Trigonometric functions : problems-solving approach

TRIGONOMETRICFUNCTIONS(Problem Solving Approach)

A. PanchishkinE. Shavgulidze

MirPublishersMoscow

TRIGONOMETRIC FUNCTIONS

A. A. Ilaasmmraa, E. T. Illasrvnanse

TPHI'OHO~1ETPl1qECRHE<DYHRUI11JB 3A,L(AqAX

A.PanchishkinE.Shavgulidze

TRIGONOMETRICFUNCTIONS(Problem-SolvingApproach)

MirPublishersMoscow

Translated from Russianby Leonid Levant

Pirst published 1988Revised from the 1986 Russian edition

Ha aH3AUUCnOM, R8bl1£e

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 examinations to those entering universi ties, engineering colleges,and teacher-training insti tutes.

The aim of this study aid is to help the student to master 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. Mordkovich, 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 independently (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 precollege 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 mathematics (trigonometry included) for those preparing topass their entrance examinations (for instance, Problems

6 From the Authors

at Entrance Examinations in Mathematics by Yu.V. Nesterenko, S.N. Olekhnik, and M.K. Potapov (Moscow,Nauka, 1983); A Collection of Competition Problems inMathematics with Hints and Solutions edited by A.I. Prilepko (Moscow, Nauka, 1986); A Collection of Problems inMathematics tor Pre-college Students edited by A.I. Prilepko (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 borrowed 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 statement.

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" mechanicomathematical faculty of Moscow State University.

From the Authors

Contents

5

Chapter 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 149

5.1. Proving Inequalities Involving TrigonometricFunctions 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 Circle

1. The first thing the student should have in mind whenstudying trigonometric functions consists in that thearguments of these functions are real numbers. The precollege 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, trigonometric 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 sexagesimal) 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 measurement 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 3600

is equal to 2n radians. Consequently, to 180° there correspond 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 between 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 57°17'44.S". ~Jt

Example 1.1.2. How many degrees are contained in

I 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 arcof 1U84°?

If Jt radians ~ 1800,

and y radians == 19840,

1t ·1984 496Jt 1then y==: 180 ==~~1145Jl· ~

2. Trigonometric Circle. When considering ei ther 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 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 direction 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 RealAxis 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 coordinate 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 aforesaid,


round the circle in the positive direction. We denote theterminal point of this path by P t and associate the nUIDbar 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 number t E R, that is, F == Ph then this point also corresponds 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 negative 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 formulat ion: the full inverse image 1)-1 (P t) of the point P t

coincides with the set

{t + 2nk: k EZ}.

Remark. The number t is usually Identified with thepoint 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 dropped 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 numbers

5: + 2nk, k EZ, and only they. ~Example 1.1.5. Find all the numbers corresponding to

the 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).<llllIII The vertices of a regular N -gon divide the trigonometric circle into N equal arcs of length 2:rt/N each. Consequently, the vertices of the given N-gon coincidewith the points A, = P 2nl, where l = 0, 1, ° • 0' N - 1.

1+rrTherefore the sought-for numbers t E R have the form


1 -I- 2;k ,where k EZ. The last assertion is verified in the

following 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 integer

k 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 trigonometriccircle 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 there

corresponds 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- == 3·2n +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 trigonometric 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 corresponding 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 quadrants. 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 approximate 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 inequalities. 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 trigonometric 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 consists of four quadrants. Making use of the upper estimate

n·6 91t < 3.~, we find that T < .6 for k = 6; at the same

· n(k+t) n·7 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' n·5 3.2·5 8 d rt-fare va l( or te=:: , Since -2- < -2- == an -2->3.~.6 = 9.3. Consequently, the point e, lies in the

second .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!

<lIllIII Let 11S find an integer k such that

nk/2 <-Vs- vr< rr (k-t- 1)/2. (1.5)To this end, we use the inequalities

2.2 < Y5< 2.3,3-

1.9< ~/7<2,

whose validity is ascertained by squaring and cubing bothsides of the respective inequality (let us recall that ifboth sides of an inequality contain nonnegative numbers,then raising to a positive power is a reversible transformation). Consequently,

/- 3V--4.3<~-V5- 7<-4.1. (1.6)

Again, let us take into consideration that 3.1 < 3t < 3.2.Therefore the fol lowing inequali ties are fulfilled:

1t (-3)/2 < -4.65 < -4.3, (1.7)1t (-2)/2 > -3.2 > -4.1. (1.8)

From inequalities (1.6)-(1.8) it Iol lows that (1.5) isvalid for k === -3, consequently, the point P - ~/5-V7

lies in the second quadrant, since the remainder after thedivision of the number -3 + 1 by 4 is equal to 2. ~

2-01644


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 number t and is denoted bycos t.

Let us drop perpendiculars from the point P t onthe coordinate axes Ox andOy. Let X t and Y t denotethe feet of these perpendiculars. 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 trigonometric 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 that

P3n/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 circle 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 particular 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). Consequently, 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 negative 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 third

quadrant; therefore

sin (V:5 +V7) < 0, cos (YS+V7) < O. ~

Note for further considerations that sin t = 0 if andonly 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 numbers 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 coordinate 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 definition of tho function tan t. Note that the point of intersection is absent exactly for those values of t for which

P t = B or D, that is, for t == ~ + nn, nEZ, when the

function tan t is not defined.Now, draw a tangent BB' to the trigonometric circle

through the point B and consider the point of intersectionW t of the 1ine 01) t and the tangent. TIle a bscissa of W t

is 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 cotangent, 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, _ and cos (l!s+ ¥7) < 0, therefore cot (V 5 +V7) >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 trigonometric 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 numbel' 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 allest 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 number 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 t

and 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 reasoning 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 since

f (t + 2n) == cos4 (t + 2n) -f- sin (t + 2n)= cos" t + sin t.

No positive number T, smaller than 2rf., is the period

of ~he function f (t) since f ( - ~ ) =F f ( - ~ + T ) ::00

f ( T ). Indeed, the numbers sin ( - ~ ) and sin ~


are dist.iuct from zero and have unlike signs, the

numbers 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 xV5t V5, x V5, x V5 -t- t ~r5 belong to the domain of defini tion of the function f (x). The defini tion of the Iunction 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 fundamental 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 Iunction f (x) has the fundamental period 1', then the Iunction 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.<llIIIIII Suppose that T is the period of the function f (x).'Ve take a positive x satisfying the equality sinj/' x,. ~--, 1.rrlteJl sin Vx -t._]' sill VX-~- '1 ~ hence Vx -1- l' - V' x:-~2Jln tn EZ). But Vx -t- T > 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 of

A a odd functions: f(.7;) == -II X3,f (x) = 2x5 + nx3

•

Note that many functionsare 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 trigonometric 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 interval 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 function f is said to be nondecreasing 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 monotonicity. 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 function 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 0

Fig. 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 interval (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 <1£,12, the inequality tan t] < tan t 2 holds.

Consider three cases:(1)O~ t1 < t 2 < n/2. Then, by virtue of Theorem 1.4,

O~ sin t l < sin t 2 , cos t1 > cos t 2 > 0,

whencesin 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,

thereforesin it ~ sin t2cos t 1 <, COg t 2 '

that is, tan t l < tan t 2 •

The proof for the function cot t is carried out in a similar way. ~

Note that the monotonicity properties of the basictrigonometric functions on other intervals can be obtainedfrom the periodicity of these functions. For example,on the closed interval [-nI2, 0] the functions sin t andcos t increase, on the open interval (n/2, 331/2) the function tan t increases, and on the open interval (n, 2n)the function cot t decreases.

Example 1.3.5. Prove that the functions sin (cos t)and cos (sin t) decrease on the interval [0, n/2J.~ If t 1 , i« E [0, n/2], where t l < t 2 , then, by Theorem 1.4,sin t l < sin t 2 , and cos t 2 < cos t l • Note that thepoints on the trigonometric circle, corresponding to thereal numbers sin t l , sin t 2 , cos t 1 , cos t 2 , are in the first

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 argument 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 is

possible 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 i

corresponding to the real number t, lies.Example 1.3.6. Compute cos t if it is known that

tan t =cc 152and 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 quadrant. In the third quadrant, cos t is negative: consequently, 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 cornpute 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 necessary 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. righthand 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 theline 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 the

equation 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 integer 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<3.2). Thus,

t === 3n - 10 == arcsin (sin 10). ~

2. Solving the Equation cos t == m, Arc Cosine. Tosol ve the equation cos t === m, it is necessary to find allreal numbers t such that the abscissa of the point P,is equal to In. For this purpose, we draw a straight linex === m and find the points of its intersection with thetrigonometric circle. A point of intersection exists if, m I~ 1. One of the points of intersection (the point Ein Fig. 19) necessarily lies in the upper half-plane, wherey~ 0, and this pain t can be written in the form

E === P l o' where O~ to~ n,

Definition. The arc cosine of a number m is a number to,lying on the closed interval [0, n], such that its cosine isequal to m. '-rlte arc cosine of the number In is denoted byarccos ni (or cos -1 m).

:~ -01B44


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

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

the point F = P <! «' Therefore the genera') solution of the., equation cos t == m, m. E

!I [-1, 1] has the form

t == + arccos m + 2Jtk, k EZ.

Note also that the following equality holds:

A arccos (--In)---T---+--~"

x = Jt -- arccos m.Example 1.4.3. Find the

value of arccos (cos 5).~ Let t == arccos (cos 5),then O~ t ~ n, andcos t ~ cos 5~ The point p~

Fig. 20 I· . h f 1 dies In t e ourt 1 qua ran t(Fig. 20) since the i nequal

ities 3n/2 < 5 < 2n hold, therefore the point P i7 lyingin the upper half-plane, must coincide wi th the pointp -5 which is symmetric. to IY 5 with respect to the axisof abscissas, that is, t == -5 + 2Jtk. The conditiono~ t~ rt wil l be fulfilled if we take k == 1, thereforearccos (cos S) = 2Jl - 5. ~

Example 1.4.4. Prove that if ()~ x~ 1, thenarcsin x == arccos V1 - x2 •

...- Let t:=:: arcsin x, Then 0~ t~ rt/2 since sin t ==

.T~ O. Now, from the relationship cos" t + sin? t :=:: 1we get: I cos t I =-.= ·V 1 - x 2 • But, bearing in mindthat t E [0, rt/2l, we have: cos t == V1 - x2 , whencet =-:: .u ccos Vi 1 - x2-. ~ •

3. Solving the Equation tall t == m. Arc Tangent,1'0 solve the equation tan t == m, it is necessary to findall real numbers t, such that the li ne passing through theorigin and point j)t intersects the line AB': x >» 1 ata point Zt with ordinate equal to m (Fig. 21). The equation of the st.raight line passing through the origin and P t

is 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 number 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 introduced 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 is

t == 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 conditions 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 <n/2. Let us prove that sin t == x/V 1 -1- x2 • For th ispurpose let us note that cos 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 Iunctions, it suffices to rememher the above defini tions andbasic trigonometric formulas.

PROBLEMS

1. t. Locate the points by indicating the quadrants:(a) t'«. (b) P10-~/2' (c) PV26+V2•1.2. 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 vertices the following points lie: (a) PIO' (b) P -11' (c) ])12.

1.3. Given a regular heptagon inscribed in the trigonometric 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 regular 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 andan 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 54

is 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 numbers, 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 numbers, is periodic if and only if the ratio alb is a rationalnumber.

1.25. Prove that the function y == tan 5x -t- cot 3x -l4 sin x cos 2x is odd.

1.26. Prove that the function y = cos 4x + sin" ~ X

tan 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 decreasing 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 ~/__ = arccosy 1+x2 Y1+x2

for 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 three

other inverse trigonometric functions.

1.49. Express: (a) arccos ( - ~ ), (b) arctan ( - 2: )I

(c) arccot ( - ;4) in terms of values of each of the

three 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 Formulas

There 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 formula 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 obtained 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 fundamental 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 transIormations 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 formula 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-+S

of 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 following. 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 1

3, P i

4be points on the

coordinate 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 following

identity 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 images of the point Po whenthe plane is rotated aboutthe origin through anglesof t, t + s, and -8 radians(~-"ig. 23).

By the definition of thesine and cosine, the coordinates 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 1

2 == (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 identity (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 have

cos (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 cosine 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 another 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 deriving 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 a

t=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 - O·sin 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 do

this as an exercise.

Example 2.1.2. Derive the reduction formula for

cos ( ~1t - t ) .

~ Using formulas (2.6) and (2.3), we have

cos ( ~1t - t ) = cos (31 + ( ~ - t ) ) = - cos ( ~ - t )

==-sint. ~.

Corollary 4. The reduction [ormulas for tangent andcotangent 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).

• 2.1. Addition Formulas 47

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 sign2±t

the 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 + t

by 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 difference (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 formulas (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 lefthand sides of (2.13). Indeed, the left-hand side is defined

for all the values of t and s except t + s === ~ -+- sin,

nEZ, while the right-hand side only for the values of t

and .': mentioned in Corollary G. Thus, for t = ~ and

S == : the left-hand side is defined, while the right-hand

side is not.Prooj. Let us apply the definition of tangent and for

mulas (2.11) and (2.1):

tan (t _~_ s) ==:-: sin (t·t 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. ~

1-4

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 ) 2tan2s=---1 == - -1 = -cos2 S 5 5

(we have used formulas (1.9) and (1.10»). Therefore

tan 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 53 12

1----4 fi

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

24·5+12·724. ·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 ) . 12arcsin -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-. 5

5. 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 denominator 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 S

sin 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 property of cotangent. ~

6. Formulas of the Sum and Difference of Like Trigonometric 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 righthand 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 ill

t+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 identity (2.17), and, subtracting (2.20) from (2.19), we get identity (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 precedingcorollary. 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 -- scos t == cos -2- cos -2- - SIn -2- SIn -2- , (2.23)

t+s t-s . t+s . t-scos 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, except

t = ~ + 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 Trigonometric Functions into a Sum. These formulas are helpfulin many cases, especially in finding derivatives and integrals 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 number b in the following way:

(2.35)b<O.for

for

arp == - arccos "II

JI a2 +b2

The quantity arccos y a is defined since y Ial ::::;; 1a2 +b2 a2 +b2

(see Item 2, Sec. 1.4). By definition, we have:

cos q> ==cos (I q> I) = cos (arccos Va) = Va.a2 +b2 a2 +b2

From the fundamental trigonometric identity it followsthat

2sin2 rn ~ 1- cos- en =:-.: 1 _ _ a__

-r 't' a2+b2

b2

or [sin cp J = Ib I . Note tha t the sign of sin rp co-Va2+b2

incides with the sign of <p since I {P I~ rr. The signs ofthe numbers rp and b also coincide, therefore

. bSIn q> =--= ---Ya2 +b2 •

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 get

a 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 identities 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 Formulas). The formulas considered here are of great importance 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 ~2

Remark. The domains of permissible values of thearguments 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 trigonometric identity (1.9), we have:

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

2 sin !!:- cos !:..2 2

cos2~+sin"~2 2

By 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- ~ , whence

a2tan 2

t+tan2~2

2 sinacos a.

sin2 5!:2

sin a = ------

1-1---cos- :!-

2

We now apply (2.37) and the fundamental trigonometric identity and get

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

sin2~1-

2

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

~-cos2~ +sin2~ sin2~ 1+tan2 !:..

