

t-l- s t - sfollowing form: t ~~ +-2-' , and tho uum her s illt+s t-s I I 11the form s === -2-..- -2-, and app y Iormu as (2. )

and (2.12):. t + s t - s t- t+ s . t - s (? 19)Sin t ~ Sin -2- cos -2-- cos -2- SIn -2-' -.J.' t+s t-s t+.~. t s- (2 20)

sm s:=: SIn -2- COR -2-- cos -2- SIn -2- .

Adding equalities (2.19) and (2.20) termwise, we get iden-tity (2.17), and, subtracting (2.20) from (2.19), we get iden-tity (2.18). ~

Corollary 11. For any real numbers t and s the [olloioingidentities hold:

t+s t-scos t + cos s ~~ 2 cos -2- cos -2- , (2.21)

t 2 t+s t-s (2 22)cos -coss:::::: - sln-2-sIn-2 - . .The proof is very much akin to that of the preceding

corollary. First, represent the numbers t and s as

t - t+s + t-s t+s t-s--- 2 2' 8=-2---2-'

and then use formulas (2.1) and (2.2):t + s t - s . t + s t -- s

cos t == cos -2- cos -2- - SIn -2- SIn -2- , (2.23)t+s t-s . t+s . t-s

cos S = cos -2- cos -2- + sm -2- SIn -2-. (2.24)Adding equalities (2.23) and (2.24) termwise, we getidentity (2.2'1). Identity (2.22) is obtained by subtracting(2.24) from (2.23) termwise.

Corollary 12. For any real values of t and s, exceptt = ~ + uk; s = ~ +nn (k, n EZ), the following iden-tities are valid:

tan t + tan s == sin (t+s)C09 t cos s '

tan t- tan s === sin (t-s) .cos t cos s

(2.25)

(2.26)

2.1. Addition Formulas

Proof. Let us use the definit.ion of tangent:t t + t sin t + sin.')an ans=----

cos t cos s

sintcoss+sinscost _ sin (t+s)cos t cos s - cos t cos s

53

(2.30)(2.31)

(we have applied formula (2.11) for the sine of a SUDl).Similarly (using formula (2.12, we get:

sin t sin stan t- tans= -----cos t cos s

sin t cos s-sin s cos t _ sin (t-s)cos t cos s - cos t cos s ~

7. Formulas for Transforming a Product of Trigono-metric Functions into a Sum. These formulas are helpfulin many cases, especially in finding derivatives and inte-grals of functions containing trigonometric expressions andin solving trigonometric inequalities and equations.

Corollary 13. For any real values of t and s the followingidentity is valid:

sin t cos s =+(sin (t +s) +sin (t -s. (2.27)Proof. Again, we use formulas (2.11) and (2.12):

sin (t + s) = sin t cos s + cos t sin s,sin (t - s) = sin t cos s - cos t sin s.

Adding these equalities termwise and dividing both sidesby 2, we get formula (2.27). ~

Corollary 14. For any real numbers t and s the followingidentities hold true:

1cos t cos s = 2 (cos (t + s) + cos (t- s)), (2.28)sintsins= ; (cos (t-s)-cos (t+s. (2.29)

The proof of these identities becomes similar to that ofthe preceding corollary if we apply the formulas for thecosine of a difference and a sum:

cos t cos s + sin t sin s = cos (t - s),cos t cos s - sin t sin s = cos (t + s).

54

2. Identical Transformations

Adding (2.30) and (2.31) tcrmwise and dividing both sidesof tho equality by 2, we get identity (2.28). Similarly,identity (2.29) is obtained from the half-difference ofequalities (2.30) and (2.31). ~

8. Transforming the Expression a sin t + b cos t byIntroducing an Auxiliary Angle.

Theorem 2.2. For any real numbers a and b such that a2 +b2 =#= 0 there is a real number qJ such that for any real valueof t the following identity is valid:

a sin t + b cos t ==Va2 +b2 sin (t + cp). (2.32)For rp, we may take any number such that

cos cp == a/Va2 + b2 , (2.33)sin q> == blVa2 + b2 (2.34)

