trigonometric

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TRIGONOMETRIC FUNCTIONS

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A. PanchishkinE.Shavgulidze '

TRIGONOMETRICFUNCTIONS(Problem-50lvingApproach)

MirPublishersMoscow'

Translated from Russianby Leonid Levant

First published 1988Revised from the 1986 Russian edition

Ha anaAuiic1:0M nsune

Printed in the Union of Soviet Socialist Republics

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/1I

rI

ISBN 5-03-000222-7 1I3p;aTeJIhcTBo HayKa., I'aaanaapenasnaa .pHaHKO-MaTeMaTHQeCKOiinareparypu, 1986

English translation, Mir Publishers,1988

FroD1 the J\uthors

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 universities, engineering colleges,and teacher-training institutes.

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 theoremsfrom 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 why 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.K. 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 hooks 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-Iepko (Moscow, Nauka, 1986); A Collection of Problems inMathematics for 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.1. 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 91.1. Radian Measure of an Arc. Trigonometric

Circle 91.2. Definitions of the Basic Trigonometric Func-

tions 181.3. Basic Properties of Trigonometric Functions 231.4. Solving the Simplest Trigonometric Equations.

Inverse Trigonometric Functions 31Problems 36

Chapter 2. Identical Transformations of TrigonometricExpressions 41

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

and Half Arguments 552.3. Solution of Problems Involving Trigonometric

Transformations 63Problems 77

Chapter 3. Trigonometric Equations and Systems ofEquations 80

3.1. General 803.2. Principal Methods of Solving Trigonometric

Equations 873.3. Solving Trigonometric Equations and Systems

of Eqnations in Several Un knowns 101Problems 109

Chapter 4. Investigating Trigonometric Functions 11:14.1. Graphs of Basic Trigonometric Functions 11:14.2. 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 1495.2. Solving Trigonometric Inequalities 156Problems 162

Answers 163

Chapter 1

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 1, cos 15), cos (sin 1), and 110cannot 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 that the degree (or sex age-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 in which it is the central angle; also calledcircular measure. Since the circumference of a circle ofa unit radius is equal to 2n, the length of the arc of 360is equal to 2n radians. Consequently, to 180 there corre-spend n radians. 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 many degrees are contained inthe arc of one radian'?.... We write the proportion:

If rr radians = 180,and 1 radian = x,

then x= 181)0 ~ 57.295780 or 5717'44.8". ~n:

Example 1.1.2. How many degrees are contained in1 f 35n: di "t 18 arc or 12 ra iansr

If n radians = 180,

d 35n: d'an t:r ra ians = x,

( 35n: )1then x = 12-.180 I Jt= 5250 ~Example 1.1.3. What is the radian measure of the arc

of 1984?

then

If rr radians = 180,and y radians = 1984,

st 1984 496n: 1y= 180 =45= 1145"Jt. ~

2. Trigonometric Circle. When considering either thedegree 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 initial point Al 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 arc clockwise is said to be negative(Fig. 1b).

We 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 with the

A2

Fig. 1

o

3. Winding the RealAxis on the TrigonometricCircle. In the theory of lJ(O,-1)trigonometric functionsthe fundamental role is Fig. 2played 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 API' taking the point A = Po as theintitial point 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.Then 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~lve~y. 8(0,/)

The trigonometric ClI'- ----.--cle will be denoted by S. ~According to the afore-said, ct-1,0)

--:---'-f---------::-t----+-'---S = {(x, y): x2 -+ y2 = 1}.


round the circle in the positive direction. We denote theterminal point of this path by P t and associate the num-ber 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 1t I. 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 P: R-+S consists in that the positive semiaxis is wound ontoS in the positive direction, while the negative semi axisis 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 = P f, then this point also corre-sponds to the numbers t + 2n, t - 2n: F = P t +2 n =]J t -2n 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 F,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 P have the form t + 2nk,where k is an arbitrary integer. Or in a briefer formula-tion: the full inverse image p_l (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 consideration.

Example 1.1.4. Find all the numbers t E R correspond-ing to the point F E S with coordinates (-V2l2,- V2/2) under the mapping P.

~ The point F actually lies on S, since


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, andf:::,XFO is a right isoscelestriangle, LXOF = 45 =n/4 radian. Thereforethe magnitude of the arcAF is equal to n + ~ = 5;,and to the point F therecorrespond the numbers5: + ze, k EZ, and only they. ~

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

y

Fig. 4

metric circle so that one of the vertices coincides with thepoint PI (see Fig. 4 in which N = 5)..... The vertices of a regular N -gon divide the trigonomet-ric circle into N equal arcs of length 2n/N each. Con-sequently, the vertices of the given N-gon coincidewith the points A, = P 2111, where l = 0, 1, .. 0' N - 1.. 1+1VTherefore the sought-for numbers t E R have the form


1 + 2~k , where k E Z. 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,l EZ, l being the remainder of the division of the integerk by N. It is now obvious that the equality 1 + 2~k =1 + 2~l + 2Jtm is true since its right-hand side con-

c

y

8=Pf3n/z

E=P..,sn!1

A=/l,x

Fig. 5

tains the numbers which correspond to the points P 2nl1+N

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

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

3n 3 3n~(a) 2=7;2:11, therefore, to the number 2 therecorresponds the point D with coordinates (0, -1), sincethe are AD traced in the positive direction has the mea-sure equal to i of a complete revolution (Fig. 5).

, 13n, n 13n(b) -2- = 32Jt +2' consequently, to the number 2"""there 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 Arc 15

then covering a quarter of revolution (n/2 radian) in thesame directiou.

(c) Let us represent the number -15n/4 ill the form2nle + to, where k is an integer, and to is a number suchthat O~ to < 2n. To do so, it is necessary and sufficientthat the following inequalities he fulfil led:

2nk~ -15n/4~ 2n (Ie + 1).

Let us write the number -15rt/4 in the form -3i rt=-4n + ~ , whence it is clear that k= -2, to = rt/4,and to the number t =!-15n/4 there corresponds apoint E = P 11/~ such that the size of the angle EOA isn/4 (or 45). Therefore, to construct the point P -1511/4' 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 - 17n =-2 ~rt=-2n-~--3n+'6 6 6 -

~ , and in order to reach the point F = P -1711/6 (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 = P11/2' C =P 11' D = P 311/2 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) PIO' (b) P 8' (c) P -8'....To answer this question, one must know the approxi-mate value of the number rr 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: rr ~ 3.141 592653589793238462643).

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 3 or that :n: < 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

~k < 10 < n (k:; 1) (1.4)Then the number of the quadrant in which the point PIO islocated will be equal to the remainder of the division ofthe number k +- 1 by 4 since a complete revolution con-sists of four quadrants. Making use of the upper estimate

n6:n: < 3.f , we find that "2 < 9.6 for k = 6; at the same. n (k+1) n7 0time, :n: > 3.1 and 2 ="2 > 3.13.5 = 1 .85.

Combining these inequalities with the obvious inequality

9.6 < 10 < 10.85,


we get a rigorous proof of the fact that inequality (1.4)is fulfilled for 1 = 6, and the point PIO lies in the thirdquadrant since the remainder of the division of 7 by 4is equal to 3.

In similar fashion, we find that the inequalities~


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 EH. is defined with the aid of a trigonometric mapping

P: H.~ S.