2 2 1+ 2 2

COS2~2

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+tan22 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

2Corollary 4. For any real numbers a) except ex === sin

(n EZ), the identity

(2.45)1-tan2 ~

2a

2tanTcota~-----

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

2ex.

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 !!..2

Example 2.2.2. Evaluate cos (2 arctan (-7».a

<lIIIIIII Let us denote ex == 2 arctan (- 7), then tan 2 ::=: -7

and -n<a<O. Using formula (2.43), we get

1--tan2 ~2 1-49 24

cos a == -t-+-ta-n-2

- a- == 1+49 = - 25 · ~2

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 and2

Jt2" < 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 consideration 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 2

tan~==--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 · ~

aex; cosT

cot-==---2 .ct 2'2a

SID 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 2cot'!!:' ==---

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) with

ex == ~ 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 2

I i - COS ~sin~=l 4

8. 212=172

2

Further,

tan ~:=8

• 1CSIn"8 V2=l!2

-1C 2+ V2cos R

I (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 <ex < nand ~ lies in the first quadrant. Now, using

62 2. 1den tical Trans formations

formula (2.16) with a positive radical, wo get

(1 (1 )) .. / 1+ cos a

C08 2" arCC08 -10 :=:: V 2

- /-==-./~==~V 21 ~ 11 ~ .

4. Trigonometric Formulas of Triple Argument.Theorem 2.4. For any real numbers a the following

identities are valid:

sin 3ex = 3 sin a - 4 sin" a, (2.53)

cos 3ex = 4 cos" a - 3 cos cx. (2.54)

Proof. From (2.11), (2.36), (2.39) and the fundamentaltrigonometric identity it follows that

sin 3a -=-: sin (a + 2ex) == sin ex cos 2a + cos a sin 2a

== sin ex (1-2 sin2 a) + cos ex (2 sin ex cos ex)

== sin a - 2 sin" ct + 2 sin a CO~2 a:=: sin ex - 2 sin" ex + 2 sin ex (1 - sin- ex)

== 3 sin (X- 4 sin" ex.

Further, by virtue of (2.1), (2.36), (2.38), and the fundamental trigonometric identity, we have

cos 3ex == cos (ex + 2a) = cos ex cos 2ex - sin ex sin 2a

=== cos ex (2 cos'' ex - 1) - sin ex (2 sin ex cos a)

= 2 cos- ex - cos a - 2 sin? a cos a

= 2 cos" ex - cos Ow - 2 (1 - cos- ce) cos a

== 4 cos" ex - 3 cos a. ~

Example 2.2.6. Evaluate cos ( 3 arccos (-*)).~ Let a = arccos ( - ~ ) , then cos« = - ~ . We make use

of identity (2.54) to get

cos 3a == 4 cos" a - :i cos ex

= 4 ( - 217

) -- 3 ( - i )= 1 - 2; = ;~. ~

2JJ. Trigonometric Trtms lormntions

Example 2.2.7. Evaluate sin( 3 arcsin t) .L · 1 tl . 1 W~ et a == arcsin 5"' len Sill a == 5 · e use

ty (2.53); whence there follows:

. 1 1 71SIn 3a == 3· 5 - 4. 12fi =: 125. ~

identi-

n Jt

j[ rr rt ) tan T- t an 6tan12=tan (4--6 = :It :It

1+tanTtanT

2.3. Solution of Problems InvolvingTrigonometric Transformations

1. Evaluation of Trigonometric Expressions. A numberof problems require the value of a trigonometric functionfor values of the argument that have not been tabulated.nor can be reduced to tabular values by reduction orperiodicity formulas, that is, they do not have the formnk/6 or nk/4, k EZ. Sometimes, the value of the argument can be expressed in terms of such tabular valuesand then apply the addition formulas for half or multiplearguments.

Example 2.3.1. Evaluate tan~.

~ Since ~ = ~ - ~ , we may apply formula (2.14) for

the tangent of a difference:

. 11+1'--rJ 3

\/3-1 (V3-1)2 y-V3+1 = (;/3+1) (:/3-1) = 2- 3. ~

Example 2.3.2. Evaluate cot ~~ .

.... Noting that ~~ = ~ + ~ , we apply formula (2.15)

for the cotangent of a sum:11, j[

5n (3t 3't) cot T cot 6-- 1cot 12 ~ cot 4 -1- 6 == n Jt

cot T+c.ot T

1'l/ 3- 1 (/3-1)2 =2-Y3. ~1+l"'3 3-1

64 2. l dentical Transiorm.attons

In the cases when the argument is expressed in terms ofinverse trigonometric functions, we have to transform thegiven expression so as to take advantage of the definitionof inverse trigonometric functions.


A t ( 3n 1 · ( 4 ) )-:= an T-4arcsln -5 ·

~ Using first identity (2.50) for the tangent of half argument and then the corresponding reduction formula,we get

i-cos ( 3; -i arcsin ( -i-))A == 3rt 1 4

sin ( 2--2" arcsin (- 5 ) )

1+sill ({ arcsin ( --} ) )

-cos ( ~ arcsin ( -- :)) •

We set a,-··arcsin ( _..!), then sin a==: - 54, -~<\ 5 2

ex < 0, cos? ex = 1- sin2 ex .~ 1 - ( - ~ ) 2 = 21)5. Conse-

quently, cos a = i, arul , by virtne or (2.47) and (2.46)

for the sine and cosine of half argument, we get

r--3

sin~ ==--.: _ ... / 1 - cos a == _ V1- 5 1_2 V 2 2 - VS'

!!:- _ .. /' 1+cos a _ V1 +- ~ __2 _cos 2 --- V 2 -- 2 -- V5 ·

The signs before the radicals have been determined

from the condition - ~ < ex < O. Finally,

2.8. Trigonometric Transformations 65


. 3 + . 12arcsin 5 arcsin 13 ·

......Lt· 3 A • 12 Tl . 3....... e a::=: arcsin 5' )J ~ arcsrn 13 • len SIn ex ==""5 .,

n . 12 nO<a<T' slnB~=13' O<~ < T and O<a+~<n'i.e. the number ex + ~ lies in the domain of values ofarc cosine. Since the points P a and Pfl lie in the firstquadrant,

Cos a -== V1-sin2a == 4/5,

cos ~ == V1 - sin2 ~ == 5/13,and we may apply formula (2.'1) for the cosine of a sum:

cos (a + ~) ~ cos a cos ~ - sin a sin ~

4 5 3 12 16=== 5 · 13 - 5 · 13 == - 6fl '

whence it follows tha t a +- ~ = arccos ( - ~:). ~When solving such problems, a common error consists

in that the magnitude of the argument a + ~ is notaken into account. They reason like this: by the formulaor the sine of the sum of numbers, we may write

sin (ex + ~) :== sin a cos ~+ cos a sin ~

3 5 4 12 63===5'[3+5·13==65'

tand then conclude incorrectly that a + ~ = arcsin :~,although the number a + ~ does not belong to the do

main of values of are sine since a + ~ > ;.Example 2.3.5. Evaluate arctan 4-arctan 5.

~ Let a == arctan 4, ~ ~ arctan 5, then tan a = 4,

O<a< ~ ,tan:p=5, 0 </3 < ~ and - ; < a-/3 <n2- Let us use formula (2.14):

tan cc-e-tan Btan(a-~)=== 1+tanatan~

a-016~~

4-5 11+4.5 = -2[-

66 2. 1d.entical Trnnsjormnttons

Consequeul.l y, a-fJ-- arctan ( -- ;1 ) since the number

ex - f1 lies ill the doruaiu of values of arc tangent. ~Example 2.3.6. Evalualc si u Hl" -sin ;:1Uu -siu 50° -sin 70°.

~Let us make use of the reduction formuJa (2.3) andmultiply the given expression by cos 10°; we get

cos '1 (}o si n '10° si n :30° cos 40° c-os 20°.

Now, we m av apply Iormu lu (2.3H) for the sine of (1 doubleangle for three ti IHOS, numclv:

cos 10° ~jn ;(.c sin 3(,°cos 20° cos 4uo

= ~ (2 cos 10° sin 10°) ~ cos :wo cos 4()0

1 1 1 (~ . 200 , 20·0) -. 40~==: 2 . '2 '"2 ~ sm cos cos

_ 1 . 1 (9 lo:' f.00 I. ()O)_ '1 ,,' SOl")-8" "2 ...... m « COS l _.- W sin .

Since the initial expression was multiplied by cos 10° ~1

sin 80°, it is now obvious that it is equnl to 16. ~Example 2.3.7. Prove the eq uality

1 tf3_ 4sin 10° - cos 10° - ·

~Multiply both sides of the equality hy the numbersin 10° cos 10° which is not zero. This transformationis reversible, and the equality being proved now has theform

or

2 ( ~ cos 10° - V~3 sin 10°) = 2 sin 20°,

2 sin (30 0 -10°) == 2 sin 20°.

We have used formulas (2.36) and (2.12) and have shownthat both sides of the equality are equal to the samenumber 2sin20°. ~

The examples considered above show that in computational problems it is often convenient first to carry outsimplifications with the aid of known trigonometric for-


mulas. For instance, it is useful to single out the expression of the form sin" ex -t- cos" a which is equal to 1 byvirtue of the fundamental trigonometric identity.


• 't. n; -L 4 3n L • 4 531 + 4 'lnsin 8 I'COS 8-. S1D 8 cos-sj.

~ A long solution would be first to compute the

I . n 3n. 5n d 7n ·th thva ues SIn 8' cos -8-' SID 8' an cos 8 WI e

aid of half-argument formulas and then to raise theterms to the fourth power and to add them together.However, we first use reduction formulas (2.3), (2.8), and(2.5) according to which

3n: (n 1t ) • 1tcos 8 = cos 2 - 8 == SIn 8 '

. 5rt . (n; + 1t ) 1t t«SIll 8=sln 2 8 ==cos 8. -coss·

Consequently, the given expression can be rewritten asfollows:

2 (sin4 ~ + cos! ~ )

=2 (sin2 ~ +cos2 ~ V-2·2sin2 ; cos2;

= 2·1-sin2 ~== 2 - .!. ==~4 2 2 •

Here, we have used the fundamental trigonometric identity, formula (2.36) for the sine of a double angle, and

also the tabular value sin ~ = ~2. ~Example 2.3.9. It is known that cos q> + sin q> = a,

\vhere q> and a are real numbers. Find sin'' (p + cos3 (p.

~ sin3 cp +cos3q>

= (sin <p + cos fP) (sin2 cp- sin <p cos (P + cos2 q»

=a { ~ (sin2 cp +cos2 cp)

- i (sin2 cp + 2 sin cp cos cp +cos2 cp) )

68 2. I dentical Transformations

=a ( : - ~ (sin rp +cos q»2) = a ( ~ _ ~2 )

3 1=2 a- 2a

3• ~


A == cos 70° cos 10°+cos BO° cos 20° .cos 68° cos 8°+cos 82° cos 22° •

~ First use the reduction formulas and then formula (2.2)for the cosine of the difference between two numbers:

A = cos 70° cos 10°+cos (90° -10°) cos (900- 70°)

cos 68° cos 8°+cos (90° - 8°) cos (900- 68°)

cos 70° cos 10°+sin 10° sin 70° cos (700 -10°)= cos 68° cos 8°+sin 8° sin 68° - cos (68° - 8°)

cos 60°= cos 600 :-=: 1. ~

Example 2.3.11. Find sin cxt~ and cos cxt~ I if

· +. A 21 + A 27 dsm c Sln p == -65' cos ce COSt' = -65' an n<ex-

~<31t.~ It follows from the two equalities that

(27)2+(21)2(cos a + cos ~)2+ (sin ct + sin ~)2 = (65)2 ,

whence we get:

(cos2 ex + sin'' a) + 2 (cos ct cos ~ + sin ex sin ~) +(cos" ~ + sin'' ~) = 18/65,

1+ 2 cos ((J.,-~) + 1 = 18/65,

1+cos (ex -~) ==: 9/65,

2 cos- a;~ = 9/65.

Consequently, by virtue of the inequalities ~ <a-p < 3n

2 2

a-~ 3cos -2- = - V130 •


Transform the sum of two sines into a product usingformula (2.17):

• • A 2· a+~ Ct-~SUl ex, + SIn tJ = SIll -2- COS -2-

= _ 3 Y2 81.n a+~V65 2·

Consequently ,

,. a+~ _ ( 21). ( 3 V2 ) _ 78111-2 - - -6"5 -;- - y65 -Yi30·

Sinrilarly , by virtue of the identity (2.21), we get

cosa+cos~=2cosat~ cos a;~

3 1;/2 a+~= - V65 cos -2-·

Therefore

• Ct+~._ ( 27). ( 3 V2 ) _ 9cos -2-- - 65 -;- - y65 - Y130· ~

An interesting method of solving problems on computingthe values of trigonometric expressions consists in thefollowing: we Iirst try to find and then apply an algebraiccondition which is satisfied by the given expression. Here

is an instructive example: sin ~3t = sin 72° = cos 18°.