Prooj. First, let us show that there is a number cp whichsimultaneously satisfies equalities (2.33) and (2.34). Wedefine the number cp depending on the sign of the num-ber b in the following way:

(2.35)b

2.2. Double, Triple, and Hal] Arguments 55

It remains to check that if the number ~ satisfies therequirements (2.33) and (2.34), then equality (2.32) isfulfilled. Indeed, we have:a sin t + b cos t

~Va2+b2( a sin r-} b cost)Va2 +b2 Va2 +b2==Va2 + b2 (sin t cos cp + cos t sin rp),

and, by virtue of (2.11), we geta sin t + b cos t ==Va2 + b2 sin (t + cp). ~

2.2. Trigonometric Identities for Double, Triple,and Half Arguments

1. Trigonometric Formulas of Double Argument.Theorem 2.3. For any real numbers ex the following

identities hold true:sin 2cx = 2 sin a cos a, (2.36)cos 2a == cos" a - sin" ex, (2.37)cos 2cx == 2 cos" a - 1, (2.38)cos 2a == 1 - 2 sin" a. (2.39)

Proof. Applying formula (2.11) for the sine of the sumof two numbers, we get

sin 2a = sin (ex + ex.) == sin a cos ex + cos ex. sin a= 2 sin a cos a.

It we apply formula (2.1) for the cosine of a sum, thenwe get

cos 2a = cos (ex + a) = cos a cos ex - sin a sin a= cos'' ex - sin" a.

Identities (2.38) and (2.39) follow from identity (2.37) wehave proved and the fundamental trigonometric identity(1.9):cos 2a == cos" ex - sin'' ex

== 2 cos" a - (cos" a + sin'' a) = 2 sin'' a - 1,cos 2a = cos" a - sin'' a

= (cos- a + sin" ex) - 2 sin" ex, = 1 - 2 sin" a. ~


Corollary 1. For any real values of a the following iden-tities are valid:

z t +cos 2acos a == 2 '

2 1-cos2aSIn a = 2

(2.40)

(2.41)Proof. Formula (2.40) follows directly from identity

(2.38). Analogously, formula (2.41) is obtained fromidentity (2.39). ~

2. Expressing Trigonometric Functions in Terms of theTangent of Half Argument (Universal Substitution For-mulas). The formulas considered here are of great impor-tance si.nce they make it possible to reduce all basictrigonometric functions to one function, the tangent ofhalf argument.

Corollary 2. For any real number ct, except a === 11: +2nn, nEZ, the [ollouiing identities are valid:

sina=ex.

2tanT

1+tan2 ~2(2.42)

(2.43)cosac=----1-tan2~

.2

1+tan2~2Remark. The domains of permissible values of the

arguments on the right-hand and left-hand sides of (2.42)and (~.43) differ: the left-hand sides are defined for allthe values of ex, while the right-hand sides only for thea/s which are indicated in the corollary.

Proof. By virtue of (2.36) and the fundamental trigono-metric identity (1.9), we have:

2 sin 5:... cos~sin ex = sin (2, ~ ) = ~ 2 2 sin !!:- cos !:..2 2

cos2~+sin"~2 2By hypothesis, cos ~ does not vanish, therefore both

the numerator and denominator of the fraction may be

2.2. Double, Triple, and II all Arguments 57

divided by cos- ~ , whencea2tan 2

t+tan2~2

2 sinacos a.

sin2 5!:-2sin a = ------

1-1----cos- :!-

2

We now apply (2.37) and the fundamental trigonomet-ric identity and get

cosa:=:.cos (2.~) :=:cos2!:.--sin2~222sin2~

1- 2cos2 !!:.- - sin2~ cos! ~ 1- tan2 !:...2 2 2 2 ~-cos2~ +sin2~ sin2~ 1+tan2 !:..2 2 1+ 2 2COS2~

2

nCorollary 3. For any real numbers cx, except ex == 2 +nk, a ~ n + 2nll (It, n EZ) the following identity holds:

tan ct :::::=

a.2 tan 2"

1-tan2 ~2(2.44)