A

!I

l'ig. 6

o

Definition. Let the mapping P associate a number t E Rwith the poi II t PI on th c trigonometric circle. Then the

ordinate y of P t is calledthe sine of the number t andis 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 sin 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 closedinterval [- 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/the coordinates(x, y) of the point P t on the trig-

onometric circle satisfy the relationship :r2 + y2= 1,and consequently cos- t + sin'' t = 1.

Example 1.2.1. Find sin t and cos t if: (a) t = 3:n:/2,(b) t - 13:n:/2, (c) t = -15:n:/4, (d) t = -17:n:/6.

1.2. Definitions 19

.... In Example 1.1.6, it was shown thatP sn / 2=D(0, -1), P 13n / 2=B(0,1),

P_15n/4=ECV2/2, V2/2), P_17n/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) = Y2/2; sin (-17n/6) = -1/2,cos (-17n/6) = - VS/2. ~

Example 1.2.2. Compare the numbers sin 1 and sin 2..... Consider the points PI and P 2 on the trigonometriccircle: PI lies in the first quadrant and P 2 in the secondquadrant since n/2 < 2< rr,Through the point P 2 , wepass a line parallel to thex-axis to intersect the cir-cle at a point E. Then thepoints E and P 2 have equalordinates. Since LAOE = c ALP20C, E = P n - 2 (Fig. 7), a:consequently, sin 2 =sin (n - 2) (this is a partic-ular case of the reductionformulas considered below).The inequality n - 2 > 1is valid, therefore sin (n - Fig. 72) > sin 1, since bothpoints PI and P n-2 lie in the first quadr!mt, and whena 'movable point traces the arc of the first quadrantfrom A to B the ordinate of this point increases from to 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 > > 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 PIO liesin the third quadrant, while the point P 8 is in the secondquadrant. The signs of the coordinates of a point on thetrigonometric 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 < 0. ..

Example 1.2.5. Determine the signs of the numberssin (V5+ V"7) and cos (V5+ V7) ... From what was proved in Example L 1.8, it follows that

n< 115+ V"7 < 3n/2.Consequently, the point P l/[;+V7 lies in the thirdquadrant; therefore

sin (115+ V7)

1.2. Definitions 21

for the values t = ~ + sik, Yk 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. cThus, the function tan t is -=+-- .xdefined on the set of allreal numbers except the

nnumbers t = 2"+nk, k EZ.The function cot t is defined ILlon the set of all real num-bers except the numbers Fig. Ilt = nk, k EZ.

Graphical representation of the numbers tan t and cot twith 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 on this axis is equal to tan t. This follows from thesimilarity of the triangles OX tP t and OAZt and the defini-tion of the 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 line OPt and the tangent. The abscissa 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 = '!tn, nEZ, and the function cot t is not defined(Fig. 9).

In this graphical representation of tangent and cotan-gent, the tangent linesAB' 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.

1J

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.

!I !I

Fig. 10

Example 1.2.7. Determine the sign of the numbercot (V5 + V7) ..... In Example 1.2.5, it was shown that sin (V5 + V7) cos t 2 > 0,whence

Consequently, tan t] < tan t 2 (2) -'Jt/2 < t] < < t 2 < 'Jt/2. In this case,tan t] < 0, and tan t 2 > 0, therefore tan t] < tan t 2(3) -'Jt/2 < t] < t 2 ::;;;; 0. By virtue of Theorem 1.4,

sin t] < sin i. 0, 0< cos t] < cos t 2 ,therefore

sin t1 .-- sin t2cos t 1

'10 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 t l , t 2 E [0, n/2lsuch that t l < t 2 , the following inequalities are valid:

sin (cos tl ) > sin (cos t 2 ) , cos (sin t l ) > 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 =1= 0), we get

1+ tan2 t = _1_ (1.10)cos2 t '

where t=l=~ +nk, kEZ. 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 Phcorresponding to the real number t, lies.

Example 1.3.6. Compute cos t if it is known thattan t o=c 152 and t E(n, 3;) ..... From formula (1.10) we find:

144169 .

12 _ 1cos t -- 1+tan2 t 25

1+ 144Consequently, I cos t I = 12/13, and therefore eithercos t = 12/13 or cos t = -12/13. By hypothesis,t E (n , 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 sin- t (forsin t =!= 0) termwise, we get

1+ 2 _ 1cot t- -.-2- ,. Sj Il t (1.11)

where t =!= stk; k E Z. Using this identity, we can com-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

xA

Fig. 16

y

.D

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 fwd all real numbers t such that the ordinateof the corresponding pointP t is equal to m, To thisend, we draw a straight liney = m. and fwd the pointsof its intersection with thetrigonometric circle. Thereare two such points if cl-__+ r:-_~I m. I < 1 (points E and Fin 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/21. Indeed, when the real axis is being wound on thetrigonometric circle, the numbers from the interval[-n/2, n/21 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 01 Trigonometric Functions

Definition. The arc sine of a number m is a number to,-n/2~ to~ nl'!., such that sin to = m. The followingnotation is used: to = arcsin m (or sin? m).

Obviously, the expression arcsin m has sense only forI m I~ 1. By definit.ion, we have:

sin (arcsin m) = m, -n/2~ arcsin m~ n/2.The following equality holds true:

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

line y = m with the trigonometric circle can be written inthe form F == P 11-1., therefore all the solutions of the equa-

tion sin t = m, I m I~ 1,!I are given by the formulas

t = arcsin m -+- 2nk, k EZ,t = rt - arcsin m 1- 2nk,

k EZ,a; which are usually united

y=-I!2 into one formula:-4r---:-::I--~II-~-:':

t = (-1)" arcsin m -+- nn,n EZ.

l"ig. 17 Example 1.4.1. Solve theequation sin 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, IOE I = 1,that is, LXOE = 30. Consequently, LAOE is measuredby an arc of -n/G radian, and E = P -11/6' Thereforearcsin (-1/2) = -nI6, and the general solution of theequation sin t = -1/2 has the form t = (_1)n+1 ~ -+-rtzz, n EZ. ~

Example 1.4.2. What is the value of arcsin (sin 10)?.... In Example 1.1. 7 iL was shown that the point PIO liesin the third quadrant. Let t = arcsin (sin 10), thensin t = sin 10 < 0 and -n/2~ t~ n12. Consequently, thepoint PI lies in the fourth quadrant and has the same or-


dinate as the point Pl O; therefore P, = P n - l o (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,

yy

.x~m x~mIml~1 1m I>'!

Fig. 18

o

Fig. 19

A

it is necessary to set k = 2. Indeed,- n/2 < 3n - 10 < 0

(these inequalities follow from the estimate 3.1 < nO, and this point can be written in the form

E = Pt., where 0:;::;;;; to:;::;;;; n,Definition. The arc cosine of a number m is a number to,

lying on the closed interval [0, rr], such that its cosine isequal to m. The arc cosine of the number m is denoted byarccos m (or cos -' m).:J-()1644


Obviously, the expression arccos m has sense only forI m I~ 1. By deiinition,

cos (arccos m) = m, 0~ arccos m.~ n.Note that the lower point of intersection coincides with

the point F = P -to. Therefore the general solution of theequation cos t = m, m E

y [-1, 1] has the formt = + arccos m. + Zstk, k EZ.Note also that the fol-lowing equality holds:

A arccos (-In),---t----e:r- = n - arccos m.