Example 2.3.12. Prove that

cos 18° = V 5+8V5 •

~ Consider the continued identity based 011 the formulafor the sine of a double argument:

cos 18° sin 18° cos 36° = ~ (2 sin H;o cos 18°) cos 36°

= ~ sin (2· 18°) cos 36° = ~ sin 36° cos 36°

= ~ (2 sin 36° cos 36°) = 1sin 72°

i 1::=: "4cos (90 0

- 720) :=: 7; cos 18° ,

70 2. Identical Trans formations

which yields the equality

sin 18° cos 36° =, 1'since cos 18° =1= O. Using formula (2.39), we write cos 3n°:==1 - 2 sin" 18° and, consequently, got the relationship

sin 18°(1- 2 sin2 18°) = 1or

8 sin" 18° - 4 sin 18° -t- 1 == 0,

that is, au algebraic equation which is satisfied by thonumber sin 18°. Setting sin 18° == x, we can solve thecubic equation thus obtained

8x3- 4x -l- 1 === 0

by factorizing its left-hand side by grouping the terms inthe following way:

8x3- 4x + 1 == 2x (4x 2

- 1) - (2x - 1)

== (2x - 1) (4x2 + 2x - 1).

Consequently, our cquat.ion r takes the form: (2x-1) X

I. 2. 1) . b · 1 -1 - 1/5(~x +2x- ~ 0, Its roots erng Xi == 2" ' x2 :.:..=.. 4 '

1"-

xa= -it l 5 . Since 0 < 18° < sin 30° = ~ (by Theo-

rem 1.4), then Xi =;6 sin 18°, X 2 :;6: sin 18°, consequently,

sin 18°= -it 1/

5 . Hence (since cos 18°> 0)

2. Slmplifylng Trigonometric Expressions and ProvingTrigonometric Identities. The usual method of provinga trigonometric identity consists in that one of its sidesis transformed wi th the aid of various trigonometric andalgebraic operations and also with the aid of the rela tion-

2.3. Trigonometric Tranelornuitions 71

ships given in the hypothesis so as to get finally the expression which represents the other side of the identity.We can make sure that the left-hand and right-hand sidescoincide, transforming them separately so as to getequal expressions. It is onJy reguired that all the transformations carried out he reversible in the domain of permissible values of the arguments of the given equality(that is, for all valnes of the argumen ts for which all theexpressions involved in the given identity make sense).This means that not only from each equality obtainedduring the process of transformations there follows a consequent equality, but also vice versa, the preceding equality itself must follow from the consequent one. This method of reasoning, of course, seems to be rather generaland is used for solving equations and systems as well asfor proving and solving inequalities.

Example 2.3.13. Prove that if

sin (x-a) _ asin (x-·~) - b '

and aB + bA =1= 0, then

cos(x-a)cos (x-~)

AIf

aA+bBaB+hA

aA+bBcos (a-~) -~-= --aB+bA ·

~ Using identities (2.11), (2.3()), .(2.17), we get

~.'~+1 Sill (x-·a) • cos (x-a) +b B sin(x-~) cos(x-~) 1

-~1-~ sin (.T.-a) _1_ cos (:t-ex)h R sin (x-~) I cos (x-~)

sin (x - ex) cos (x -_.a) +sin (x -~) cos (x- ~)

== sin (x-a) cos (x-~)+cos (x-a) sin (x-~)

'12 (sin 2 (x - a)+Sill 2 (x --- Ii))

sin ((.r. -- a) -r- (:r --~»

== sin (2x ~- (a + ~» cos (a - ~) == cos (a _~).sin (2x - (ex -r- ~»)

The last transformation consisted i It dividing both thenumerator and denominator of the fraction by the IlUln

her sin (2x - (a + ~). This transformation is reversible


. . (2 (+ A» (aB+bA)sin (x-~) cos(x-~) -I- 0SInCe SIn x - ex p ~ bB rr:

for the values of the arguments x, (x, p consideredin the problem. ~

Example 2.3.14. Simplify the expression

sin" 2a cos 6a + cos" 2a sin oa.~ Applying formulas (2.53) and (2.54) for the sine and co

'sine of a triple argument and also formula (2.11), we get

sin" 2a cos 6a + cos" 2a sin 6a

( 3 · 2 1. 6) (.== "4 Sill et- 4 SIn a cos nee

+t: cos 2a +1cos (la ) sin Ha

3_ 2 6 1.~ t:= 4 SIll a COS a-7; SIn \Jet cos va

+ : cos 2a sin 6a + { cos 6a sin 6a:

= : (sin 2a cos (ja + cos 2a sin 6a)

= : sin (2a + fla) = : sin 8a.

All the transformations carried out are reversible, andany real value of ex is permissible. ~

Example 2.3.15. Prove that the expression

2 (sin" x + cos6 x) - 3 (sin" x + cos! x)

is independent of x.~Apply the fundamental trigonometric identity:

2 (sin" x + cos" x) - 3 (sirr' x + cos! x)

= 2 ((sin 2 X)3 + (cos'' X)3) - 3 sin' x - 3 cos" X

=== 2 (sin" x + cos'' x) (sin" x - sin'' x cos'' x + cos" x)

- 3si1l 4 X - 3 cos! X

== 2 (sirr' x - sin? x cos? x + cos" x)

- 3 sin" x - 3 cos! X

= -(sin4 x + 2 sin" x cos'' x + cos! x)

== -(sin2 x + cos" X)2 == -1. ~


One should remember that in problems on transforminga trigonometric expression it is always supposed, althoughfrequently not stipulated explicitly, that the givenexpression must be transformed in the domain of itsdefinition, that is, only for the values of the argumentsfor which the suggested expression has sense.

Example 2.3.16. Prove the identity

3+4 cos 4a+cos 8a t4 ')3-4 cos 4a+cos 8a = co .(1 .

...-1-'he right-hand side of the identity is defined for numbers a such that sin 2a =1= 0, but it is not yet clear whetherthe domain of the right-hand side coincides with the domain of the left-hand side, which is defined by the condition 3 - 4 cos 40:. + cos Ba =1= O. It is not convenient tobegin the solution by determining the domain of the lefthand side. One has first to carry out formal transformations using identities (2.40) and (2.41):

3-+·4 cos 4.a+cos Sa _ 2+4 cos 4a+2 cos2 4a3-4 cos 4a+cos Set - 2-4 cos 4a+2 cos2 4a

2 (1+2 cos 4a+cos2 4a) _ 2 (1+cos 4a)22 (1-2 cos 4a +cos2 4a) - 2 (1- cos 4a)2

2 (2 cos- 2a)2== 2 (2 sin2 2a)2 == cot4 2a.

As a result of these transformations, it has been clearedup, in passing, that the denominator of the left-handside is equal to 2 (2 sin" 2a)2, and the domain of definition of the left-hand side coincides with that of theright-hand side. In the given domain of definition

sin 2a =1= 0, that is, a =1= ~k, k EZ, and all the transforma

tions carried out are reversible. ~

When proving identities involving inverse trigonometric functions, their domains of defini tion should be trea tedattentively.

Example 2.3.17. Provo that if x E [0, 1], y E [0, 1J,then the following identity holds true:

arcsin x + arcsin y ==.: arccos (V 1 - x2V1 - y2- xy) .

Let 0:. == arcsin x, ~ = arcsin y, then sin a == x> 0,O~ a~ n/2, sin ~ == y?3 0, O~ ~~ n/2, O~ a + ~~


11, and the number a + ~ lies in the range of arc cosine.Further note that

cos ex ~ -V1- x2 , cos ~ :.::.; ·V 1- y2 ,

since the points P'; and P y lie in the first quadrant. Now,we may apply formula (2.1) for the cosine of a SUIIl to get

cos (ex +~) ~ cos a cos ~ - sin ex; sin ~

=== V1- x2 V1- y2 - xy.

Hence it follows that

a + ~ ;= arccos (V 1- x2 V1- y2 - xy) ,

since ex + ~ E [0, n], ~Note that the identity being proved ceased to be valid

if the condition x E [0, 1] and Y E [0, 1] are not fulfilled.For instance, for x = -1/2, y = -1/2 the left-handside is negative, while the right-hand side is positive.

3. Transforming Sums and Products of Trlgonometr!cExpressions.

Example 2.3.18. Find the product

p = cos ex cos 2a cos 4a . . . cos 2lta

for a =1= nk, k EZ.~_Multiply the given product by the number sin a whic his nonzero by hypothesis (a =1= ttk, k EZ). ThIs transformation is reversible, and we get

P sin a. = (sin a. cos a) cos 2a ... cos 21t a

= { sin 2a cos 2a cos 4a ... cos 2" a

= ~ (2 sin 2a cos 2a) cos 4a cos 2" a

= ;3 sill 8a ... cos 2" a ,= .

_ 1 · 2· u. 2n 1. 21t+1- ¥ SIn ex cos ex = 2n +1 SIn (1,.

Consequently,1 sin 2n +1 a

p ==- 2n +1 · sin a • ~

2.9. Trigonometric Transjormations 75

Example 2.3.19. Compute the sum

S = sin a + sin 2a + ... + sin na,

~ If a =: nk (k EZ), then S == sin nk + sin Znk + . 0 • +sin nstk = 00 Let a =1= nk (k EZ). We multiply the suru

S by sin ~ =1= 0, and then use formula (2.29) for the

product of sines. We get

S · a (si + 0 2 + in nee) · a.SIn 2 = SID (X SIn a ... -+- SIn na SIn 2

· .a+02. (1..+ -l-si . (J,== SIll a SIll 2 SIn a SIn 2 · · · SIn na SIn 2

= ~ (cos ~ _ cos ~a) -+ ~ (cos 3; _cos 5; )+ + 1 ( (2n--3) a (2n-1) (1,)

• 0 • 2 cos 2 - cos 2

+ 1 (2n-1)a (2n+2)a)2" COS 2 cos 2

1( a (2n+1)C'.t). (n+1)a. no:~"2 cos 2-COS 2 = SIn 2 SIn -2-

(we have also applied formula (2.22)), \VhCllCU

(n+1) a . nasill 2 SIn -2-

s= . ~" a

SID 2

Example 2.3.20. Compute the sum

S = cos a + cos 2a -~ ... + cos na,

~ If a == 2nk, k EZ, then

S == cos 2nk -t- cos 4nk + . . . + cos nsik

= 1 + ... -t- 1 = n,

Let a =1= 2nk, k EZ, then sin ~ =f=. 0 and, using the

formulas (2.27) and (2.18), we get

S . aSIn 2""

· a -L' a. 2 I +' ex,==: SIn 2 cos ex I SIn 2 cos a.-... SIn 2 cos na


1 (. 3a . a) I 1 (. 5 . 3a)== 2 SIn 2 - SIn 2 .. - 2" SIll 2" ex- SIn -2-

+ + 1 (. (2n-~1)a . (2n-1) a)· · · 2 SIn 2 - SIn 2 .

1 (. (2n+1)a, . a) . no: (n+l)a=-="2 SIll 2 - SIll -2- === SIll -2- cos ~"';"2--

Consequently,(n+1) a

2. na

SlIl -2- cosS~------.~

. exsluT

Example 2.3.21. Prove the identity1 a, 1 a. 1 a"2tauT+t; tan T+ .. · + 2n tan 2U

1 a==21lcot ~-cota.

~ First, we prove a useful auxiliary formula:sin x cos x

tanx-cota==-- --.-cos x SIn x

cos2 x-sin2 x == _ 2 cos 2xsin x- cos X sin 2x

:=;. - 2cot 2x.

1 a IThen we subtract the number 2n cot 2n from t re left-

hand side of the identity and apply n times the formulajust derived:

1 a 1 ex, 1 ex."2 tan 2 +. 4 tan T -1- · · •+ 21t-1 tan 2n-1

1 a, 1 a.-t- 2n tan ~- 211 cot 211

-1 a 1 ex, 1 a== 2 tan T + 4 tan T + ···+ 2n-1 tan 2n- 1

1 ( ex a) 1 a 1 a+- tan --cot - == -tan--I-- tan -2n 2n 2n 2 . 2 4 4

1 a 1 a+ ··· -1- 2n-1 tan 2n - 1 - 2n - 1 cot 27H'

1 a 1 a 1 a== 2 tan 2 + T tanT -1- · · . - 2n-2 cot 2n-2

Problems

1 a 1 a 1 a== • . • == "2 tan T -+- 4 tan T - "4cot T

1 ex, 1( a a)- - tan - +- tan - - cot --2 2 4 4 4

1 ex 1 a- -tan ---cot-- 2 2 2 2

1 (a. a )=="2 tan T-cotT == -eota.

77

The given identity follows from this continued equality.Note that the domains of definition of both sides of thegiven identity coincide with the set of real numbers Ow

such that sin a =1= 0, that is, a =f= stle, k EZ.

PROBLEMS

In Problems 2.1 to 2.17, prove the given identities:

tan 2t +cot 35 tan 2t2.1. cot2t+tan3s = tan Bs '

cos2 t - cos'' 52.2. cot 2 t-cot2s= . 2 t . 2 •sin SIn s

sin 4a. cos 2a. t ( 3 )2.3. · 2 == co - n - a1+cos 4a 1+cos ex 2 •

2.4. (cos ex - cos ~)2 - (sin ex - sin ~)2

a-p= - 4 sin2 -2- cos (a -r- ~).

sin! t+cos4 t-1 22.5. sine t+cos6 t-1 - 3 •

2.6. cot t-tan t-2 tan 2t ~ 4 cot 4t.2.7. tan Bt- tan 4t - tan 2t ~ tan 6t tan 4t tan 2t.

2 8 t 3t - 3 tan t-tanSt•• ant - - 1 _ 3 tan 2 t ·

2.9. sin! t = ~ (cot 4t - 4 cos 2t + ; ) .2.10. cos 4t = Scos! t-8 cos- t+ 1.

2 tt 3-4cos2t+cos4t_t 4t• • 3+4 cos 2t+cos 4t - an ·


2.12. cos t + cos (t + s)+ cos (t+ 28).

( ns ) ( n+ 1 )cos t+ 2 sin -2- 8

+ ... + cos (t +ns) ----: .. ssiu 2

2.13. sin t + sin (t + s)+ sin (t + 28)

. sin(t+~S)sin(nt1s)+ ... +sln(t+ns) = .

ssin 2"

2.14. 8cos4 t +4 cos" t- 8 cos2 t- 3 cos t+ 17 t= 2 cos 2' t cos 2 .

2.15. sin (t + s + u) = sin t cos s cos u

-t- cos t sin s cos u + cos t cos s sin u

- sin t sin 8 sin u,

2.16. cos (t + S + u) = cos t cos s cos u

- sin t sin s cos u - sin t cos s sin u

- cos t sin s sin u:

tan2 t - tan- s2.17. 1 t 2tt 2 =tan (t+s) tan(l-s).- an ,an s

In Problems 2.18 to 2.23", simplify the indicatedexpressions.

2.18. sin2 ( T+28) - sin2( ~ - 28) .