Proof. Note tha t, by hypothesis, cos a does Hot vanish,and the condition of Corollary 2 is fulfilled. Consequently,we may use the definition of tangent and then dividetermwise identi ty (2.42) by identity (2.43) to get

2 tan ~ 1- tan~ !!:-sin a 2 2

tan a == -- == -----....cos ex. a. ex.1+tan2 2 1+tan2 "2

ex.2tanT

1- tan2 .!!:.2

58 ~. Identical Transformations

Remark. This formula can also be derived from (2.13).Indeed,

ex.2tanT

tan Cot = tan (~ + ~ ) = a .1-tan2 - 2

Corollary 4. For any real numbers a) except ex === sin(n EZ), the identity

(2.45)1-tan2 ~2

a2tanTcota~-----

is fulfilled.Proof. The values of the number a satisfy the require-

ments of Corollary 2 and sin a does not vanish, thereforeidentity (2.45) follows directly from the definition ofcotangent and identities (2.42) and (2.43):

t - tan 2 !!:- 2 tan ~cot ex, == C?S a. == 2_ 2

SUi a. a. a1+ta02 "2 1+ta02 2

ex,1=tan2 2

ex.2 tan "2

Example 2.2.1. Evaluate sin (2 arctan 5).-llIIIIII Let us use formula (2.42). If we set ex, = 2 arctan 5,then 0 < ex, < 1t and a/2 = arctan 5, and, by virtue of(2.42), we have

(J,

2 tan 2 2 tan (arctan 5) 2.5 5sin ex =::. ---- - 1+(tao (arctan 5)}2 = 26" == 13 ~

1+ tan 2 !!..2Example 2.2.2. Evaluate cos (2 arctan (-7.

a

2.2. Double, Triple, and Half Arguments 59

EX81upie 2.2.3. Evaluate tan (2 arctan i3).~Settillg a == 2 arctan 3, we get tan!!: == 3 > 1 and2Jt2" < ex < n. (Ising formula (2.44), we get

o:2tanT

tana==.;-----1-tan2 ~2

3. Trtgonometrle Formulas of Half Argument,Corollary 5. For any real number a the following identi-

ties are valid:

a + .. /1+cosacos 2"== - V 2 '

. a _. + -. / i-cos aSIn 2- - V 2 '

(2.46)

(2.47)

where the sign depends on the quadrant in which the pointp a/2 lies and coincides with the sign of the values containedon the left-hand sides of the equalities (in that quadrant).

Proof. Applying identity (2.40), we get

whence

1+cos (2,"*)2

1 +cos a2

Icos ~ I= 11 1+~os a .To get rid of the modulus sign, the expression cos ~should be given the sign corresponding to the quadrant

~ lies in; whence follows formula (2.46). Similarly(2.41) yields equality (2.47). ~ .

Corollary 6. .For any real number C't, except ex = n (2n +1) (n EZ), the following identities are valid

t a l/ i-cosaau-=2 1+cos a (~.48)


(the sign before the radical depends on the quadrant thepoint P a/2 lies in),

a sinatan 2 == 1+cos a. (2.49)

Proof. Identity (2.48) is obtained if we take into consid-eration that cos a, =t= 0 and divide identity (2.47) by(2.46) termwise. Further, by virtue of (2.36) and (2.40),we have:

. aex. SlllT

tan 2= a.cosT

2 sin s. COl .!!:-2 2 sin Ct

2 cos2..~ - 1+cos a. ~2

Corollary 7. For any real number ct, except ct == sin(n EZ), the following identity is valid:

t ex _ i-cos a.anT- sin ce (2.50)

Proof. In this case the condition sin ~ =1= 0 is ful-filled, therefore from (2.:::>6) and (2.41) there follows

2 sin2!:..2

a.cosT

. a.sln 2tan~==--2 2sin~ cos~2 2

_ 2 i-cos (2 (a/2)) == i-cos a. ~- sin (2 (a/2 sin ex. .......