Example 1.4.3. Find thevalue of arccos (cos 5)....... Let t = arccos (cos 5),then O~ t ~ n , andcos t = cos 5~ The point P"

Fig. 20 1 h Iies in t e fourt 1 quadrant(Fig. 20) since the inequal-

ities 3n/2 < 5 < 2n hold, therefore the point P t, lyingin the upper half-plane, must coincide with the pointP -5 which is symmetric to p." with respect to the axisof abscissas, that is, t = -5 + Zstk, The condition()~ t~ rt will be Iulfill ed if we take k = 1, thereforearccos (cos f)) = 2n - 5. ~

Example 1.4.4. Prove that if ()~ x ~ 1, thenarcsin x = arccos V1 - x2 4IlIII Let t = arcsin x. Then ()~ t~ n/2 since sin t =.T~ O. Now, from the relationship cos" t + sin" t = 1we get: I cos t I =~V 1 - x2 . But, bearing in mindthat t E [0, n/2l, we have: cos t = Vi - x2, whencet =, arccos V 1 - x 2-. ~

3. Solving the Equation tun t = m, Arc Tangeat.To solve the equation tan t = m, it is necessary to findall real numbers t, such that the line passing through theorigin and point P t intersects the line AB': x = 1 ata point Zt with ordinate equal to m (Fig. 21). The equa-tion of the straight line passing through the origin and P tis given by the formula y = TnX. For an arbitrary real


Fig. 21

!I

number m there are exactly two points of intersection ofthe line y = mx with the trigonometric circle. One ofthese points lies in the right-hand half-plane and can berepresented in the form E = P to' where to E(-n/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 thefollowing equalities hold:

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 m.,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 -IT lies inthe fourth quadrant, tan (t - rr) = tan t = -2/5,and the condition -n/2 < t - st < n/2 is satisfied.Consequently, the number t - n satisfies both condi-tions defining arc tangent, i.e.

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


Example 1.4.6. Prove the identitysin (arctan x) = xIV 71-+-'--X".2.

.... Let t = arctan x. Then tan t = x, and -n/2 < t 0 since -n/2 < t < n/2.By virtue of (1.10), 1Icos2 t = 1 + tan'' t = 1 + x2 orcos t = 1/V 1 + x2 , whence sin t = tan t cos t =xV 1 + x2 ~

It is dear from the above examples that when solvingproblems, it is convenient to use more formal definition ofinverse trigonometric functions, for instance: t = arccos Inif (1) cos t = In, and (2) 0~ t~ rr. To solve problemson computation involving inverse trigonometric Iunc-tions, it suffices to rememher the ahove definitions andbasic trigonometric formulas.

PROBLEMS

1.1. Locate the points hy indicating the quadrants:(a) Pn, (h) P 10-t' 2' (e) PV26+V 2'1.2. Given a regular pentagon inscribed in the trigono-

metric circle with vertices zl , = P 2 n k / r"k = 0,1,2,3,4.Find on which of the ares joining two neighhouring ver-tices the following points lie: (a) PIO' (h) P -11' (e) P12'

1.3. Given a regular heptagon inscribed in the trigono-metric circle with vertices B k = P 2+2 n k/ 7 ' k = 0, 1, 2, 3,4, 5, 6. Find on which of the arcs joining the neighhouringvertices the point PY-29 lies.

1.4. Prove that there is a regular N-gon inscribed inthe trigonometric circle such that its vertices include thepoints PY-2Hn/ll' PY-2+7n/13' Find the least possiblenumber N.

1.5. Prove that any two points of the form P nx, 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 PIX and P['J to he vertices of one and the same regu-lar N-gon inscribed in the trigonometric circle.

1. 7. Compare the following pairs of numhers:

(a) sin 1 and sin ( 1.+ ~n ) ,

Problems 37

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

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

l- (a>O, a=#=1),

oga

(b) ( V3-1 _ )-1?V5-~/3

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

. 1001:n: 123:n: .(a) Sill -6-' (b) cos -4-, (c) Sill (-117n/4),(d) cos (- 205n/6).1.12. Determine the sign of the number tan H.1.13. Evaluate:

l011n 101l1:n:(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-1

~ cos ( a + 2~k ) = 0.k=O

. llt 13t1.15. Prove that the Junction f (t) = tan 34+cot 54is periodic and find its fundamental period.

1.16. Is the function f (t) = sin (2x + cos CV2 x))periodic?

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

1.18. Prove that the function tan (V2x) + cot (V3x) isnot periodic.


1.19. Prove that the function f (t) = sin 3t + cos 5tis periodic and find its fundamental period.

1.20. Provo that the function f (x) = cos (V Ixl z) isnot periodic.

1.21. Find the fundamental period of the function:(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 + cot 3x +4 sin x cos 2x is odd.

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

1.27. Represent the function y = sin (x + 1) sin" (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 + b sin t is odd.In Problems 1.30 to 1.32, without carrying out com-

pu tations, determine the sign of the given difference.1.30. (a) sin 2; - sin 1(~ , (b) 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 [-n/2, OJ?

1.34. Is the function sin (cos t) increasing or decreasingon the closed interval ln, 3n/2J?

1.35. Prove that the function tan (cos t) is decreasingon the closed interval [0, n/21.

1.36. Is the function cos (sin (cos t)) increasing or de-creasing on the closed interval [n/2, rt]?

In Problems 1.37 to 1.40, given the value of one fUIIc-tion, find the values of other trigonometric functions.

1.37. (a) sin t = 4/5, n/2 < t < n ,(b) sin t = -5/13, :It < t < 3n/2,(c) sin t = -0.6, -n/2 < t < 0.1.38. (a) cos t = 7/25, < t < n/2,(b) cos t = -24/25, n < t < 3n/2,(c) cos t = 15/1?, 3n/2 < t < 2n.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, n < t < 3n/2,(b) cot t = -5/12, 3n/2 < t < 2n.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 = -112 be-

longing to the closed interval [-2n, 6nl.1.43. Solve the equations:(a) tan t = 0, (b) tan t = 1, (c) tan 2t = V3,(d) tan 2t = - y3, (e) tan (t - ~ )- 1 = 0,(f) V3 tan ( t + ~ ) == 1.1.44. Compute:(a) arcsin 0+ arccos + arctan 0,

. 1 V3 V3(b) arcsin z+arccos -2-+ arctan -3-'V3 '-3 V3(c) arcsin _.-2- +arccos ( - ~ )-arctan ( - -3-) .

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) =

1IV1 + .:t2


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

and arccos x =x1.47. (a) arcsin x = arctan -Y--=-=1-=X=--2x

arccot y for O~x< 1,1-x2

Y1-x2(b) arcsin x = arccot for < x~ 1,x

1(c) arctan - = arccot xx

. 1 x= arCSIn ./ . arccos

v 1+x2 y1+x2 '1

arccot - = arctan xx

. x 1= arCSIn /: = arccos

v 1+x 2 Y 1+x 2for x> o.

. 3 12 51.48. Express: (a) arcSIll5" , (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 ( ~ ;4 ),(C) arccot ( ~ ~) in terms of values of each of thethree other inverse trigonometric functions.

1.50. Find sin a if tan a = 2 and n < a < 3n/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 from 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 formulas 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 sum. 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-formations of trigonometric expressions.