2.19. cos- (t+ 2s) {-sin" (t-2s) -1.

2.20. cos" (t- ~ )+sin6 (t- a: )- f (sin2

( t + ~ )- cos2( t + a: ))2 •

2.21. sin2 ( 3: _2t ) _ sin2 ( 7: - 2t )

. t3n ( 11'1 )-SID 12cos 12-4t .

Problems 79

2.22. sin" 2t cos 6t -1- cos" 2t sin 6t.2.23. 4 (Si1l4 t -~ cos' t) - 1 (sin" t --j-- COSO t) - 1.In Problems 2.24 to 2.28, transform the given expres-

sions into a product.2.24. sin 6t - 2V3 cos'' 3t + va.2.25. tarr' t - 4 tan" t + 3.2 26 . sin2 (t+s)-sin2 t-sin2 s

• • sin 2 (t+s)-,cos2 t-cos2 S

2.27. sin2 (2t - s) - sin'' 2t - sin" s.2.28. cos 22t + 3 cos 18t + 3 cos 14t + cos 10t.In Problems 2.29 to 2.32, check the indicated equalities.2.29. cot 70° + 4 cos 70° = va.2.30. sin2 ( arctan 3 - arccot ( - ~ )) = { .

2.31. sin (2 arctan ~ ) + tan ( ~ arcsin ~) = f .2.32. sin (2 arctan i- )- tan ( ; arcsin H) = ~ .

In Problems 2.33. to 2.36, compute the given expres-sions.

2.33. arccos (cos (2 arctan <V2 - 1»).

2.34. sin2 ( arctan { - arctan ( - ~ ) ) •

2.35. cos ( ; arccos : -- 2 arctan (- 2) ) •

2 36 sin 22° cos 8°+cos 158°cos 98°• • sin 23° cos 70 +cos 1570 cos 97° •

2.37. Find tan ~ if sin t + cos t = 1/5.2.38. Knowing that cx, p, and V are internal angles of

a triangle, prove the equality

· · ~+ · 4 a ~ ySIn ex+ SIn t' sm y == COR T COS 2" cos "2.

Chapter 3

Trigonometric Equationsand Systems of Equations

3.1. General

Written entry examinations include, as a rule, problemson solving trigonometric equations. This is partly explained by the fact that there is no general method whichwould be applicable for solving any trigonometric equationsand in each concrete case the search for a solution require,a certain of ease in carrying out identical transformationsand the knack of finding and applying the proper trigonometric formula. In most cases, the transformationsused to solve such problems are generally aimed at thereduction of a given equation to several simple equationsto be solved in a regular way, as it was described inSec. 1.4.

It is of importance to note that the form of notation ofthe roots of trigonometric equations often depends on themethod applied in solving a given equation. To prove thefact that two different notations of the answer are equivalent is sometimes an interesting problem in itself,although the examination requires to solve the givenequation using only one method, rather than to transformthe answer into other notations.

When solving trigonometric equations and systems ofequations the student has frequently to deal with rathercomplicated expressions composed of trigonometric functions. Competition problems often involve expressionsthat, along with trigonometric functions, contain othertypes of functions (inverse trigonometric, exponential,logarithmic, rational, fractional, etc.). The student mustremember that in most cases such expressions are notdefined for all values of the variables in the given expression. Frequently, identical transformations result insome simplifications, however, the equation (or system)obtained may have another domain of permissible values

8.1. General 81

of the unknowns. This can lead to "superfluous" or"extraneous" roots if the domain of permissible values isbroadened as a resul t of a certain transformation or evento a loss of roots if as a result of some transformation thedomain of permissible values is narrowed. A good methodto avoid such troubles is to watch for the invertibili ty ofthe identical transformations being carried out on thedomain under considertion. Whenever the transformationsare not invertible a check is required. Namely, first, findand check the possible values of the roots which mighthave disappeared as a result of the reduction of thedomain of permissible values, and, second, make sure thatthere are no extraneous roots which do not belong to thedomain of permissible values of the initial expression.

Example 3.1.1. Solve the equation

sin x + 7 cos x + 7 = O.

~One of the methods for solving this equation (thoughnot the best) consists in using the universal substi tutionIormul as (2.42) and (2.43):

x. 2 tan 2"~lnx= x '

1+lan2 -2

1-tan2~2

cosx~-----

1-1- tan 2 -=2

As a result, we get the following equation:

2 tan ~ I 7 ( 1- tan2 ~ )

----- T x + 7 :-:= 0,1+ lan 2 ~ 1+tan2 -2 2

which is equivalent to tan; = -7, and, consequently, to

x == 2nk - 2 arc tan 7, k EZ. However, the above reasoning is erroneous, since the universal substitution formu-

las are applicable only to the x's for which tan ~ is de

fined, that is, x =/== rt 1- 2rtk, k EZ. Therefore, as a resultof the transformations, the domain of definition of thefunction on the left-hand side of the equation is reducedand in order to get a correct answer, it is necessary tocheck whether there are roots among the numbers which

fi - 01644

82 H. Trigonometric Equations and Sust cm«

are left outside the domain of definition of the expressionobtained, that is, to chock whether or not the numbersx == n -f- 2nk, k EZ, arc also roots of this expression.J1"Ol' these values of x we get sin x == 0, cos x == -1, andthe equation turns into an identity (we mean the originalequation), therefore the correct answer is:

x == -2 arctan 7 -+- 2nk, x == n + 2nk, k EZ.

This problem can be sol ved in a shorter way, by usingonly invertible transformations. For this purpose, it isnecessary to use the method of introducing an auxiliaryangle (Theorem 2.2). We divide both sides of the equa-tion by the number 'V 12 + 72 == Y50, transpose thenumber tfy 50 to the right-hand side, and then rewritethe equation as follows

sin (x + fP) = -7IVSO,

where (p == arcsin .:_. Thus, the original equation hasy 50

been reduced to a simple trigonometric equation of theform sin t === a considered in Sec. 1.4, and the generalsolution is wri tten in the following form: -

u: -1- rp == (_1}H arcsin (-7)fV' 50) + nn, nEZ,or

x = - arcsin :. + (-1)"+1 arcsin ./_ -+- nn, n EZ.l 5f) y 50 .

Thus, for an even n == 2k, If EZ, we get a series of solutions

2 . 7 I 2 Ix ::- - arcsui V'S!) .-: - n r,

and for an odd n = 2k + 1 another series of solutions

x = n + 21tk, k EZ.

The equivalence of the two notations of the answernow fol lows from the equal ity sin (arctan x) ==- x/V 1+ .r2

(see Example 1.4.6) or arctan x = arcsin y x ,1+x2

1 .7 7w renee arcsin r- == arctan . ~1 50

3.1. General 83

Example 3.1.2. Among tho roots of the equation

CO~:~Jtx :_-:: 0'1+ V3 tan nx

find the one which has the least distance from the number'V 8" on the number line.

~The function tan nz is defined if nx =f= ~ + nk,

i.e. x=t= ~ +k, kE Z.

In order to find the domain of defmi tion of the function on the left-hand side of the equation, it is necessary,in addition, to bear in mind that the denominator mustnot vanish, that is, 1 + V3tan nx "* 0, or tan nx =1=-1I"V3, which is equivalent to tho condition nx =t= - ~+

sik, that is, .r =I=- - ~ -+- k, k EZ. Thus, the domain of

tl f t.i COR 3nx . t f 1 1I ie nne Ion I consists or rea num iers x

1 -t- l 3 tan nx1 1

such thatr* 2 + k and x =1= - 7\ + k for k E Z. In the

given domain of permissible values the equation is equivalent to the following:

cos 31tx = 0

whose solution has the form:

o n + · t t n EZdnx ~ 2 sin; i.e x ~ "'if - -3 ' n ·

Let us now find out which of the obtained values of x (10

not belong to the domain of definition. We have:1 n 1ff + "3 :=: 2" + k for n == 1 + 3k,

1 n tfi+3"== -6-~lc for n~ -1+:1k.

Note that n = 3k + 1, k E Z, is the set of all the integers not divisible by 3. Consequently, the formula for xdescribing the solutions of the given equation must (',011

tain only n d ivisi ble by 3. Setting n =-= 31£, we get x :-:-=

:i + k . k E Z. Further, from these solutions we choose

tJ. Trigonometric Equations and Sustems

the one which satisfies tho add itional condition of theproblem (with the least distance Irom the number Vs=2"V2). Since 1.41 < y2 < 1.42, we have 2.82 <2 Y2< 2.84, and it suffices to consider the t\VO numbers

1x==--·~-3,6

which are the closest to the number V·S. The problemhas been reduced to comparing the two numbers

12 ~ - 2 y2:1 = 2 y2: - 2 ~

and

/3 ~ -2 V2/=3 ~i --2 V2.The following inequality holds trne:

1 y- ,r- 136"-22<2 y2-2(fe

Indeed, it is equivalent to the inequality 4V2" > 5 ~which is proved by squaring both members since both of

them are positive. Therefore the number x = 3i is the

sought-for solution of the given equation. ~If in this example the domain of permissible values is

treated inaccurately, for instance, one forgets about thecondition 1 + Y3tan nx =1== 0 or that tan nx must bedefined, then a wrong answer is obtained.

Even these two examples show the importance of taking into account the domain of permissible values whensolving trigonometric equations. However, it is not a1ways convenient to write out the domain of permissiblevalues in an explicit form, that is, to indicate explicitlyall the values of the unknown belonging to this domain.It suffices to write the conditions wherefrom this domaincan be found. Thus, in-Example 3.1.2, these conditionswere cos nx =1= 0 (i.e. tan Jtx is defined) and 1 -t..Y3 tan 1tX =1== O. Sometimes, one succeeds in solvingsuch conditions with respect to x in a simple way (as in

3.1. General 85

Example 3.1.2), but in most cases this is a cumbersomeproblem in itself. I-Iowever, when checking individualvalues of the unknown which may turn out to be roots,one may also successfully use au implicit representationof tho domain of permissible values, checking whetherthe indicated conditions are met for the val lies undercousideration.

Example 3.1.3. Solve the equatioJl

2 sin (3x + ~ ) =V1 + 8 sin 2x cos- 2x . (3.1)

<llIlIIIThe domain of permissible values of the given equation is specified implicitly by the conditions: 1 +8 sin 2x cos" 2x > 0, "sin"(3x+ ~) > O. If we squareboth sides of the equation, then, on the given domain,the original equation (3.1) is equivalent to the followingequation:

4 sin2 (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 original equation (3.1) are also the roots of equation (3.2),but all the roots of (3.2) will not necessarily be tho footsof (3.1). Therefore on finding all the roots of (3.2) wehave to choose those which will be the roots of the original equation. Applying formula (2.41), we get

i-cos (6x+ ~) 1lsin2 (3x+ ~) = =2"(1+sin6x),

2

using formulas (2.36), (2.27), we have

8 sin 2x cos" 2x == 4 cos 2x (2 sin 2x cos 2x)=== 4 cos 2x sin 4x = 2 (sin 6x + sin 2x).

l'herefore equation (3.2) may be rewritton as Iol lows

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

.2 1SIn x ==-2 ·

(3.3)

86 3. Trigonometric Equations and SY8tems

For a further invostigat.iou, the solutions of this equationshould be conveniently written ill the form of two seriesof solutions (bu t not uni ted, as usual, into oue; seeSec. '1.4):

11: 5n Zx-~12-t--nn, x~"12.~-nn, nE·

Since equation (:-3.3) is equivalent to equatiou (3.2), wehave to check whether all of its solutions are the solutions 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 been

fulfilled. For x == ~ -t- sin, nEZ, the left-hand side of

the 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 numbers

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 second 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 trigonometric identities, to express all the trigonometric functionsin the equation being considered in terms of one trigonometric function.

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

is,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 formulas 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


cos 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. ~


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 x

i-tan T tan x

H 1+tan x .. ence, tan x + 1 t == - 2. Setting y ~ tan x, we- anxget an algebraic equation:

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

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

3.2. Methods of Solving Equations ., ~H

consequently, tau x = + V~ that is,

1tx·::=. + T+nn, nE Z.

Both series of solutions belong to the domain of permissible values of the original equation which was not reduced under transformatio-n.


(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 consider 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 x

cos 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 · ~


~ 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 form

8 sin x cos x + 3 cos 2x - 5 == 0,or

4 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 x

and designate y = tan x. Then the equation is transformed into an algebraic one:

8y 3-3y2

1+y2 .+ 1+y2 =5,

or 8y + 3 - 3y2 === 5 + 5y2, whence

2 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 the

left-hand side of (3.5), which is equivalent to the original 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 particular, 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.


sin'' x + cos" 3x === 1.

~Applying identities (2.40) and (2.41), we get

~ - ~ cos 2x+{+ ~ cos 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 equations:

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. Consequently, 4x == rui or:

x ~ :nn/4, n EZ. ~


sin x + sin 2x -t- 2 sin x sin 2x == 2 cos x + cos 2x.

<lllIIIIBy virtue of identity (2.2n) for the product of sines,we have

sin x + sin 2x + cos x - cos 0X == 2 cos x + cos 2x,

92 3. Trigonometric Equation." and Systems

orSill 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 get

sin 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 == 0

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 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' - ,

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

sin 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 also

sin fix == 0, and the equality sin ~ == 0 means that

sin x == 0 (sin 2nn === 0, n EZ). Consequently, the domain 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 expression

cos (2 (~ + n;t) )= cos n(21~t:.!~ takes on val ues eq ual to

1/2, 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 or

n = 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 nand check whether the constraints cos 2.1: =1= 0 and

sin 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 to

see that the expression sin (fi (~ + 2~n)) does not vanish

either for any integral values of n, Thus, we have found

all the solutions of the original equation: x = ~ -+ Jl;,1t 21tn

X = n + 9' n E z. ~


5 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 identity and (2.54), we rewrite the given equation in the form

5 (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 Inllowingthree equations:

(1) cosx=O,

(2) sinx-==1,

nx~2+nn,

n , 2X~2T sin,

nEZ,

nEZ

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

(3) 6 sin" x + sin x - 2 == O.

Setting y = sin x we get an algebraic equation6y2 + y - 2 == 0,

whose roots are Yl == -2/3 and Y2 == 1/2, and it remainsto consider two cases:

(a)'sinx=--}, x=,(_1)n+'arcsin ;+nn, nEZ,

(b) sin z e- ~ , x=(-1t ~ +nn, nE Z.