Corollary 8. For any real value of ct, except a == 2nn,nEZ, the following identities hold true:

cot~ === + 1//" 1+~os a.2 - Y i-cos a (2.51)

(the sign before the radical depends on the quadrant thepoint P a/2 lies in)

a sin cecot T == i-cosa (2.52)

Proof. Since by hypothesis sin ~ * 0, formula (2.51)is obtained after terrnwise division of (2.46) by (2.47).

2.2. Double, Triple, and Half Arguments 61

From identities (2.36) and (2.41) it also Iol lows that

1+t:osctsina ~

2' a asui 2: cos 2 sin a

== i-cos a ~a

ex; cosTcot-==---

2 .ct 2'2aSID 2 SIn 2:

Corollary 9. For any real number a, except a === tin, nEZ,the following identity is valid:

cota~ i~eosa .SIll ex.

Proof. By virtue of (2.36) and (2.40), we have:a 2 cos2 .!!..-cosT 2

cot'!!:' ==---2 . ex. 2 sin !:- cos 5::..

SIll "'2 2 2

Example 2.2.4. Evaluate sin ~ , cos ~ , tan ~ .~Let us first apply identities (2.40) and (2.47) withex == ~ and take the radical to be positive since ~belongs to the first quadrant. We have:

V j(1+cos T -V 2+ 1/2cos ~ == 2 2I i - COS ~sin~=l 48. 2 12=1722

Further,

tan ~:=8

1CSIn"8 V2=l!2

-

1C 2+ V2cos RI (2- y"2)2 -

= 1 (2+ )/"2) (2- V 2) =-0 V2 -1. ~Example 2.2.5. Evaluate cos ( ~ arccos ( - +d )

... Let IX = arccos (-*), then cos IX = - io, 0 0 has beenfulfilled. For x == ~ -t- sin, nEZ, the left-hand side ofthe original equation is equal to

2 sin ( 3x + :) = 2 sin ( ~ +2nn ) = 2 cos sin,If n is an even number, then 2 cos sin === 2, and if n isodd, then 2 cos nn === -2. Hence, from the first seriestho solutions of the original equation are only the num-bers

J'tX == 12 -1- Znk; k EZ.

For x = ~~ + sui, nEZ, the left-hand side of the orig-inal equation is equal to

2 sin ( 3x +- ~ ) = 2 sin ( 3; + 3nn) = -- 2 cos sin:If n is au even number, then -2 cos tin == -2, and if nis odd, then -2 cos sin. ~ 2. Consequently, from the sec-ond series, the Iol lowiug numbers are the solutions ofthe original equ a lion:

x",-= ~~ +(21.+1)31, kEZ.Answer: x 0= ;;, +2nk, x = ~~ + (2k + 1) 31, k EZ. ~

When sol ving this problem, most errors occur owing toincorrect underst.a uding of the symbol V. As ill algebra,

3.2. Methods of Solving Equation.., H7

in trigonometry this radical sign means an arithmeticsquare root whose value is always nonnegative. Thisnote is as essential as the requirement that a nonnegativeexpression stand under the radical sign of an arithmeticroot. If for some values of the arguments these coudi tionsare not Iulfrlled , then the equality under considerationhas no sense.

3.2. Principal Methods of Solving TrigonometricEquations

1. Solving Trigonometric Equations by Reducing Themto Algebraic Ones. This widely used method consists intransforming the original equation to the form

F (f (t === 0, (3;4)where F (x) is a polynomial and f (t) is a trigonometricfunction; in other words, it is required, using trigonomet-ric identities, to express all the trigonometric functionsin the equation being considered in terms of one trigo-nometric function.

If Xl' x 2 , , tIm are roots of the polynomial F, thatis,

F (Xl) = 0, F (x2) = 0, ... , F (Xm ) == 0,then the transformed equation (3.4) decomposes into msimple equations

f (t) = Xl' f (t) == X 2, ., f (t) == x 1n For instance, if the original equation has the form

G (sin t, cos t) = 0,where G (x, y) is a polynomial of two variables x and y,then the given equation can be reduced to an algebraicequation with the aid of the universal substitution for-mulas by getting rid of the denominators during the p.ocessof t.ransformation. As it was stressed in Sec. 3.1, such 8.reduction requires control over the invertibility of allthe transformations carried out, and in case of violationof invert.ibil ity a check is required.