1. The Cosine of the Sum 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 /

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.

To 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 fa to i; that is, such that I t 1 - t2 I =I ., ----: t 4 I, ~nd let PI" e.; PI., P I. be points on thecoordinate circle corresponding to those points. Then thearcs PI,p t and PI1Pt~ are congruent (Fig. 22). Hence it

2.1. Addition Formulas 43

follows that the chords P t,P t. and P t,P t. are also con-gruent: I P t,Pt, I = I Pt,P t I

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

cos (t + s) = cos t cos s - sin t sin s, (2.1)

Fig. 23

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

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

= (cos s, -sin s).

Proof, Consider the point Po of intersection of the unitcirc]e and the positive direction of the x-axis, Po = (1,0).Then the corresponding points Pit P t+s' and P -s are im-ages of the point Po whenthe plane is rotated aboutthe origin through anglesof t, t + s, and -s radians(Fig. 23).

By the definition of thesine and cosine, the coordi-nates ofthe points Pit P t +s'and P -s will be:

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+s: IPoP t+ s i2 = (1-eos (t + S))2 + sin'' (t + s) =1 - 2 cos (t + s) + GOS2 (t + s) + sin" (t + s) or, hyvirtue of the fundamental trigonometric identity,


On the other hand,I P -SPt 12 = (cos S - cos t)2 -+- (-sin s - sin tr

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

Here we have used the fundamental trigonometric identi-ty (1.9) once again. Consequently, from the equalityI PoP t + s 12 = I P-sP t 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 -+- (-sand apply Theorem 2.1:

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

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

cos (t - s) = cos t cos s + 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 -+- sin 0 sin t = cos t,that is, relying on the addition formula (2.2). When 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 ~ = cos 15.

2.1, Addition Formulas 45

.... 8 rr rr Jt f' f Illlce - = - -- - we cet . l'OmOrlllU a12 3 4' '"t = n/3 and S = 'JTFl:

(2.2) for

rr (Jt Jt) Jt Jt I Jt . rtcos 12 = cos "3 - 4"" = cos "3 cos 4"" -1- sin "3 SIll 4""

1 yi V3 v,.2 ~/2+ ,/7;="'2'~+-2-'-2-= 4 . ~2. Reduction Formulas.Corollary 2. For any real number t we hare

(2.7)

(2.4)

(2.3)

(2.5)(2.6)

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

cos (rt - t) =. - cos t,cos (n + t) = -cos t..

Proof. Let 11S use formula (2.2):( Jt ) Jt . Jt . tcos "'2 - t = cos "'2 cos t + SIll :2 sin

= Ocos t +- 1 sin t = sin t,cos (rr - t) = cos n cos t + sin rt sin t

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

( Jt ) Jt . st .cos "'2-1- t = cos "'2 cos t - SIll "'2 SIll t=Ocost-1.sint= -sint,

cos (n + t) = cos n cos t - sin n sin t= (--1)cos t-Osin t=-cos t.

These equalities prove identities (2.4) and (2.5). ~Corollary 3. For any real number t we have

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

sin (n- t) == sin t,sin (n +- t) = - sin t.

(2.8)(2.9)

(2.10)


tan ( ~ + t ) = - cot t, t =/= stk,cot ( ; + t ) =~ - tan t,

t =/= ; -t- sck,

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

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

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

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

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

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

obtain reduction formulas for cos ( 3; + t) , cos (2n- t),sin (~n + t), sin (2n - t). The reader is invited to dothis as an exercise.

Example 2. J.2. Derive the re(!llCtion formula forcos ( ~n - t) ..... Using formulas (2.6) and (2.3), we have

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

Corollary 4. The reduction formulas for tangent andcotangent are

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

tan (n-t) == -tan t ; t =/= ; + stlc .cot (n- t) = --cot t, t =/= sik (k E Z).


Proof. All these formulas can be obtained from Lhedefinition of the functions tan t and cot t applying theappropriate reduction formulas. ~

Note that the reduction formulas together with theperiodici 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, :rt/4l.

To facilitate the memorizing of the reduction formulas,the following mnemonic rule is recommended:

(1) Assuming that 0 < t < ~ , find in which quad-rant the point P 11k ,k EZ, lies, determine the sign

Tt

the given expression has in this quadrant, and put thissign before the obtained result.

(2) When replacing the argumentrt + tor 2:rt - t by i,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 following

identities are valid:sin (t + s) = sin t cos s + cos t sin s, (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+8)) = cos (( ~ -t) -8).We now appl.y the formula for the cosine of a differ-

ence (2.2):sin (t + 8) = cos ( ~ - t) cos s + sin ( ~ - t ) sin 8

= sin t cos 8 + cos t sin 8(in the last equality, we have used the reduction formu-las (2.3) and (2.7)). To prove formula (2.12), it suffices

48 2. 1dentical Transformations

to represent the difference t - s as t -+ (-s) and use theproved identity (2.11) and the properties of the evennessof cosine and oddness of sine:

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

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

n n nt=T+nk, s=-~T-+nm, H-s"=T,1-nn (k, m, nEZ),the following identity holds:

tan (t -+ s) = tan t+tan s1-tan t tan s (2.13)

Remark. The domains of permissible values of thearguments t and .'I 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 -+ .'I = ~ -+ nn,nEZ, while the right-hand side only for the values of tand s mentioned in Corollary G. Thus, for t = ~ and.'I = ~ the left-hand side is defmed, while the right-handside is not.

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

t (t l ) __ sin (t+s) sin t cos s+cos t sin sanT .'I - ----7--;--7--cos (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 .'I, then

tan (t-\-s)-jsin t -+ sin scos t cos s

1 __ sin t sin scos t cos s

tan t+tan s- 1 -- tan t tan s

Corollary 7. For any real t and .'I, except for t =n k n nT-+:n: , s=T+:n:m, t-s=:r-+nn (k, m, nEZ), the


following identity holds:tan t-tan s

tan (t-s) = 1+tanttans (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 t1-tan ttan ( ~ + t) =

:n;tanl;+tan t

:n;1-tanT tan t

31+1;

--""'3- = 7. ~1-1;

Example 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)2tan2 s = - - - 1 = - -1 = -coss 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. Tdentical Trans iormattons

the tangent of a sum:

tan t+ tan 8tun(t+s)= 1-tanttan8

15+4820-3R

Example 2.1.5. Evaluate

tan ( arccos ( - ;5 )+ arcsin ( - ~~ ) ) .~ We 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 = - arcslIli3 '

245+1272412-- 5 7-tan (t-:-s) =-

rl 7 . 12an set t = arccos 2"5' s -r-: arcslIli3 to get

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

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

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

24 127+5

24 121---7- ' 5

5. Cotangent of a Sum (a Difference).Corollary 8. For any real numbers t and s, except t = silt,

s = :rrm, t s = :rrn (k, m, n EZ), the following identityholds true:

cot t cot 8-1cot (t+s)= +cot t cot .~ (2.15)


The proof follows from the defmition of cotangent andformulas (2.1) and (2.11):

. t (t+ )= cot (t+s) = costcoss-sintsinsco s . ( ') . + ..Sill t- T S Sill t COR S COS t sin s