Thus, all the solutions of the original equation aredescribed by the formulas

x= ; +:rrn, x=(-1t+larcsin; +nn,

X :-=: (-1)11 ~ + nn, n EZ. ~

96 3. Trigonometric Equations and Sustems

1sin 2x '


col2x+3tan3x-==2tanx+ .2 .sln4x

~Let us represent the given equation in the form

(cot 2x + tan x) +:3 (tan 3x - tan z) ~ ~4sin X

and use the Iol lowing identity:

cos 2x sin xcot 2x + tan x ~ sin 2x --1' cos x

cos 2x cos x+~in 2x sin x cos xsin 2x cos x sin 2x cos x

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

_._1_ -1- ~ sin 2x, 2sin 2x ,cos 3x cos r sin ~.(. ·

The domain of permissible values for the given equationcan he specified by the following two conditions: sin 4x =I=0, that is, x =I=- sinl/«, nEZ, and cos 3x =1= 0, that is,

x =1== ~ + n;, n E Z. We transform the obtained equation

on the given domain:

sin 2x ~

3 sin 2xcos 3x cos x

sin'2x3 3cos x cos x

2sin 4x

1sin 2.xcos 2x

1

1sin 2x '

~ sin 2x i-cos 2.r.i) cos 3x cos x == sin 2.1' cos 2:r '

Gsin x cos x 2 sin2 xcos 3x cos x 2 sin' x cos x cos 2x •

Since sin x =1= 0 and cos x =1= 0, w~ have

Geos xcos 3x

1cos 2x •

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

f) cos x cos 2x =-= cos 3x,


1

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

6 cos x cos 2x == 4 cos" X - 3 cos ~1:,

whence

6 cos 2x === 4 cos" X - 3 (cos x =1= 0),

Gcos 2x == 2 (t + cos 2x) - 3,

6 cos 2x == 2 cos 2x - 1,

4 cos 2x == -1, cos 2x == -1/4,

whence x = ± ~ arccos ( - ~ ) + sin, n EZ.

It is easy to check that the found values of x satisfy theconditions specifying the domain of permissible valuesin which all the transformations carried out were invertible. ~


1"2 + cos x -t- cos 2x + cos 3x + cos 4x = O.

~ Note that x == 2rtn, 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 t· cos 3x + cos 4x

.1

x (2 cos x sin ; +2 cos 2x sin i2.1n 2

+2 cos 3x sin ; +2 cos 4x sin ; )

1 (. 3 . x+. 5 . 3SIn - X-SIn - SIn - X-SIn-x2' x 2 2 2 2

SIll T

+ . 7 .5+.9 .7)SIn 2 X-SIn "2 x SIn "2 X-SIn 2" x

9

(. 9 . X) sin "2x 1

- -2-'--X - SIn "2 X-SIn"2 = 2' x -"2'Sin T Sin 2"

7-01644

98 .1. Trigonometric Equations and Systems

Consequently, the original equation can be transformed tothe form

DsiJl:T x

x2sin T

=0 or. 9

SlOT x == 0,

h - 2:tlr 1. Z . 1 1 ') 1 . Iw ence x - -9-' u. E , prov ir e( x =t= _rrn W uc 1 is

equivalent to 2;k =1= 2nn 01' k =1= 9n (11 EZ). ~

3. Sol V lug Trtgonometrtc Equations Using the Properties of 'I'rlgonometrlc Functions. Frequen tl y, we have todeal wi th equations of the form t (t) == g (t), where fand g are some functions containing trigonometric expressions such that enable us to investigate the domains ofvalues .E (/) and E (g) and to prove that these domainseither do not intersect or have few points in common. Insuch cases, the sol utions of the equation t (t) = g (t)should be sought for among such t's which satisfy (simpler) equations f (t) === a, g (t) = a, where a is a real number such that a EE (I) and a EE (g), that is, a EE (I) nE (g).

Example :i.2.11. Solve the equation

sin2 4x -f- cos" x = 2 sin 4x cos! :1..

~Let us write the given equation in the form

sin'' 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" X

or(sin 4x - cos! ·X)2 = -cos2 ,7; (1 - cos" x).

The left-hand side of the equation is nonnegative, whilethe right-hand side is nonpositive (cos'' x ~ 0, 1cos" X ~ 0), consequently, the equality will be valid onlywhen the following conditions are fulfilled simultaneously:


{

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

(sin 4x-cos' X)2= 0.

The first equation decomposes in to two:(1) cos" x == O· or cos x = 0,

whence x=- ~ + sen, n EZ. The ohtained values also

satisfy the second equation since

sin (4 ( ~ + nn ) ) = sin (231 +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 consists

of the numbers x= ~ ~nn, nEZ. ~


sino x + cos" X = p,

where p is an arbitrary real number.<lIlIIIII Note that

sin6 x +cos" X

==- (sin2 x + cos2 x) (sin' x - sin2 x cos2 x +cos' x)

= sin' x - sin2 x cos2 x -l-cos- x

== (sins x +. 2 sin2 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== B cas 4x+ 8 '

and the given equation takes the form

3 L 5 4 8p-5gcos&Jx+g=p or cos x=-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 fundamental trigonometric identity and formula (2.11) for the sineof the sum of two numbers. We get

cos: sinx-2sin2x+cosx+sin : cosx-2cos2x=O,

that is,

sin(x+ ~ )+cosx-2(sin2x+cos2x)=O,

or

· 5x 2sin T+cosx= .

Note that the sum in the left-hand side of the obtained

equation will equal 2 only if sin 5: = 1 and cos x = 1simultaneously, that is, our equation is equivalent to thesystem of equations:

{

· 5x 1slnT= ,

cos x== 1,

whence

{5: = ~ +2nn, nEZ,

x=21tk, kEZ,

. 2,,; 8nand the equality 2nk = 5+ 5 n must hold, whence

k = 1~4n • Since kEZ, we have n=5m+1, mEZ

(since for the remaining integral n's., that is, n = 5m!

3.3..Equations in Several Unknowns 10t

n = 5n~ + 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 certain 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 encountered 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 different cases, to check the validity of auxiliary inequalities, etc. In some cases, when solving systems of equations 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 functions as new unknowns. As in solving trigonometricequations in one unknown, we can try to carry out identical 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

{Vsinxcosy=O,

2 sinZ x- cos 2y- 2 === O.

~ It follows from the first equation thatsin x> 0, twocases are possible here: if sin x == 0, then the equationturns into an identity,and if sin x ;» 0, then the equationimplies that cos y = O. Consequently, the system is equiv-

t02 3. Trigonometric Equations and SNstems

alent to the collection of two systems:

{sinx=O,

2 sin2 x - cos 2y - 2 = 0and

{cos y = 0,

2 sin2 x -cos 2y-2 = 0, sin x > o.The first system has no solutions (cos 2y -t- 2 =1= 0),while the second is equivalent to the system of two simpleequations

{cosy=O,

sin x= V2/2.

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

( (- 1)k ~ + nk, ~ + nl), k, l EZ. ~


3+2cos{x-y) __ 1/3+2 2 '02 x-y ~ sin2(x-y)

2 - V x-x cos -2-'- 2

<lllIIIII Using formula (2.40) for reducing the power of cosine, we get

3+2cos(x-y)2

= V3+2x-x2 1 + CO~(X-y) + Sin2~-y)

or

1-sin2 (x -y) + (2 - V3-t- 2x- x2) cos (x- y)

-l- 2- V3 +2x- x2;..= 0,

cos- (x- y) + (2- V3 +2x-x2) cos (x- y) + 2

- V3+2x-x2 == O. (3.6)

Let us set t == cos (x - y) and a = 2 - V3 + 2x - x 2,

then equation (3.6) can be rewritten as follows:

t2 + at + a == O.

3.3. Equations in Several Unknowns 103

The given equality can be regarded as a quadratic equation with respect to t which has solutions only if its discriminant is nonnegative, hence,

a2- 4a ~ 0 or a (a - 4) > 0,

whence a ~ 0 or a > 4. I-Iowever, according to the introduced notation, a == 2 - V3 + 2x - x2 == 2 V4 - (x - 1)2, and therefore 0 == 2 - V4~ a ~ 2.Consequently, a == 0, whence 2 - V4 - (x - 1)2 === 0or x === 1. Then equation (3.6) takes the form cos" (1 - y)= 0 or cos (y - 1) = 0, whence

y-1==~+JTn, nEZ.

Thus, all the solutions of the original equation are

pairs of numbers (x, y) of the form (1, 1 + ~ + nn) ,nEZ. ~


{cot x -~- sin 2y == sin 2x,

2 sin y sin (x + y) ~~ cos x.

~ Using formula (2.29), we can represent the secondequation as follows:

cos x - cos (x -r- 2y) = cos x,

Consequently, cos(x+2y)==O, whence x+2y~: ~ +J1k,

x = ~ --2!1+nk, kEZ. Note that cotx=~cot( ~ -2!1) =

tan 2y and sin 2.1: = sin (n - 4y) == sin 4y. Therefore thesubstitution of :£ into the first equation yields:

tan 2y + sin 2y = sin 4y,or

sin 2y (1 + cos 2y - 2 cos" 2y) == 0,

sin 2y (cos 2y -1) (cos 2y +f)-= o.The last equation deCOITlpOSeS into three equations:

sin 2y = 0, cos 2y === 1 and cos 2y ::=: -1/2.


From the equation sin 2y ==0 we get y = n;, nEZ,

and therefore x ==T + sil, l EZ. From the equation

cos 2y == 1 we find 2y == 2nn, y ~ tin, n EZ; hence, x==;

~ - 2nn -t.. tik == ~ + til, l EZ. We see that the set of

solutions of the second equation belongs to the setof solu lions of the first equation. Finally, from the

equation cos 2y = - ~ it follows that 2y = + 2; +2nn,

and therefore y:.~ ± ; + nn, n EZ; hence, x == ~ +2n n 2n IT- 21(n+ sclc =-~ 2 + 3 + stl , l EZ. The fina answer:

( ~ + ttl; n;), (~ + 2; + stl; + ~ +nn) •

n, lE z. ~


{IX!+~:} =3,SIn -2--1.

~From the second equation it follows directly thatnx2 1t-2- =-=- 2"+ Zsck ; that is, x2 =--:..: 1 + 4k, k EZ, whence Xi ==- V4k +1, x 2 === V4k + 1. By virtue of the firstequation of the system, we have I x I~ 3. Consequently,It: may take on only the valnes 0, 1, 2. Thus, there are sixvalues of x, namely: +1, + V5, +3. If x :=:: +1, thenI y I == 2, that is, y === +2; if x == + V5, then I y I =3 - V5, that is, y == +(3 - V5); finally, if x == +3,then I y I == 0, that is, y == o.

Answer: (+1, +2), (+5, +(3 - V5)), (+3, 0), allcombinations of sign being possible, that is, the systemhas ten solutions. ~

Example 3".3.5. Find out for what values of a thesystem of equations

{sin xcos 2y == (a2 - 1)2+ 1,

cos x sin 2y ::::: a + 1has a solution. Find all the solutions.

3.3. Equations in Several Unknowns 105

~ Since the left-hand sides of the equations do not,obviously, exceed 1, the given system may have a solution only for the a's such that

{(a2-1)2+1~1,

la+ 11~1.

Solving the first inequality, we get a = +1, the secondinequality being satisfied for a = -1.

Thus, the original system of equations has a solutiononly for a == -1 and, consequently, takes the form

{sin x cos 2y == 1,cos x sin 2y == O.

Adding and subtracting the equations of the systemtermwise, we get the system

{sin x cos 2y + cos x sin 2y == 1,sin x cos 2y - cos x sin 2y == 1,

which is equivalent to the given system, whence (by formulas (2.11) and (2.12»

{sin (x + 2y) !=~ 1,

sin (x - 2y) ::= 1.

Obviously, the last system yields an algebraic system

{

X + 2Y: ~ + 2nk, Ii, EZ,

x-2Y--T+ 2nn, nEZ.

Solving this system, we getn n(k-n)

x~:2-l-(k+n) n, y:--= 2 •

Thus, (~ +(k+n}n, ~(k-n)), k, nEZ, are all

the solutions of the original system. ~Example 3.3.6. Find all the solutions of the system

{Isinxlsiny==-1/4,

cos (x + y) + cos (x - y) == 3/2,

such that 0 < x < 231, It < Y < 2Jt.

f06 3. Trigonometric Equations and Systems

~Consider the case when sin x ;;:: 0, then, by hypothesis; 0 < x ~;rt, and

[sin xl sin y = sin x sin y = ~ (cos (x- y) -cos (x + y».

Then the system takes the form

{cos(x-y)-cos(x+y)~ -1/2,

cos (x + y) + cos (x - y) =.:: 3/2,

whence

{

COS (x - y) == 1/2,

cos (x + y) == 1.

By subtracting and adding the inequalities 0 < x ~ nand Jt < y < 2n we get -21£ < x - y < 0 and It <X + y < 3n. Thus, we have two systems of equations

{

11: { 5Jtx-Y=-3' x-Y=-3'

x+y==21t, x+y~21t,

whose solutions are (5n/6, 7n/6) and (n/6, 11n/6), respectively.

Similarly, if sin x < 0 we get

{

COS (x - y).:- cos (x + y) ~ 1/2,

cos (x + y) + cos (x - y) == 3/2

and rr < X < 2n, whence

{

COS (x - y) == 1, .cos(x+y)=1/2, -n<x-y<n, 2n<x+y<4n,

whence

{x-y=O, and {X-Y=-.;O,x+y == 7:n/3, x+y == i'ln/3.

The solutions of these systems are (7n/6, 7n/6) and(1111/6, 111t/6), respectively. Finally, the original systemhas the following solutions:

(5n/6, 7n/6), (n/6, 1in/6),

(71t/6, 7n/6), (i1n/6, 11n/6). ~

1 12 y cos -iL-SI-I12 y cos lli -71+124 Y cos lli +13 = /11- y s-i-n-1t-(X---3~2Y--~1) ,

2 (x 2+ (y-a)2) -1 = 2 ·Vx2+ (y -a)2- : .