88 3. Trigonometric Equations and Systems

Example 3.2.1. Solve the equationcos 2t - 5 sin t - 3 == o.~By formula (2.39), we have 1 - 2 sin'' t - 5 sin t -3 == 0, or

2 sin" t + 5 sin t + 2 == O.We set x == sin t; then the original equation takes theform of an algebraic equation:

2x2 + 5x + 2 = O.Solving this equation we get Xl == -112, X 2 == -2. Allthe transformations carried out are invertible, thereforethe original equation is decomposed into t\VO simpleequations:

sint= - ~ and sint=2.The second equation has no solutions since I sin t r ~ 1,therefore we take sin t == -1/2, that is,

t=(_1)n+l ~ +nn, nEZ. ~Example 3.2.2. Solve the equation

tan x + tan ( ~ + x ) = - 2.~By formula (2.13) for the tangent of the sum of twoangles, we have:

n(~ ) _ tan T+tan x _ 1+tan xtan 4 + x - - 1 t 11: -an xi-tan T tan x

H 1+tan x .. ence, tan x + 1 t == - 2. Setting y ~ tan x, we- anx

get an algebraic equation:

.. + 1+y --2y 1-y-- ,or

y (1 - y) + 1 +.Y == -2 (1 - Y), y::;=; + V3,

3.2. Methods of Solving Equations ., ~H

consequently, tau x = + V~ that is,1t

x::=. + T+nn, nE Z.

Both series of solutions belong to the domain of permis-sible values of the original equation which was not re-duced under transformatio-n.

Example 3.2.3. Solve the equation(1 - tan x) (1 -t- sin 2x) = 1 + tan x - cos 2x.

... Note that the numbers x = -]- + nk, I: E Z, are notsolutions of the given equation, therefore \VO may con-sider the given equation on a smaller domain of per-missible values specified by the condition x =1= i -t- sik ,k EZ, and use the universal substitution formulas (2.42)and '(2.43) which are reversible transformations in thegiven domain:

. 2 tan xSI n 2x =~----1+tan2x '

1-tan2 xcos 2x == 1+ tan 2 x

We set y = tan x, then the given equation is reduced toan algebraic one:

( 2y ) 1-!l2

(1- y) 1 + 1+ y2 == 1+ y - 1+ y2 Since 1 -1- y~ =1= 0, this equa tion is equivalent to

(1 - y) (1 -t- y2 + 2y) = (1 + y) (1 + y2) - 1 + y2,whence, by successive invertible transformations, we get:

(1 - y) (1 -1- y)'J == (1 + y) (1 .1- y2) + (y - 1) (y + 1),(1 - y) (1 + Y)~ == (1 + y) (1 + y'J + y - 1),,(1 - y) (1 -t y)2 === (1 + y)'!.y,(1 -t- y)2 (1 - 2y) = O.

The roots of the obtained eqnation are: Yl =:-:: -1 andY2== 1/2. Consequently, the original equation is brokeninto two simple equations: tan x :='-1 and tan x == 1/2in the sense that. tho set of sol utions of tho original equa-

9J- 3. Trigonometric Equations and Systems

tion is a union of the sets of solutions of the obtainedequations, and we get

1t . t 1 -t EZ .....x=-= -4 Inn, x==arc an "2 -Jtn, n ~Example 3.2.4. Solve the equation

~ si 11 4x + 1{j si n 3 x cos x + 3 cos 2x - 5 == O.~Note that, by~ virtue of formulas (2.36) and (2.38),

2 sin 4x = 8 sin x cos x cos 2x== 8 sin x cos x - 16 sin" x cos x,

and the equation takes the form8 sin x cos x + 3 cos 2x - 5 == 0,

or4 sin 2x + 3 cos 2x = 5. (3.5)

Let us make use of the universal substitution formula. 2tanx 1-tan2x

Sin 2x == 1. +tan2 x' cos 2x = 1+tan2 xand designate y = tan x. Then the equation is trans-formed into an algebraic one:

8y 3-3y21+y2 .+ 1+y2 =5,

or 8y + 3 - 3y2 === 5 + 5y2, whence2 1 -0Y -Y+ 4 -

Consequently, y == 1/2 or tan x = 1/2, whence x =arctan ~ + nn, n EZ. It remains to check that no rootsare lost during the process of solution. Indeed, only thosex's might be lost for which tan x has no sense, that is,x === i- + nk, k E Z. Substituting these values into theleft-hand side of (3.5), which is equivalent to the origi-nal one, we get

4 sin (rr + 21tk) + 3 cos (n + 2nk) = -3.Consequently, besides x = arctan ~ + sin, nEZ, theequation has no other roots. ~

3.2. Methods 0/ Solving Equations 91

2. Other Methods of Reducing Trtgonometrfc Equationsto Several Simple Equations. Basically, we mean hereapplica tion of the formulas for transforming the sum ordifference of trigonometric functions into a prod uct (seeSec. 2.1, Item 6). However, it is often necessary first tocarry out additional identical transformations. In parti-cular, the left-hand sides of formulas (2.17), (2.18), (2.21),(2.22), (2.25), (2.26) contain the basic trigonometricfunctions (in the first power), therefore to use them, it is,for instance, useful to Pply formulas (2.40), (2.41) whichreduce the power of trigonometric functions in the givenexpression.

Example 3.2.5. Solve the equationsin'' x + cos" 3x === 1.~Applying identities (2.40) and (2.41), we get

~ - ~ cos 2x+{+ ~ cos 6x = 1,1

or "2 (cos 6x - cos 2x) = 0, whence, by virtue of iden-tity (2.22) we get

-sin 4x sin 2x = O.

The original equation has been broken into two equa-tions:

sin 4x = 0 and sin 2x == O.Note that the solutions of the equation sin 2x == 0 (x =nk/2, k E Z) are solutions of the equation sin 4x = 0(since sin 4 (nk/2) = sin 2nk = 0, k E Z), therefore itsuffices to find the roots of the equation sin 4x == O. Con-sequently, 4x == rui or:

x ~ :nn/4, n EZ. ~Example 3.2.6. Solve the equationsin x + sin 2x -t- 2 sin x sin 2x == 2 cos x + cos 2x.

92 3. Trigonometric Equation." and Systems

or

Sill x -t-- sin 2x - cos x - cos 2x - cos 3x =-= 0.sin x --+- sin 2x - (cos x -1- cos 3x) - cos 2x == O.

Applying formulas (2.2'1) and (:~.~()), we getsin x + 2 sin x cos x - 2 cos 2x cos x - cos 2x === 0,

or

sin x (1 + 2 cos x) - cos 2x (1 + 2 cos x) = 0,(sin x - cos 2x) (4 -~- 2 cos x) = o.

Thus, the original equation has been decomposed into twoequations:

(1) 1 + 2 cos x == 0 or cos x = -1/2,for which x = +~ + 2nn, nEZ,

(2) sin x - cos 2x == O.By formula (2.39), we transform this equation to theform

sin x - 1 + 2 sin~ x == 0and set y == sin x:

2y2 + Y - 1 == O.The quadratic equation thus obtained has two roots:Yl == -1, Y2 == 1/2. In the first case

sin x == -1,

In the second case

nor x:=:: - T-t- 2nn, nEZ.

sin x == 1/2, that is, x == (_1)1t ~ -1- sin, n EZ.Thus, the solutions of the original equation are writtenin the form of three series

x=+ 2; +2nn, x==-~+2nn,2

3.2. Methods of Solving Eqn ations

Example 3.2.7. Solve the equa tion

t . 4) 1 . t .+ 1au-..Jx+-.- -- co x -'-5-.SID X SlIl .. x

~Transform the given equation:sin 2x cos x 1 1 _ 0ros 2x sin x -t- sin:r - sin ~.1' - ,

or

sin 2x sin x-cos 2:r eos x _1_ sill 5.'X-sin x - 0sin x cos 2.r . r sin x sin ~).x --.