Since the product sin t sin s is not equal to zero since t =/=silc, s =/= sim; (k, m EZ), both the numerator and denomi-nator of the fraction may be divided by sin t sin s. Then

cot t cot s-1cot t+cot s

COS t . cos s -1sint sins

C?S t + C?S SSill t sin s

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

Corollary 9. For any real values of t and s, except t = stk,s = nm, and t - s = sen (k, m, n EZ), the followingidentity is valid:

t(t-)- cottcots+1co s -- cot t-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 followingidentities are valid:

. . 2' t+s t-sSill t + Sill S = sm -2- cos -2- ,

. t . 2' t-s t+sSill - sin s = sm -2- cos -2-

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

52 2. ldentical Transformations

f II . f t -1- s + t - s d I J .o owing orm: t = -2- --2-, an t 10 uum lor Ii 111t+s t-s I 11the form s = -2-- -2-, and apply formu as (2. )

and (2.12):. t . t+s t-s 1- t+s. t-s (2 19)SIn = Sin -2- COS -2-- COS -2- SIn -2-' ... . t+s t-s t+s. t-s (2 20)SIn S=SIn -2- COS -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 followingidentities hold:

t+s t-s 1cos t + cos s zr-: 2 cos -2- COS -2- , (2.2 )

t 2 t+s . t-s (2 22)cos - cos s = - SIn -2- slll-2- .The proof is very much akin to that of the preceding

corollary. First, represent the numbers t and s ast = t+s + t-s t+s t-s

2 2' s=-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- - sm -2- sm -2- , (2.23)

t+s t-s . t+s . t-s I.cos s = cos -2- cos -2- +sin -2- Slll -2-. (2.2*)

Adding equalities (2.23) and (2.24) termwise, we getidentity (2.21). Identity (2.22) is obtained by subtracting(2.24) from (2.23) termwise.

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

sin (t+s)tan r-j-tan s e- costcoss 'tan t- tan s = sin (t-s)

cos t cos s

(2.25)

(2.26)

sin (t+s)cos t cos s

2.1. Addition Formulas

Proof. Let us use the definition of tangent:t t + t sin t + sin san ans=----

cos t cos s

sin t cos s+sin s cos tcos t cos s

53

sin (t-s)cos t cos s

(2.30)(2.31)

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

sin t sin stan t - tan s = -----cos t cos s

sin t cos s-sin s cos t- 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) termwise 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+ b2sin (t+ rp). (2.32)For qJ, we may take any number such that

cos qJ= a/Va2 +b2 , (2.33)sin rp = b/Va2 +b2 (2.34)

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

(2.35)b

2.2. Double, Triple, and Half Arguments 55

It remains to check that if the number !jJ 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 t+ b cos t)Va2+b2 Va2+b2= Va2 + b2(sin zcos q.+ cos t sin !jJ),

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

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 2ex = 2 sin ex cos ex, (2.36)cos 2ex = cos" ex - sin'' ex, (2.37)cos 2ex = 2 cos- ex - 1, (2.38)cos 2ex = 1 - 2 sinex. (2.39)

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

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

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

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

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

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

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


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

I+cos 2acosza= 2 (2.40)

I-cos 2asinza = 2 (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 since they make it possible to reduce all basictrigonometric functions to one function, the tangent ofhalf argument.

Corollary 2. For any real number a, except a = n +2nn, nEZ, the following identities are valid:

(2.42)a2tanT

1+ tan2 !!--2sina=-----

(2.43)I-tan2 -.::..

2

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

arguments on the right-hand and left-hand sides of (2.42)and (2.43) differ: the left-hand sides are defined for allthe values of a, while the right-hand sides only for thea

ls which are indicated in the corollary.Proof. By virtue of (2.36) and the fundamental trigono-

metric identity (1.9), we have:2 sin -.::.. cos-'::"

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

By hypothesis, cos ~ does not vanish, therefore boththe numerator and denominator of the fraction may be

2.2. Double, Triple, and lJ all Arguments 57

IX2 tan "2

1+ tan2 .!!:.-2sin GG = ------

IXdivided by COs2 T' whence2 sin IX

cos IX

1-1----C082 !!'-

2

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

COSGG=cos (2, ~) =cos2 ~ -sin2 ~sin2~

1- 2

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

1+2 2

cos2!!:--2

Corollary 3. For any real numbers GG, except GG = ~ +sik; GG = rr+ 2nn (It, n EZ) the following identity holds:

(2.44)IX

2tanT

1-tan2~2

tan GG =-----

Proof. Note that, by hypothesis, cos a does not vanish,and the condition of Corollary 2 is fulfilled. Consequently,we may use the definition of tangent and then dividetermwise identity (2.42) by identity (2.43) to get

2 tan ~ 1-tan2 ~sin IX 2 2tan GG = -- = ----~cos IX 1+ tan2 ~ 1+ tan2 !!:--

2 2IX

2tanT

1- tan2 !!:--2

58 :!. Identical Trans iorm.ations

Remark. This formula can also be derived from (2.i3).Indeed,

IX2tanT

tan ex: = tan ( ~ + ~ ) = -----l-tan2 .!!:.-2

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

(2.45)l-tan2..!!:.-2

IX2tanT

cotex:=-----

25-2"(3

2 tan (arctan 5)1 + (tan (arctan 52

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

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

l-tan2 .!!:.- 2tan..!!:.-cot ex: = c~s IX = 2_ 2

srn IX IX IX1+tan2 T 1+tan2 2

IX1=tan2 T

IX2tanT

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

IX2tanT

sinex:=-----1+tan2..!!:.-2

Example 2.2.2. Evaluate cos (2 arctan (-7)).IX

.... Let us denote ex: = 2 arctan (- 7), then tan "2 = -7and -n

23t)

2.2. Double, Triple, and Halt Arguments 59

Example 2.2.3. Evaluate tan (2 arctan 3).~Setting a = 2 arctan 3, we get tan'!!: = 3 > 1 and2

~ < a < rt. Using formula (2.44), we geta2tan 2tana=-----

a1-tan2 23. Trigonometric Formulas of Half Argument.Corollary 5. For any real number a the following identi-

ties are valid:

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

. a _ + .. / 1-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

COS2~2

1+cos a2

Icos ~ I= V 1+~os aTo 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 a, except a = 11: (212 +1) (12 EZ), the following identities are valid

a / 1-cosatan 2= + l 1+cosa (2.48)


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

a sinatan-2 = i+ cosa (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(2.4G) termwise. Further, by virtue of (2.3G) and (2.40),we have:

. aa smT

tanT= acosz

2' a asmT c0l2

-

2 cos'' s.2

sin ai+cosa . ~

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

t a i-cos aanT=-s""'in-a- (2.50)

Proof. In this case the condition sin!!:.... =1= 0 is ful-2filled, therefore from (2.36) and (2.41) there follows

. as1ll2

acos2

2 sin2!!:....2

2 . a a8m 2 cosT

_ 2 i-cos (2 (a/2))- sin (2 (a/2)) = i--:cosa . ~sm

(2.51)

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

cot !!:.... = + 1/ i+cos a2 - Y i-cos a

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

cot!!:- _ sin a2 - i-cosa (2.52)

Proof. Since by hypothesis sin ~ =1= 0, formula (2.51)is obtained after terrnwise division of (2.46) by (2.47).