3.3. Equations in Several Unknowns f07

Example 3.3.7. Find all the val ues of a for which theindicated system of equations has a solution:

r

\~The left-hand side of the first equation has the form

F( ycos lli), where F(z)=I12~-SI-I12z-71+

124z+ 131· Bearing in mind that Z= V cos i > 0,

we test the function F (z) for z~ 0 using the method ofintervals. For this purpose, it is necessary to mark 011

the number line all the points at which the expressionsinside the modulus sign change sign and to consider ourfunction separately on each of the intervals thus obtained,thus getting rid of the modulus sign. Then, we get:(1) 0 ~ z ~ 5/12, F (x) = (5 - 12z) -1- (12z - 7)

-~- (24z -~- 13) = 11 + 24z,(2) 5/12 ~ z ~ 7/12, F (z) = (12z - 5) + (12z - 7)

+ (24z + 13) = 48z + 1,(3) z ~ 7/12, F (z) = (12z - 5) - (12z - 7)

+ (24z -1- 13) == 24z -t- 15.Thus, F (z) is an increasing function for z~O, its

least value being F (0) == 11. The right-hand side

of the first equation takes on values ::;;; 11 (since

V sin II (x-~!/-l) > 0) . Cousequent ly. the first equa

tion of the original system is equivalen t to the system

{

COS 1ty :::.:: 0,

sin II(X-;y-:) -c,O,or

{y == 2t + 1,

2 1 3t, v EZ. (3.7)

x- y- == v,


Consider now the second equation of the original

system. We substitute u = V x2+ (y -a)2- 14

,and get

2 (u2+ ~ )-1=2u or u2-u+ 1=0, whence u=1/2,

that is, x2+ (y-a)2- ~ :::..-= 1' or

x2+ (y - a)2= 1. (3.8)

Since x is an integer, only the cases :c == +1, x == 0 arepossible. Let us consider them separa tely.

Case (a) x === -1. From system (3.7), it follows that

2y == -3v - 2, y == 2t + 1, v, t EZ.

Rewriting these equalities in the form 2 (y + 1) == -3v,y -1- 1 == 2t + 2, we conclude that y -1- 1 is divisibleby 6. In addition, from (3.8) for x = -1 we have(y - a)2 == 0, hence y == a == 6k - 1, k E Z.

Case (b) x == 1. Here again from (3.8) and (3.7): y == a,2y == -3v, y === 2t + 1 for some integers v and t, whence2 (a - 3) == -3 (v + 2), a - 3 == 2 (t - 1). Hence,a - 3 is divisible by 6. In this case a == fik --1- 3, If; EZ.

Case (c) x == O. Here, according to (3.8), (y - a)2 === 1,y === a + 1, and system (3.7) takes the form

2y == -3v - 1, y === 2t -f-- 1, or

2 (y - 1) == -3 (v + 1), y - 1 =-= 2t, v, t E Z.Consequently, y - 1 is divisible by 6, and since a ==y + 1, we have a == 6k or a == 6k + 2, k E Z. Puttingtogether all the results, we get the final answer: the system has a solution if a takes on the values 6k - 1, 6k,6k -t- 2, 6k + 3, k Ez. ~

Example 3.3.8. Find all the solutions of the equation

V2 - I y I (5 sin" x - 6 sin x cos x - U cos" x + 3 V33)

= arcsin- x + arccos" x - ~ ,..;2.

~ Let us prove that the left-hand side of the equation isalways nonnegative, while the right-hand side is non-positive. Indeed, V2 - I y I~ O. Now, we transformthe expression in the parentheses:5 sin2 x - 6 sin x cos x - 9 cos" X + 3 V33

Problems 109

= 6 sin" X - (sin" X + 6 sin x cos x + 9 cos" x) + 3~ 33= 6 sin" x - (sin x + 3 cos X)2 + 3V33= 6 sin'' x + 3V33 - 10 sin" (x + cp),

where q> = arcsin 31V10 by Theorem 2.2. We then notethat 3V33 > 10 (this is checked by cubing the positiveright-hand and left-hand sides) and also that 10 sin" (x ---!rp) ~ 10, therefore the expression in the parentheses isalways positive, that is, the left-hand side of the equationis nonnegative and vanishes only in the case V2 - I y 1=0, that is, y = +2.

As to ' the right-hand side, we use the identi ty

arcsin x + arccos x = :rrJ2.

(If we rewrite this identity in the form arccos x =

~ -arcsin x, then its validity follows from t.he fact

that cos ( ~ - arcsin x) = sin (arcsin z) = x and that.

~ - arcsin x E[0, n], that is, belongs to the range of

values of arc cosine since x E[- n/2, n/2].) Let us sett ==arcsin x. Then the right-hand side of the equation

takes the form t2 + ( ~ - t ) 2 - : n2 = 2t2 - nt - n2 ,

where t E [-n/2, n/2l. The greatest value of this quadratic function is attained at the end point t = -n/2 of theclosed interval [-rt/2, n/2J and equals zero: 2 (-rt/2)2 n (-n/2) - n 2 = O. Thus, the right-hand side of theoriginal equation is nonpositive and vanishes forarcsin x = -11,12, that is, for x = -1.

Consequently, the equation has two solutions: (-1, -2)and (-1, 2). ~

PROBLEMS

In Problems 3.1 to 3.36, solve equations.

s.r. sin (~ +2x) cot3x+sin(n+2x)- V2cos5x=O.

3.2. sin x cos 2x +cos xcos4x = sin (~ +2x)

X sin ( ~ - 3x ) .

110 H. Trigonometric Equations and Systems

3.3. sin 2x = cos" ; - sin! ~ .

3.4. (1 + cos 4x) sin 2x == cos- 2x.3.5. sin" 2z + sin" 3z + sin" 4z + sin" 5z = 2.3.6. sin 2x sin 6x - cos 2x cos f3x = V2sin 3x cos 8x.3.7. sin 3x cos 3x = sin 2.T.3.8. cos 2x - 5 sin x - 3 = O.3.9. 3 sin 2x + 2 cos 2x = 3.

3 10 (3n ) 2 1+ cos 2x 0

• • cot T - x + cot x + sin2 x =.3.11. 6 sin" x + sin x cos x - cos'' X = 2.3.12. cos 7x + sin 8x = cos 3x - sin 2x.3.13. sin" x - 2 sin x cos x = 3 cos'' z,

3.14. cos 5x + cos 7x = cos (n + 6x).

3.15. 4sinxcos (~ -x)+4sin(n+x)cosx

+2 sin (3; - x) cos (n+x) = 1.3.16. sin x - sin 2x + sin 5x + sin Bx = O.3.17. 2 sin z - cos z = 2/5.

3.18. cos ( ~ +5x) + sin x=2 cos 3:1:.

3.19. (1+sinx)tan (~ -;)= co~x -cosx.

3.20. cos x - V3' sin x = cos 3x.3.21. sin 2z -1- 5 (sin z + cos z) + 1 == O.

3.22. sin3 2t + cos" 2t + ~ sin 4t = 1-

a.23. tan z tan 2z = tan z + tan 2z.sin 3 z + cos3 Z

3.24. 2 . = cos 2x.cos X--Sln x

3~25. cot24t + ~o2t 4t = O.

sin t SID t3.26. tarr' x = 36 cos" 2x.a.27. cot x - tan x - 2 tan 2x - 4 tan 4x + 8 = O.3.28. 4 sin" x cos 3x + 4 cos" x sin 3x = 3 sin 2x.

3.29. 2 cos zsin3 ( a; - z) - 5 sin 2 z cos2 z

+sin z cos! ( 3; + z) = cos 2z.

Problems 111

3.30. siu 2x Sill fix cos 4x + ~ cos 12x = O.

3.31. COS5~ X + Sill40 X = 1.3.32. logcosx sin x === 1.3.33. cot (sin x) == 1.3.34. tan 5.1: - 2 tan 3x = tan 5x tan" 3x.

" cos 2x3.35. cot x - 1= tan x +1 •

3 36 2 3 + 1 2 - 3 4., • cos x ,cos X - cos x COS x.

3.37. Find all the solutions of the equation

1 + cos x + cos 2x + sin x + sin 2x + sin 3x,

which satisfy the condition ~ < 13x- ~ j::::;:rr.3.38. Find all the solutions of the equation

2 - V3 cos 2x + sin 2:c = 4 cos2 3x,

which l'111tisfy the inequality cos (2x- ~ ) >0.

:l.39. Solve the equation

"I f 1-4COS2 4x2n

. = cos (2x- ~ ) .V 8 cos (2X - 3"") )


2+cos ; x+V3sin ; x=4sin2 ~ l

sat.isfying the condition sin ( ~ + ~ )> 0.

3.41. Show that the equation

1cot 2x +cot 3x+. . 2 . 3 == 0

SiD X sin x Rill x

has no roots.3.42. Find all the solutions of the equation

4 (3 V4x-x2sin2 ( xt y) -+- 2 cos (x+ y})

== 13+4 cos2 (x + y)

in two unknowns x and y.

3.45.

3.48.

112 .1. Trigonometric Equations and Systems


-V x2 - 4 (3 sin" x + 10 sin x cos x + 11 CO~2 x - 2 V301)

== 5rt2 - 4 arcsin- y-4 arccos- y.

In Problems 3.44 to 3.50 1 solve the given systems ofequations.

{

COS x Vcos 2x == 0,3.44. (n )

2sin2x--cos 2Y-T -==0.

{

X + y::=: rt/6,5 (sin 2x -t- sin 2y) :::--= 2 (1 -t- cos- (x - y)).

{

X - y :-=: 5n/3,3.46. sin x = 2 sin y.

{sin x cos y =-= 1/4,

3.47. 3/4Rin yeaR X:=:· •

{X-Y== -1/3,COR2 nx - sin2 nz == 1/2.

{

3.51. Find all the values of a for which the system ofequations

16 ycos ll: -51-11-6~/cos 11: I

+ 112 Vcos 7 + 1 1= 5 -,-sin2 n: (yt-; 2x) •

V . 810-9 (x2+ (y - a)2==: 3 x2+ (y-a)2_-

l 9

has a solution.

Chapter 4

Investigating Trigonometric Functions

4.1. Graphs of Basle Trigonometric Functions

To construct graphs of functions, one must know how touse all basic properties of these functions, such as periodicity, evenness or oddness, increase or decrease of afunction on an interval, as well as the arrangement ofpoints of extremum. However, constructing the graph ofa function cannot be a substitute for a rigorous proof ofthe properties of the function. Nevertheless, a graphillustrates vividly the properties of a function. At anexamination the student is usually asked to constructgraphs of functions composed of elementary functions,and most often such a graph is constructed by translatingor changing the graphs of elementary functions.

This section deals with the construction of graphs ofbasic trigonometric functions using the properties ofthese functions which were considered in Chapter 1. Usingthe formulas introduced in Chapter 2, it is possible toconstruct similar graphs for many other trigonometricfunctions.

First of all, let us recall that the graph of a function fwith domain D (/) is defined as a set of points on thecoordinate plane with coordinates (x, y) such that y ==f (x). This definition should be always referred to whenproving the properties of graphs of functions and whenconsidering operations with graphs.

1. Properties and Graph of the Function! (x) == sin x.(1) The domain of definition D (/) == R, the range of

values E (/) == [-1, 1].(2) sin x is a periodic function. Any number of the

form 2nk, k EZ, is a period of this function, 2n bei.ngits fundamental period (see Sec. 1.3, Item 1). Consequently, when constructing the graph we can first confine ourselves to the construction of it for the closed interval

8-01644

114 4. lnuestigating Trigonom.etrtc Functions

[-Tt, n] of length 2n, and then translate this section by2nk, k EZ, along the z-axis. This is because all pointsof the form (x -too '2nk, Sill x) == (x -t- Znk, sin (x -l- 2nk))have the same values as the value of the point (z', sin x)on the graph.

(3) sin x is au odd Iunction, therefore its graph is symmetric with respect to the origin. Indeed, for any point(x, sin x) of the graph the point (-x, -sin x) = (-x,sin (-x)), which is obtained by the application of central symmetry to the point (x, sin x), also lies on thisgraph (see Sec. 1.3, Item 2). Consequently, to constructthe graph of the function on [-1[, n], it suffices to construct it on [0, n], and then to map it central-symmetrically with respect to the origin.

(4) On the interval [0, n] the graph has two points(0, 0) and (rt, 0) in common with the x-axis. In general,the equality sin x == ° is equivalent to that x = nk,k EZ.

(5) The Iuncl.iou sin x increases on the interval [0,n/2] and decreases on [n/2, rt], this means that if 0 ~Xl < X 2 ~ rt/2, then sin Xl < sin x 2 , and if n/2 ~Xl < X 2 ~ J"t, then sin Xl > sin x 2 (see Sec. 1.3, Item 3).Hence it follows that (nI2, 1) is a point of maximum ofthe function sin x.

The graph of the function sin x is constructed now inseveral steps. Firstly, we construct the graph on the interval [0, n]. This can be done by compiling a table ofvalues of the function sin x for some points of the interval [0, rt}, for instance,

n 11 Jt n 2rt 3rc 5n(x) l}

6 T 3""" "2 -3- -4- -6- n

1 V2 va V3 V2 tsin x () 2 -2- -2- 1 -2- -2- "2 0

and, on plotting on the coordinate plane points of theform (x, sin x), where x are numbers from the table, wejoin these points with a smooth line. By mapping central-symmetrically the constructed section of the graphwith respect to the point 0 and then applying a series

4.1. Graphs 115

of translations along the z-axis by Znk, k EZ, we get thegraph of the function sin x which is called the sinusoidor sine curve (Fig. 24).

Another method of construction of the graph whichdoes not require the computation of individual values ofthe function sin x consists in using the trigonometriccircle. For this purpose, we mark the interval [0, n/2)

Fig. 24

on the z-axis and on the same drawing construct a circleof unit radius centred on the extension of the x-axis. Toconstruct the graph of the function sin x, we shall "unwind" the circle on the number line marking the ordinates of points P t corresponding to real numbers t. Toconstruct sufficiently many points belonging to the graph,we may bisect the interval [0, n/21 and the subintervalsthus obtained, as well as the corresponding arcs on thetrigonometric circle. Note that after marking the pointx = nl2 on the x-axis, all further constructions are carried out with the aid of a pair of compasses and a ruler(Fig. 25).

Note that the function sin x increases from -1 to 1 on

any interval of the form [ - ~ + 2Jtk, ~ + 2JtleJ. Ie EZ,and decreases from 1 to -ion any interval of the form

[; + 2JtIe, 323t + 2JtleJ, Ie EZ. The maximal value

sin x = 1 is attained at points x = -i + 2nk, k EZ,

while the minimal value sin x = -1 at points x ==- ~ + 2nk, k EZ.