Applying identities (2.1) and (2.18), we get- cos 3x 2 cos 3x si n 2x

----:-+ :=: 0,sill X cos 2x sin x sin 5x

or

('.083x (sin 5x - 2 sin 2x cos 2x) === 0sin x cos 2x sin 5x '

_ cos 3x (sin 5x - sin 4x) :=: 0sin x cos 2x sin fix ..

and, again hy formula (2.18),3 2 x 9xcos X' SIn 2"" cos 2

. 2' r.: ==0.Sin x cos x sm ,)X

Note that if sin .x === 0 (i.e, x == nn, n E Z), then alsosin fix == 0, and the equality sin ~ == 0 means thatsin x == 0 (sin 2nn === 0, n EZ). Consequently, the do-main of permissible values of the given equation can bespecified by t\VO conditions:

cos 2x =F 0 and sin 5x =1= 0,and on this domain the original equation is decomposedinto two equations:

(1) cos 3x == 0,(2) cos ; x = O.Let us solve equation (1). We have 3x == ~ + nn

(n E Z), or x = ~ -f- ~~n, and we have to check whether

fl4 3. Trigonometric Equations and Systems

the constraints specifying the domain of perrnissihlevalues are met. For n == 3k, k EZ, the expressioncos (2 (~ + n;t) )= cos n(21~t:.!~ takes on val ues eq ual to1/2, for n = 3k + 1, k E Z, values equal to -1/2, and forn == 3k + 2, k EZ, values equal to -1, that is, cos 2x =1=o for all these val ues of x. Further, for n == Gk orn = 6k + 2, k E Z, the expression sin (!J( ~ + ~;)) = (5n 53tn) / 6kSIn "6+3" takes on values equal to 1 2, for n ~ ,-t-

1, k EZ, values equal to 1, for n == 6k + 3 or n == 6k -t-- 5,k E Z, values equal to -1/2, and for n == 6k + 4, k E Z,values equal to -1, that is, sin 5x =1= 0 for the indicatedvalues of x, and all of them belong to the domain ofpermissible values.

Consider now equation (2). We have:9 n + x :==~ + 21tn" nEZ,-2 x=-2 sen, or -9 n

and check whether the constraints cos 2.1: =1= 0 andsin 5x =I=- 0 are met. Notethatthe expression cos (2(; +2:n)) takes on one of the following nine values: cos ~qn,

1 iOn 14ft 22n 26n iOn- 2' cos 9' cos 9,1, cos 9' cos 9' cos T,cos ~n, none of them being zero. Similarly, we check tosee that the expression sin (fi (~ + 2~n)) does not vanisheither for any integral values of n, Thus, we have foundall the solutions of the original equation: x = ~ -+ Jl;,

1t 21tnX = n + 9' n E z. ~

Example 3.2.8. Solve the equation5 cos 3x -+_. 3 cos x = 3 sin 4x.~ Let us first note that if we apply twice formula (2.::lG) forthe sine of a double angle, we shall get the identity sin 4.T =-=2 sin 2x cos 2x == 4 sin x cos x cos 2x. Using this iden-tity and (2.54), we rewrite the given equation in the form5 (4 cos" x - 3 cos x) -t- 3 cos x == 12 sin x cos x cos 2x,

or

,'1.2. Methods of Soloint; Equations 95

cos x (20 cos" J~ - 15 -i-- ;) - 1~ sin x (1 - ~ Sill~X == 0,cos x (20 (1 - sin 2x) -12 - 12 sin x (1 - 2 sin 2x == 0,cos x (20 - 20 sin2:J~ - 12 - 12 sin x + 24 sin3x) == 0,

(6 3 r.::. 2 3 + 'J) - 0cos x . sin x - ~) sin x - .. Sf I] x _ -- ,cos a (0 sin2 x (sin x-i) + sin x (sin x - 1)

- 2 (sin x - 1 = 0,cos x (sin x - 1) (6 sin'' x + sin x - 2) == o.

Thus, the given equation decomposes into the

trigonometric functions problem solving approach

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