2.2. DOllble, Triple, and Half Argllment.~ 61

From identi ties (2.36) and (2.41) it also follows that

1+cos asina . ~

sin ai-cos a . ~

2sin ~ cos~2 2

2 . a a8l1lT cos T

. aslllT

acos-

a 2cot-=---

2 .a 2'2aSill "2 sin T

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

cota= 1+cosasin a .

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

Example 2.2.4. Evaluate sin ~ , cos ~ , tan ~~Let us first apply identities (2.46) and (2.47) withex = ~ and take the radical to be positive since ~belongs to the first quadrant. We have:

... ( 11:V 1 + cos 4 V2+ l/2cos ~ = 2 2

1 1 - COS ~sin2= 1 !18 2

12=1722

Further,

tan~=8 11:

Sill 811:

cos sI (2- Y2)2 -

= 1 (2+Y2) (2-l/2) =V2 - 1. ~Example 2.2.5. Evaluate cos ( ; arccos ( - *)).

... Let ex = arccos ( - *), then cos ex = - *' 0 0, "sin' (3x + :) > 0. If we squareboth sides of the equation, then, on the given domain,the original equation (3.1) is equivalent to the followingequation:

4 sinZ (3x+ ~ ) = 1+8 sin 2x cos- 2x. (3.2)However, if one does not take into account the domain ofpermissible values, then, although the roots of the origi-nal equation (3.1) are also the roots of equation (3.2),but all the roots of (3.2) will not necessarily be the rootsof (3.1). Therefore on finding all the roots of (3.2) wehave to choose those which will be the roots of the orig-inal equation. Applying formula (2.41), we get

i-cos (6x+ ~) 1Jsin- (3x+ :) = ='2(1+sin6x),

2

using formulas (2.36), (2.27), we have8 sin 2x cos- 2x = 4 cos 2x (2 sin 2x cos 2x)

= 4 cos 2x sin 4x = 2 (sin 6x + sin 2x).Therefore equation (3.2) may be rewritten as follows

2 + 2 sin 6x = 1 + 2 sin 6x + 2 sin 2x,or

. 2 1sin X= 2' (3.3)

86 3. Trigonometric Equations and Systems

For a further investigation, the solutions of this equationshould be conveniently written in the form of two seriesof solutions (but not united, as usual, into one; seeSec. 1.4):

it 5it Zx-=12+ nn , x~12 f-nn, nE .

Since equation C-3.3) is equivalout to equat.iou (3.2), wehave to check whether all of its solutions are the solu-tions of the original equation.

Substituting the found values of x into the right-handside of the original equation, we get the number 2, thatis, the condition 1 + 8 sin 2x cos'' 2x > 0 has beenfulfilled. For x = n+ nn, 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 sin = -2. Hence, from the first seriesthe solutions of the original equation are only the num-bers

itx=T2-j-2nk, kEZ.

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

2 sin ( 3x +T) = 2 sin ( 3; + 3nn) = -- 2 cos sin:If n is an even number, then -2 cos sui = -2, and if nis odd, then -2 cos sen = 2. Consequently, from the sec-ond series, the following numbers are the solutions ofthe original equation:

5itx cc= 12 -+- (2k + 1) rt, k EZ.

Answer: x c= 1~ -+ 2:nk, x = ~~ -1- (2k + 1) n, k EZ. ~When solving this problem, most errors occur owing to

incorrect underst.anding of the symbol V. As in algebra,

3.2. Methods of Solving Equations R7

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 conditionsare not fulfilled, 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, , .Tm 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 m 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 pocessof transformation. As it was stressed in Sec. 3.1, such 1\reduction requires control over the invertibility of allthe transformations carried out, and in case of violationof invertibility 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 = 0.We set x = sin t; then the original equation takes theform of an algebraic equation:

2x2 + 5x + 2 = 0.Solving this equation we get Xl = -1/2, X 2 = -2. Allthe transformations carried out are invertible, thereforethe original equation is decomposed into two simpleequations:

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

t=(_1)n+1 ~ +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:

tan ( ~ + x) =n:

tanT+tanxn:

i-tan T tan x

i+tan xi-tan x

H i+tan x .ence, tan x + 1 t = - 2. Settmg y = tan x, we- anx

get an algebraic equation:..

Y + i+Y = _ 2i-y ,or

y (1 - y) + 1 +.y = -2 (1 - y), y = + V;r,

3.2. Methods of Solving Equations .. ~H

consequently, tan x = + Y3, that is,n

x=+T-+Jtn, nEZ.

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

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

... Note that the numbers x = -i -+ stk; lc EZ, are notsolutions of the given equation, therefore we may con-sider the given equation on a smaller domain of per-missible values specified by the condition x =f=. i -+ stk .le E Z, and use the universal substitution formulas (2.42)and (2.43) which are reversible transformations in thegiven domain:

. 2 tan xSIn 2x == -:-,......,...-~1+tan2x '

1-tanZ xcos2x= 1+tan2x'

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

( 2y ) 1- y2(1- y) 1 -+ 1+ y2 = 1 -+ y - 1+ y2

Since 1 -+ y2 =f=. 0, this equation is equivalent to(1 - y) (1 -+ y2 -+ 2y) = (1 -+ y) (1 -+ y2) - 1 -+ y2,

whence, liy successive invertible transformations, we get:(1 - y) (l -+ y)2 = (1 -+ y) (1 + y2) -+ (y - 1) (y + 1),(1 - y) (1 -+ y)2 = (1 -+ y) (1 -+ y2 + y - 1),(1 - y) (1 -+ y)2 = (1 -+ y)2y,(1 -+ y)2 (1 - 2y) = O.

The roots o] the obtained equation are: YI = -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 solutions of the original cqua-

9;)- 3. Trigonometric Equations and Systems

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

:n: , t 1 -j EZ ...x= -T Inn, x=arc an "2 -JTn, n . ~Example 3.2.4. Solve the equation

2 sin 4x + 1f) sin" 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 formula2 tan x 2 1- tan> x

sin 2x = cos x = -,-..,.-----,,--1+tan2x' 1+tan2xand designate y = tan x, Then the equation is trans-formed into an algebraic one:

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

or By + 3 - 3y2 = 5 + 5y2, whence1y2_ Y+ "4= O.

Consequently, y = 1/2 or tan x = 112, whence x =arctan ~ + sen, 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 (n + 2nk) + 3 cos (n + 2nk) = -3.Consequently, besides x = arctan ~ + sen, nEZ, theequation has no other roots. ~

3.2. Methods of Solving Equations 91

2. Other Methods of Reducing Trigonometric Equationsto Several Simple Equations. Basically, we mean hereapplication of the formulas for transforming the sum ordifference of trigonometric functions into a product (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

1 1 1 1z-zcos2x+ 2+ zcos 6x= 1,

1or 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 =Jtk/2, k E Z) are solutions of the equation sin 4x = 0(since sin 4 (Jtk/2) = sin 2Jtk = 0, k E Z), therefore itsuffices to find the roots of the equation sin 4x = O. Con-sequently, 4x = Jtn or:

x = Jtn/4, n E Z. ~Example 3.2.6. Solve the equationsin x + sin 2x + 2 sin x sin 2x = 2 cos x + cos 2x.~By virtue of identity (2.29) for the product of sines,we have

sin x + sin 2x + cos x - cos 3x = 2 cos x + cos 2x,

92

or

3. Trigonometric Equations and Systems

sin x +- sin 2x - cos x - cos 2x - cos 3x = O.sin x -+- sin 2x - (cos x + cos 3x) - cos 2x = O.