8*

116 4. lnuestigating Trigonometric Functions

2. Properties and Graph of the Function f (x) == cos a:The graph of cos x is best of all constructed by usi ug the

identity sin (x + ~) = cos x (the reduction formu

la (2.28). It follows from this identity that the graph ofthe function cos x is obtained from the graph of the function sin x by translating the latter leltwards along the

x

-1

Fig. 25

x-axis by :rrJ2. Indeed, for every point (x, sin x) of the

graph of sin x the point (x - ~ , sin x) lies on the graph

f . (n). ( 11.+1t) .o cos x, SInce COS x - 2" == SIn x - 2 2" == SIn x.

The converse is also true: for any point (x, cos x) of the

graph of cos x, the point (x + ~, cos x) lies on the

graph of sin x since sin (x + ~) = cos x,Practically, it is more convenient first to construct the

sine curve y == sin x, and then to translate the y-axis tothe right by rr/2 (see Fig. 26, where the old y-axis is drawnin a dashed line and the new axis in a continuous line).

Consider the properties of the function cos x = j (x).(1) D (/) === R, E (f) == (-1, 1].(2) cos x is a periodic function. Any number of the

form 2nk is a period of the function (k EZ), 2n being itsfundamental period (see Sec. 1.3, Item 1).

(3) cos x is an even function, and its graph is symmetricabout the axis of ordinates: if a point (x, cos x) lies onthe graph of cos x, then the point (-x, cos x) = (-.'C,cos (-x)) also lies on this graph.

4.1. Graphs 117

(4) cosx=O for x= ; +nk, kEZ.

(5) cos x decreases from 1 to -ion any interval of theform [2:rtk, rt -f- 2nkl, k EZ, and increases from -1 to 1on any interval of the form [-j[ + 2nk, 2nkl. For x =n -~ 2nk, k EZ, the function cos x takes on the minimal

Fig. 26

value -1, while for x == 2nk, k E Z, the maximalvalue 1.

Example 4.1.1. Find the least and greatest values ofthe function f (x) = sin (cos (sin x)) on the closed inter-val [n/2, n], . .~Let n/2 ~ Xl < X 2 ~ rt , then 0 ~ sin X 2 < sin X t ~1, and the points P sin Xl' P sin XI lie- in the first quadran tsince 1 < n/2. Since the function cos x decreases on theinterval [0, n/2], we have

o< cos (sin Xt) < cos (sin x 2) ~ 1.

But the points Pcoslsln x.) and Pcostsm XI) also lie inthe first quadrant and the function sin x increases onthe interval [0, n/2l, therefore

o< sin (cos (sin Xt)) < sin (cos (sin x 2 ) ) < 1,

that is, the function f (x) = sin (cos (sin x)) is increasingon the interval [n/2, nl, consequently, the minimal valueof t (.T) on this interval is equal to f (n/2) === sin (cos 1),whi Ie the maximal value to f (rr) == sin (cos 0) === sin 1. ~

Example 4.1.2. Compare the numbers sin (cos 1) andcos (sin 1).

118 4. Investigating Trigonometric Functions

~According to the reduction formula (2.7), cos (sin i):=.;

sin ( ~ - sin 1) , and it suffices to compare the num-

bers cos 1 and ~ - sin 1 from the interval [0, n/2]

(by virtue of the monotonicity of the function sin x 011

this interval). Note that

cos 1+ sin 1 = V2 sin ( 1 + ~) ,consequently, the inequality

cos 1 + sin 1 ~ ·V2< n/2

is true, whence, since the function sin x increases on theinterval [0, n/2], we have

sin (cos 1)< sin ( ~ - sin 1 ) = cos (sin 1). ..

3. Properties and Graph of the Function f (x) = tan e.(1) The domian of definition is the set of real numbers

except for the numbers of the form ~ + nk, k EZ, the

range of values E (I) = R.(2) tan x is a periodic function. Any number of the

form nk, k EZ, may be a period of tan x, 11: being itsfundamental period (see Sec. 1.3, Item 1).

(3) tan x is an odd function, and, consequently, itsgraph is symmetric with respect to the origin (see Sec.1.3,Item 2).

(4) tan x = 0 for x = nk, k EZ.

(5) On any interval of the form ( - ~ + nk, ~ -+- rtk) ,

k EZ, the function tan x increases from -00 to +00.The graph y = tan x (tangent curve) can also be con

structed using a table of values, for instance,

1t 1t 1tX 0 6 T 3

tan x "ltra

1 V3"3

4.1. Graphs

r

Fig. 27

119

which was compiled only for the values x E [0, n/2) byvirtue of Properties (2) and (3) (Fig. 27).

Another method of constructing the graph (with the aidof the trigonometric circle and the line of tangents)makes it possible to plot arbitrarily many points for thetangent curve without computing individual values ofthe function tan x (Fig. 28).

Example 4.1.3. Find the greatest and least values ofthe function tan (cos x) on the interval [n/2, 11].

~Let n/2 ~ Xl < x 2 ~ n, then -~ < -1 ~cos X 2 < cos Xl ~ 0, since the function cos x decreasesfrom 0 to -1 on the interval [n/2, n]. Therefore thepoints P r oR X

1and P('os x, lie in the fourth quadrant, and

-tan 1 ~ tan (cos x 2) < tan (cos Xl)~ 0,

since tan x increases in the fourth quadrant. Consequently,the function f (x) = tan (cos x) decreases OIl the interval


!IIIII

----j------

IIII /----,---- --

P.E I~---T----o

Fig. 28

1/1 a:12IIIIIIIIII

[,,;/2, a], its greatest value is t (n/2) = tan (0) = 0, theleast value being f (n) = tan (-1) == -tan 1. ~

4. The Graph of Harmonic Oscillations. Harmonicoscillations are defined as rectilinear motions of a pointgoverned by the rule s = A sin (rot + ex), where A > 0,co > 0, and t denotes the time coordinate. Consider themethod of constructing graphs of such oscillations.

Example 4.1.4. Construct the graph of the functionf (x) = sin 4x.~We first represent the graph of the function sin x. Ifthe point M (a, b) lies on this graph, that is, b == sin a,then the point N (a/4, b) lies on the graph of the function

f (x) = sin 4x since b = sin (4'1-). Thus, if we takea point on the graph of sin x, then a point having thesame ordinate and whose abscissa equals one-fourth of theabscissa of that point will lie on the graph of the function

.ain 4x. Consequently, the graph of sin 4x is obtained from

4.1. Graphs 121

the graph of sin x by contracting the latter along the axisof abscissas four-fold (Fig. 29).

To construct the graph of the function y = sin 4.£, .itsuffices to do the following. From the equation sin 4x === 0we find the points of intersection of the graph with the

Fig. 29

x-axis: x = nk/4, k EZ. Further, sin 4x = 1 for x ==1t nk . 1t nk8 + 2' k E Z, and SIn 4x = -1 for x = -8 + 2'k EZ. In addition, the fundamental period of the function f (x) = sin 4x is equal to :rt/2:

t (x + ~) = sin (4 (x + ~))= sin (4x ± 2n) = t (x)

(see Sec. 1.3, Item 1). Now we can construct the graph:mark the points (0, 0), (n/4, 0), (n/2, 0) and also (n/S, 1),(3n/8, -1), join them with a curve resembling the sinecurve (on the interval [0, nI2]), and then apply transla-.tions along the axis of abscissas, the origin going into thepoints of the form (nk/2, 0). ~

In the same way, we can construct the graph of anyfunction of the form sin WX, where w > O. To this effect,we have to draw the graph of the function sin x andthen to contract it co times along the axis of abscissas,that is, to replace each point Al (a, b) of this graph bythe point N tau», b). The following is of importance: ifo< (j) < 1, then, instead of contracting, the graph isstretched 1/co times: the points will move away from theaxis of ordinates.

It is not difficult now to construct the graph of anyfunction of the form A sin (wx + ex), A > O. Using the


equality A sin (wx -+ a) = A sin (w(x -+:)), we are

able to construct a graph in several steps.(1) Construct the graph of the function sin rox using

the method described above.

(2) The graph of the function sin (w (x -+ ::) )is ob

tained from the graph of the function sin wx by translating the latter along the axis of abscissas by -aJroo Indeed, for every point (x, sin cux) of the graph of sin (Ox

the point (x - :' sin wx) lies on the graph of the

function sin ( CJ)( x -+ : )), since sin ( CJ)( x -: -+ ~ ))=sin oox. Usually, the graph of the function sin (Ox isconstructed in a simpler way, the line x = a/ro beingtaken for the new axis of ordinates.

(3) To obtain the graph of the function sin ( CJ)( x -+: ))from the graph of the function A sin ( CJ){ x -+ : ) ) it re

mains to multiply the ordinate of each point of thegraph by A, that is, to stretch this graph A times alongthe axis of ordinates (if 0 < A < 1, then, instead ofbeing stretched, the graph is contracted 1/A times).

In practice, we may proceed as follows: to find thepoints of intersection of the graph of the functionA sin (wx + a) with the axis of abscissas by solving the'equation sin (wx + a) = 0, and, from the equationssin (wx + a) = 1 and sin (rox + a) = -1, to find thepoints of extremum of the function A sin (roz + a) (at thepoints of maximum the function takes on the value A,at the points of minimum the value -A); then join thefound points to obtain a curve of sinusoid type. Notethat the function y = A sin (wx + a) is periodic with

the fundamental period 2nlCJ) since A sin ( CJ)( x -+ ~1t) -+ex) = A sin (wx -f- a + 2n) == A sin (wx + ex).

Example 4.1.5. Construct the graph of the function

4 sin (3x - ~).~ The fundamental period of the function is 2n/3. Findthe points of intersection of this graph with the axis of

4.1. Graphs 123

abscissas: sin (3x - ~) = 0, that is, 3x - ; = nk,

k EZ, or x = ~ + v k EZ. Further, solving the

equation sin (3x - ~) = 1, we get 3x - ~ = ~ + 2nk,

g

g:45in(J.x-~ )

Fig. 30

that is, x = -i + 2;k, k EZ. At these points r: func

tion takes on the maximal value equal to 4;/equation sin (3x - ~) = -1 we find 3x - //

. 21tk / i2nk, that IS, x = 3' k EZ. At these pr !

/'


tion 4 sin ( 3x - ~) takes OIl the minimal value equal

to -4. Mark the found points on the coordinate planeand join them with a smooth line to get the requiredgraph (Fig. 30). ~

Example 4.1.6. Graph the function sin 2x + 'V'3 cos 2x.~ The most convenient technique here is to reduce this

Fig. 31

n nkx=-= -6+ 2' kEZ.

of the function:

x:== ;2 + nk ; k EZ,

( ~ +nk, 2), kEZ,

function to the function A sin (roz + a) using the method of introducing an auxiliary angle (see Theorem 2.2):

sin 2x+ V3 cos2x = 2 ( ~ sin 2x+ ~3 cos 2x)

= 2 sin (2x + ~ ) .The fundamental period of this function is 1t.

(1) The points of intersection of the graph with theaxis of abscissas:

sin (2x+ ~) =0, that is,

(2) The points of maximum

sin (2x+ ~ ) = 1, that is,

therefore the points of the form

Iie OIl the graph of the f unction.

4.1. Graphs f25

(3) The points of minimum of the function:

sin (2x+ ~) = -1, that is, X= - ~~ +nk, kEZ,

therefore the points ( - ~~ + sik, - 2 ), k EZ, lie on

the graph.The graph of the function y = sin 2x + )/3 cos 2x is

shown in Fig. 31. ~

of

Fig. 32

Example 4.1.7. Graph the function

3 /,l(x)=Tsin2x+2cos2x. ,/

~Using Theorem4

2.2, we get /1(xl

where cp == arcsin ""5' 0 < qJ < «n.qualities are valid: )/2/2 < 4/5 «:the function sin x increases 9Y

We get n/4 < arcsin ~ < n/'f

Note that the fundament/'equals rt , /


(1) The points of intersection of the graph with theaxis of abscissas:

sin (2x + cp) ,= 0, that is, x = - i + :It;, k EZ.

(2) The points of maximum of the function:

. (2 + )- '1 h t · - rr <P I 1,. k ZSin x rp -, t a IS, x _.- T - "2 -, :J1h" L E ·(3) The points of minimum of the function:

sin (2x - fP) == -'1, that is, x ~.7: - ~ - i + nk, leE z.With the aid of these points we construct the required

graph as in the preceding example (Fig. 32). ~

~.2. Computing Limits

The theory of limits underlies the important notions ofthe continuity and differentiability of a function andthe finding of derivatives and integrals. We confineourselves to solving problems on finding limits at certainpoints of functions represented- by trigonometric expressions. To solve these problems, one should know well thedefinition of the limit of a function at a point a E R, basic properties of limits (the limit of a sum, product, andratio, as well as Theorem 4.2 on the first remarkablelimit.

Definition. Let a function / (x) be defined on the setD (f) c= R, and let a point a be such that any of itsneighbourhoods contains infinitely many points of D (/)(an accumulation or limit point of the set D (/). Thenthe number b is said to be the limit of the function j(x) atthe point a if for any positive number 8 there is a positive number 6, dependent on 8, such that for any pointxED (/) satisfying the condition 0 < I x - a I < {,there holds the inequality J f (x) - b 1< E. Written:

b ~~ 1im f (x).X~(l

Definition. A function f (x) is said to be continuous at apoint a ED (f) if lim f (x) = f (a).

x~a

A function is continuous on a set X c:: D (/) if it is con-tinuous at each point of this set. The sum, difference, and

4.2. Computing Limits 127

product of two functions continuous on one and the sameset. are also continuous on this set. If the denominator ofa Iractiou does not vanish on a set, then the quotient oftwo functions continuous on this set is also continuous.

The following statement plays' a key role in testingtrigonometric functions for continuity and computingvarious limi ts:

For all real numbers x, satisfying the condition 0 < I x 1<n/2, there hold the inequalities

I sin x 1'< I x I < I tan x I·The proof will be given later on (see Example 5.1.1),

and now we are going to deduce the continuity of basic.trigonometric functions.

Theorem 4.1. The functions sin x, cos x, tan x, cot xare continuous in the domains of their definition.

Proof. Let us prove, for instance, that eosine is continuous throughout the number line, that is, show thatfor any X o E R ...

lim cos x = cos xo.X~XO

Indeed,

[cos x- cos xol = /2 sin xo;-x sin xotx I

= 21 sin xo;-x II sin xotxI~21 sin xo;-x I~21 xo;-x 1= Ix- xul·

Therefore, setting 6 == e, for any e > 0 for 0 < I x Xo I < 6 we have

I cos x - cos Xo I~ I x - Xo I < B.