Applying formulas (2.21) and (2.3(j), 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 - cos2x) (~ +- 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 = +2; +- 2nll, II EZ,

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

sin x - 1 +- 2 sin2 x = and 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 z e, -1,

In the second case

:n:or x= -2+ 2nn,

sinx=1/2, that is, x=(-1)" ~ +-nn, nEZ.Thus, the solutions of the original equation are writtenin the form of three series

_ + 2:n: -1-2x~- _ 3 ,nn, x= -~+-2nn,2x=(-1)" Z ~,-nn, nEZ. ~

3.2. Methods of Solving Equations

Example 3.2.7. Solve the equation

tan 2x +_.1_ cot x + _._1_ .sin x Sill 5x~Transform the given equation:

sin2x _ eosx +_1 1_- 0I'OS 2x sin x sin x sin 5:1' - ,

orsin 2x sin x - cos 2:1: cos x

sin x cos 2.1''. r.::

I Sill ;:J.T- Slfl X1- sin z sin Sr o.

or

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

--;-----0:-- + = 0,sin .T cos 2x sin x sin 5x

cos 3x (sin 5x - 2 sin 2x cos 2x) = 0sin x cos 2x sin 5x '

_ cos 3x (sin 5x - sin 4x) = 0sin x cos 2x sin 5x '

and, again hy formula (2.18),3 2 x 9xcos x- sm 2 cos 2

. 2'~ =0.sin x cos x Sill ,)XNote that if sin x = 0 (i.e. a: = sin; n E Z), then also

sin .5x = 0, and the equality sin ~ = 0 means thatsin x = 0 (sin 2nn = 0, n E Z). Consequently, the do-main of permissible values of the given equation can bespecified by two conditions:

cos 2x =1= 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 EZ), or x = ~ + ~;, and we have to check whether

n4 3. Trigonometric Equations and Systems

the constraints specifying the domain of permissiblevalues are met. For II = 31, k E Z, the expressioncos (2 (~ + ~t) )= cos n(21t3-1::.!~ takes OIl val ues eq ual to1/2, for n = 31 + 1,1 E Z, values equal to -112, and forn = 31 + 2,1 EZ, values equal to -1, that is, cos 2x =1=o for all these values of x, Further, for n = 6k orn = 61 + 2, 1 E Z, the expression sin (S( ~ + ~qn)) =

. ( 5n 5nn ) / 61SIll () +3' takes on values equal to 1 2, for n = . +1,1 EZ, values equal to 1, for n = 61 + 3 or n = 61 + 5,1 E Z, values equal to -1/2, and for n = 61 + 4, 1 E Z,values equal to -1, that is, sin 5x =1= 0 for the ind icatedvalues of x, and all of them belong to the domain ofpermissible values.

Consider now equation (2). We have:9 n x _ _n -t 2nn , Z-2 x = """"2 + sin, or - n E '-",9 9

and check whether the constraints cos 2.1: =1= 0 andsin .5x =1= 0 are met. Note that the expression cos (2 (~ +

2~n )) takes on one of the following nine values: cos ~n,1 10n 14n 22n 26n 1nn

- 2' cos 9' cos 9,1, cos 9' cos 9' cos 3,cos 3;n, none of them being zero. Similarly, we check tosee that the expression sin (5 (~ + 2~n) ) does not vanisheither for any integral values of n. Thus, we have foundall the solutions of the original equation: x = ~ + ~;,

n 2nnx = !J + 9' n E Z.

Example 3.2.8. Solve the equation.5 cos 3x -+ 3 cos x = 3 sin 4x.

.... Let us first note that if we apply twice formula (2.:iG) forthe sine of a double angle, we shall get the identity sin 4.1: =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) + 3 cos x = 12 sin x cos x cos 2x,

..

or

.'1.2. Methods of Solntng Equations 95

cos x (20 cos! J: - 15 +- 3 - 12 sin x (1 - 2 sin2x = 0,cos x (20 (1 - sin 2x) -12 - 12 sin x (1 - 2 sin2x = 0,cos x (20 - 20 sin- J: - 12 - 12 sin x + 24 sin-z) = 0,

(6 . 3 t::. 2 3 + 'J) cos X . Sin x - oJ sin x - < sin x ~ = ,cos x (fJ sin" x (sin x - 1) + sin x (sin x - 1)

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

Thus, the given equation decomposes into the followingthree equations:

(1) cosx=O,

(2) sin x =1,

nx=T+ nn,

n I 2x=T' nn,

nEZ,

(we see that the solutions of equation (2) are at the sametime solutions of equation (1,

(3) 6 sin" x +- sin x - 2 = 0.Setting Y = sin x we get an algebraic equation

6y2 + y - 2 = 0,whose roots are Yl = -2/3 and Y2 = 1/2, and it remainsto consider two cases:

(a) sinx= -f, x=,(-1t+1arcsin f+ nn , nEZ,(b) sinx=-~, x=(-1)n ~ +nn, nEZ.Thus, all the solutions of the original equation are

described by the formulas

X== ; +nn, x=(-1t+larcsin; +nn,

96 3. Trigonometric Equations and Sustems

Example 3.2.9. Solve the equation.)

cot 2x+ 3 tan 3x= 2 tan x+~4 .sin x

.. Let us represent the given equation in the form

(cot 2x + tan x) + 3 (tan 3x - tan x) = ~4sm x

and use the following identity:

t ? +t ~ cos 2x ... si n xco za: an z -.. sin 2x i cos xcos 2x cos x +llin 2x sin x

sin 2x cos xcosx

sin 2x cos x1

sin 2x '

whose domain of permissible values is specified by thecondition sin 2x =1= 0, since in this case also cos x =1= O.Applying this equality and formula (2.26) for the dif-ference of tangents, we get

_._1_ +:1 sin 2x __2_sm 2x cos 3x cos :c sin 4." .

The domain of permissible values for the given equationcan he specified by the following two conditions: sin 4x =1=0, that is, .X =1= Jtn/4, nEZ, and cos 3x =1= 0, that is,x =1= ~ + J[3n , n E Z. We transform the obtained equationon the given domain:

3 sin 2x 2 1cos 3x cos x sin 4x sin 2x .

3 sin2x 1 1cos 3x cos." sin 2x cos 2x sin 2x ,

') sin 2x i-cos 2x.J

cos 3x cos x sin 2." cos 2." ,

6 sin x cos x 2 sin" xcos 3x cos x 2 sin x cos x cos 2x

Since sin x =f=and cos x =1= 0, we have(j cos xcos 3x

1cos 2x .

Thus, in the domain of permissible values the originalequation is equivalent to

(j cos x cos 2x = coos 3x,

3.2. Methods of Solving Equations 97

1

and, by virtue of identity (2.54) for the cosine of a tripleangle, we have

6 cos x cos 2x = 4 cos" X - 3 cos x,whence

6 cos 2x = 4 cos" X - 3 (cos x =1= 0),Gcos 2x = 2 ('1 + cos 2x) - 3,6 cos 2x = 2 cos 2x - '1,4 cos 2x = -'1, cos 2x = -'114,

whence x = + ~ arccos ( -1) + stn; n EZ.It is easy to check that the found values of x satisfy the

conditions specifying the domain of permissible valuesin which all the transformations carried out were inver-tible. ~

Example 3.2. to. Solve the equation12" + cos x + cos 2x + cos 3x + cos 4x = 0.

.... Note that x = 2nn, nEZ, are not solutions of thegiven equation, therefore we may assume that x =1= 2nn.Then sin ~ =1= 0, and the following equalities hold:

cos x + cos 2x +cos 3x + cos 4x,1 x (2 cos x sin ~ + 2 cos 2x sin -i

21lll 2""

+2 cos 3x sin ~ +2 cos 4x sin ~ )1 (. 3 . x+, 5 . 3SIll - X - SIll - SIll - X - SIll - x

2' x 2 2 2 2sm 2""

+ . 7 ,5+,9 .7)SIn - X-SIll - X SIn - X-SIn - x2 2 2 2, 9

sm "2x

2 ' xsin 2""7-01644

98 S. Trigonometric Equations and Systems

Consequently, the original equation can be transformed tothe form

oSill;rX

x2sin 2""""

=0 or . 0SlllT x =0,

2nk .whence x = -9-' 1,EZ, provided x =1= 2nn which isequivalent to 2;k =1= 2nn or Ii; =1= gn (n EZ). ~

3. Sol V lug Trigonometric Equations Using the Proper-ties of Trigonometric Functions. Frequently, we have todeal wi th equations of the form t (t) = g (t), where tand g are some functions containing trigonometric expres-sions such that enable us to investigate the domains ofvalues E (f) and E (g) and to prove that these domainseither do not intersect or have few points in common. Insuch cases, the solutions of the equation t (t) = g (t)should be sought for among such t's which satisfy (sim-pler) equations t (t) = a, g (t) = a, where a is a real num-ber such that a EE (I) and a EE (g), that is, a E E (I) nE (g).


sin2 4x + cos" X = 2 sin 4x cos! z.

~ Let us write the given equation in the formsin 2 4x - 2 sin 4x cos! x = -cos2 x.

Adding cos" x to both sides of the equation, we get

sin" 4x - 2 sin 4x cos! x + cos" x = cos" x - cos" Xor

(sin 4x - COS4X)2 = -cos2 .'C (1 - cos" x).The left-hand side of the equation is nonnegative, whilethe right-hand side is nonpositive (cos" x > 0, 1-cos? x> 0), consequently, the equality will be valid onlywhen the following conditions are fulfilled simultane-ously:

8.2. Methods 0/ Solving Equations 99

{- COS2 X (1-cos6 X) = 0,

(sin 4x - cos- X)2 = o.The first equation decomposes in to two:(1) cos'' X = 0 or cos x = 0,

whence x = ~ + sen, n EZ. The obtained values alsosatisfy the second equation since

sin (4 ( ~ + nn) ) = sin (2n +4nn) = O.(2) 1 - cos" X = 0 or cos x = 1,

whence x = nn, n EZ. Substituting these values of xinto the second equation, we get (sin 4nn - cos" nn)2 = 0or (0 - 1)2 = 0 which is wrong.

Thus, the solution of the original equation consists3t

of the numbers x=2~nn, nEZ. ~Example 3.2.12. Solve the equation

sinG x + cos6 X = p,where p is an arbitrary real number.

- .... Note that

sin6 x +cos" X= (sin2 x +cos- x) (sin4 x - sin2 x cos2 x +cos! x)= sins x- sin2 x cos2 X + COS4 x== (sin- x +2 sin- x cos2 x + cos- x) - 3 sin2 x cos2 x= (sin- x +cos2 X)2 - ~ (2 sin x cos X)2=1- : sin22x=1- ~ (1-cos4x)

3 5= gcos4x+g,

and the given equation takes the form3 5 8p-5gcos4x+g=p or cos4x=-3-'

,.

tOO 3. Trigonometric Equations and Systems

The equation has the solution x = + 1arccos 8p~5 -+n; , nEZ, for -1::;;;;(8p-5)/3~1 or 1/4~p~1. ~


(cos: -2sinx).sinx-+(1-+sin ~ -2cosx)cosx=O.... Remove the parentheses and then use the fundamen-tal trigonometric identity and formula (2.11) for the sineof the sum of two numbers. We get

cos ~ sinx-2sin2x+cosx+sin ~ cosx-2cos2x=0,that is,

sin ( x + ~ ) + cos x - 2 (sin" x + cos- x) = 0,or

. 5x 2sin T + cos x = .

Note that the sum in the left-hand side of the obtainedequation will equal 2 only if sin 5: = 1 and cos x = 1simultaneously, that is, our equation is equivalent to thesystem of equations:

{. 5x 1

slnT= ,

cos x=1,whence

{5: = ~ + 2rtn, n EZ,x = 2rtk, k EZ,

. 2n 8n Iand the equahty 2rtk=T+ Tn must ho d, whencek = 1~4n . Since kEZ, we have n=5m+1, mEZ{since for the remaining integral n's, that is, n = 5m l

3.3. Equations in Several Unknowns 101

n = 5m + 2, n = 5m + 3, n = 5m + 4, it is obviousthat k ~ Z), and then x = 2n + 8nm, m. EZ, that is,x = 2n (4m + 1), m EZ.

3.3. Solving Trigonometric Equations and Systems ofEquations in Several Unknowns

The presence of two or more unknowns involve certa-in difficulties in solving trigonometric equations andsystems. The solution of such an equation or system isdefined as a set of values of the variables which turn thegiven equation or each of the equations of a system intoa true numerical equality. To solve a given equation orsystem is to find all such sets. Therefore, answering aproblem of this type by giving the values taken on byeach unknown is senseless. One of the difficulties en-countered in solving such problems is also that the set ofsolutions for these equations and systems, is, as a rule,infinite. Therefore, to write the answer in a correct wayand to choose desired solutions, one has to consider dif-ferent cases, to check the validity of auxiliary inequali-ties, etc. In some cases, when solving systems of equa-tions we can eliminate one of the unknowns rather easilyby expressing it in terms of other unknowns from one ofthe equations of the system. Another widely used methodis to try to reduce a trigonometric system to a system ofalgebraic equations involving some trigonometric func-tions as new unknowns. As in solving trigonometricequations in one unknown, we can try to carry out iden-tical transformations to decompose one or more ofthe equations to the simple equations of the typesin (x + 2y) = -1, tan (x - y) = V3, and so forth.

Example 3.3.1. Solve the system of equations

{ Vsin x cos y =0,2 sin2 x- cos 2y- 2 = O..... It follows from the first eqnation that. sin x> 0, twocases are possible here: if sin x = 0, then the eqnationturns into an identity,and if sin x ; 0, then the equationimplies that cos y = O. Consequently, the system is equiv-

102 3. Trigonometric Equations and Stfstems

alent to the collection of two systems:

{sin x=O,

2 sin- x - cos 2y - 2 = and

{cosy= 0,

2 sinzx-cos 2y-2 = 0, sin x > 0.The first system has no solutions (cos 2y + 2 -=1= 0),while the second is equivalent to the system of two simpleequations

{COS y= 0,sin x= V2/2.

Consequently, the set of all solutions of the original sys-tem consists of pairs of numbers (x, y) of the kind

( (_1)k ~ + stk, ~ + nl), k, l EZ. ~Example 3.3.2. Solve the equation

3+2cos(x-y) 1/3+2 Z z x-y

trigonometric

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