The continuity of sine is proved in similar fashionand that of tangent and cotangent. follows from the property of continuity of the qnotient of two continuousfunctions. ~


Example 4.2. t. Find out whether the function

{

X sin~ if x:#= 0, x ER.f (x) == 0 x

if x ~ 0,

is continuous on R.~ For x =1= 0 the gi ven function has the form of the product 11 (x) •f2 (x) of two functions /1 (x) == x and /2 (x) :=:

sin .!. The function 11 (x) is continuous and the functionx

f 2 (x) == sin-! can be represented in the form of the com-x

positionf 2 (x) == / 4 (/a (x))

of the Iunctionsj, (x) = ~, x =t= 0, and 14 (y) = sin y. ByTheorem 4.1 the function 14 (y) is continuous on Randthe function f 3 (x) is continuous for x =f:= 0 by virtue of theremark on the continuity of a fraction. Consequently, thefunction /2 (x) is continuous on "e set {x E R: x =t= O}; itremains only to check whether it is continuous for x == 0,that is, whether the equality

lim x sin J.. ~~ 0 -_:;: f (0)x-+O x

is fulfilled.For e> Oweset «5 == e, then for 0 < Ix 1< 6 we have

I:xsin ~ --OI=lxllsin ~ 1~IXI<c=l).•Recall that the i)-neighbourhood of a point a E R is

defined as an interval of the form (a - 6, a + 6), where6 >0.

Example 4.2.2. Find out whether the function

{sin ..!.. if x=¥=O, xER,

/ (x) ==-~ 0 xif x =::: 0,

is continuous on R.Note that on the set {x E R: x =t= O} the given function

coincides with the function /2 (x) from Example 4.2.1and therefore it is continuous on this set. Let us showthat f (x) is not continuous (discontinuous) at the point

4.2. Computing Limits

Xo = 0, that is, that the equality

I · · 1 °lIn SIn - ===X~O X

129

is violated. To this end, let us take E == 1/2 and considerthe sequence of points

2xn == n (1+4n) , n :::= 1, 2, 3, ... ,

which satisfy the condition f (xn ) = 1. Then I I (xn ) t (0) I = 1 > 1/2 == e, and any <5-neighbourhood of thepoint X o == °contains infinitely many such points X n ,

that is, the condition of the continuity of the function atthe point X o == 0 is violated. Thus, the function I (x) isdiscontinuous at the point X o == O. (Similarly, we canshow that no real number b is the limit of the function

sin.!. at the point X o == 0.) ~x

The computation of many limits is based on the fol-lowing theorem.

Theorem 4.2 (on the first remarkable limit):

I" sin x 1lm--=== .X40 X

Prool. Let 0 < [z] < 1'£/2. Then the inequality'sin xl < x < [tan z] holds (see Equation (4.1). Dividing both sides of this inequality by [sin xl * 0, we

get 1 < 1-.x_I < I 1 i · Taking into account thatsin z cos x

~>O, cosx>O for 0< Ixl < 1t2

, we have 1<SIn x

_._x_<_1_ and, consequently, 1> sin x >cosx.SInx cos X tt

Hence we getsin x 1 00< 1--- < -cosx=cos -cosx.

x

Since cosine is a continuous function, for any e > 0, thereis a <5 > 0 such that .

o< 10 - x I < ~ ==>- I cos°- cos x 1<8.9-01644

(b)


Therefore for the numbers x =t= 0, satisfying the inequality 0 < I x I <~, we have:

Isin x I sin x1--

x- = 1- -x-<cosO-cosx

= [cos O-cos z] < B.

This means that lim sin x == 1_ ~x-..o x

E I , 2 3 C t (a) 1° 2x (b) 110 m sin 5x ,xamp ell. •• ompu e a lID -.-3- ,x-..o Sill x x-+O x

( ) 1° tan 2x (d) 1° tan 2xc 1m , 1m ° 3 .x~O x x~o sin x

( ) I - 2x I· 2 1-lllIIIIII a Im-.-== Im-··-.--

x-e-O SIll 3x x-+O 3 sin 3x3x

2 1 2 2== 3""" • 10 sin 3x ==3 -1 = 3 '

1m --3x~O 3x

I - sin 5x 1° 5 sin 5x 5 I" sin 5x 51ffi---== lID .---= lID---=,x-+O x x-+O 5x 5x-+O 5x

(c) 1° tan 2x I" 2 sin 2x 11m --- === 1m .---.---

x-+O X x-+o 2x cos 2x

2 1° sin 2x 1° 1 2 1 1 2== 1m ---. 1m ---= · · = ,2x-+ 0 2x 2x-)oo 0 COS 2x

(d) 1° tan 2x 1° 2 tan 2x 3x1m ---== Iffi-. ---e---

x-s-O sin 3x x-+O 3 2x sin 3x

_ lim ~. sin 2x • __1_ • __1_- x-+O 3 2x cos 2x sin 3x

3x

2 1° sin 2x I- 1 i==-. lID ---. 1m --.-----3 2x-+ 0 2x 2x-+ 0 cos 2x sin 3x

lim 3X3x--+O

212==3·1-1·T=3·~

. i-cos xExample 4.2.4. Compute 11m 2 °

x-+O x~Note that, by virtue of formula (2.41),

1-cos x = 2 sin2 ~ •

4.2. Computing Limits 131

Consequently, by the property of the li.mit of a product,;1: X

1- cos x . 1 sin"2 1 sin "2lim ==llm2.---·-·--x-+O x

2x~o 2 ~ 2 -=-

2 2

xsiny

I . 2 sin 2 x+sin x-1Example 4.2.5. Compute rm 2 . 2 3' -1 t •

x-+n/G SIn x - SIn x -

~Note that for x = n/6 both the numerator and denominator of the given fraction vanish, therefore, in thiscase, it is impossible to use directly the theorem on thelimit of a ratio. In similar problems we may proceed asfollows: we single out in the numerator and denominatorof the fraction a common factor which vanishes at thegiven limit point, but does not vanish in the neighbourhood of this point, and such that after cancelling thisfactor the denominator of the given fraction no longervanishes at the point under consideration. Then we canuse the property of the limit of a ratio.

In our case:

2 sin" x -t-- sin x - 1 = (2 sin x-i) (sin x + 1),

2 sin" x - 3 sin x + 1 = (2 sin x - 1) (sin x - 1),

and, consequently,

lim (2 sin x-i) (sin x+1)x-+1t/6 (2 sin x -1) (sin x-i)

lim (sin x+ 1)~ lim s~n x+1 x--::--+_1t_/6 _

x-+1t/6 SID X -1 -- lim (sin x -1)x-+1t/6

1tsinT+1----=-3sin~-t

()

(in addition to the theorems on the limits of a quotientand a sum, we have used the fact that the function sin xis continuous). ~

9*

lim f (h)- f (0)h~O It


4.3. Investigating Trigonometric Functionswith the Aid of a Derivative

1. The basic properties of many functions can be successfully studied without the aid of the derivative, andthe properties of the derivative are a good illustration ofthe properties of the function itself. However, in manyproblems points of extremum, intervals of increase ordecrease cannot be determined by elementary means, suchproblems must be solved using derivatives. Besides,when graphing some functions, one must know moreabout the behaviour of the function, for instance, whetherits graph touches the axis of abscissas at a certain pointof intersection with this axis or intersects it and whatangle is formed. Such questions can be answered only byconsidering the derivative.

Let us first consider the rules for finding the deriva ti vesof basic trigonometric functions.. Definition. The derivative 0/ a junction / (x) at a pointXo is defined as the number

t' ( ) -I· f (xo+h) - t (xo)Xo - 1m h •

h-+O

A function having a derivative at a certain point isdifferentiable at this point. Let D 1 be a set of points atwhich the function f is differentiable. Associating eachnumber x ED 1 with the number t' (x), we get a functiondefined on the set D I • This function is called the derivative of the function t and is symbolized as /' (x).

Example 4.3.1. Show that the function

{

X sin ~ if x * 0, x ER,f (x) = x

o if x == 0,

is not differentiable at the point x = o.Recall that in Example 4.2.1 we proved the continuity

of the given function throughout the number line.~Let us prove that this function is not difierentiableat the point x = 0, that is, that no real number b can beequal to

4.3. Derivatives of Trigonometric Functions 133

Consider the numbers X n = n (1~4n)' n = 1, 2, · · ·(such that t (x n ) = x n ) and Zn = 1/nn, n = 1, 2, · · ·(such that condition f (zn) = 0). If b = 0, then for 8 =1/2 we set h = X n • Then

I !(Xn>-f(O)_OI==1>..!.-=e,X n 2

and the inequality

I f (h) -;; f (0) - bI< e

does not hold for all the numbers h = X n (any B-neighhourhood of the point h = 0 contains infinitely manysuch points). Analogously, if b =f= 0, then we take 8 =I b 1/2 and set h = Zn' Then

If (zn);:: f (0) - b1= Ibl > e

for infinitely many numbers of the form h = Zn lying inany 6-neighbourhood of the point h = O. ~

Recall that if a function f has a derivative at a pointX o, then a tangent line to the graph of f is defined, itsslope being equal to j'(xo)' The equation of the tangentline is:

Y == f (xo) + f'(x o) (x - xo). (4.2)

Example 4.3.2. Prove that the function

{Ix13/2 sin _1 if x*O, xER,t (x) ~ xo if x == 0,

is differentiable at the point x = 0, and I' (0) = O.~For a given positive number e we set 6 = e2 • Then

I f (h);; f (0) I~ Ihp/2 < e,

if 0 < I h I < 6. Consequently,

o= lim f (h) - f (0) -,- t' (0).h-+O b

131: 4. Investigating Trigonometric Functions

Note that in this case the tangent line to the graph ofj (x) at x == 0 coincides with the axis of abscissas. ~

Theorem 4.3. On the domains oj definition oj the functions sin x, cos x, tan x, cot x the following equalitieshold true:

(sin x)' ==.: cos X, (cos x)' = - sin x,

(tan x)' == 1/cos2 x (cot x)' = -1/sin2 x.

Proof. For the function sin x this theorem is deducedfrom the first remarkable limit (Theorem 4.2). Indeed, ifj (x) === sin x, then, by virtue of identi ty (2.18) for thesine of two real numbers, we have:

f (.r+h)- f (x) _ sin (x+h}-sin x11, - h

= (2cos (x+ ; )Sin+)lh.Now, using the properties of limits, we get

. h

lim .f(X+~)-f(:r) = lim cos (x+~) Sill 2h-+O ~ h-+O 2 h

2'". h

. J sloT= Jim cos (x+ +) lim -1- =eos x.

h./2-+-0 h/2-+0 _1,

2

Here, we have also taken advantage of the continuity ofthe function cos x.

For the function cos x the theorem can be proved usingthe reduction formulas (2.7) and the Iol lowing rule fordifferentiation of a composite function:

If a function g has a derivative at a point Yo == f (xo),and a junction j at a point x o, then the composite junctionh (x) == g (f (x) also has a derivative at x o, and


Let 118 now represent the function cos x in the form

sin ( ; - x ) = g (f (x)) and apply the rule (4.3) for

g (y) == sin y, / (x) == ~ - x:

(Sin ( ~ - x ) )' = g' (f (x)) f' (z) = - cos ( ; .- x )

= -sinx.

In order to prove the theorem for tan x and cot x, it suffices to use the rule for differentiating the quotient of twofunctions. Let us also recall some other differentiationrules. If the functions f (x) and g (x) are differentiable,then:

(I (x) + g (x))' = I' (x) + g' (x), (4.4)

(CI (x»' = C]' (x), C is a constant, (4.5)

(I (x) g (x»' = I' (x) g (x) + t (x) g' (x), (4.6)

(f(x)/g(x»),= .f'(x)g(x)-f(x)g'(x) (4.7)g2 (x)

for points x such that g (x) =t= O.Using the rule (4.7) and the aforeproved, we get

, ( sin x ) I sin' x cos x-cos' x sin xtan x = -- == --------cos X cos2 x

cos2 x + sin2x 1- cos2 X = cos2 X '

t ' ( cosx ) I cos' x sin x-sin' x cos xco X = sin x = sin2 x

-sin2 x-cos2 xsin 2 x

Example 4.3.3. Find the derivative of the functionsinx-xcosx

y === cos x + x sin x •

~ Using the rules (4.7), (4.4), (4.6), and the formulas ofTheorem 4.3, we get

(sinx-xcosx )' _ u'v-uv'cos x + x sin x - v2 ,

136 4. 1noesugattng Trigonometric Functions

whereu = sin x - x cos I, V = cos x + x sin x,, ., ( )'U = SIn x - x cos x

= cos x - (cos x - x sin x) = x sin x,, '+ ( . )'v = cos x x SIn x

= -sin x + (sin x + x cos x) = x cos x,

u'v - uv' = x sin x (cos x + x sin x)

- (sin x - x cos x) x cos x

= x2 sin" x + x2 COS2 X = x2,

whence, x2

y ;== (cos x +x sin X)2 ,

the derivative exists everywhere in the domain of thegiven function, that is, for x's such thaf cos x + x sin x =I=-o. ~

Example 4.3.4. Compute the derivative of the indicatedfunctions:

(a) y = sin (sin (sin x)),(b) y = sin (cos" (tan" (x4 + 1))).

~ (a) By the rule (4.3) for differentiating a compositefunction applied twice, we get

y' = cos (sin (sin x)) -cos (sin x). cos x.

(b) Let us write the function in the form of a compositefunction:

Y = f6 (15 (14 (f 3 (12 (f1 (x)))))),

where 11 (x) == x4 + 1, /2 (x) == tan x, fa (x) = x3,

f 4 (x) = COS x, f 6 (x) == x2, 16 (x) = sin x, and we get

t; (x) = 4x3, i; (x) = L/cos'' x, t; (x) = 3x2

, f~ (x) =-sin x, f~ (x) = 2x, f~ (x) = cos x. Consequently, byapplying the rule (4.3) several times, we get

y' (x) = cos (cos- (tan" (x4 + 1))). 2 cos (tan" (x4 + 1))

X (-sin (tan3(x4+1)))-3tan2(x4 + 1). 1 .4x3cos- (x 4 + 1)

=== -12x3 COS (cos- (tan" (x4 + 1)))

. (2t 3( 4+1)) sin2(x 4 + 1) ........

X SIn an x cos- (x 4 +1) • ......


Example 4.3.5. Prove the following formulas for derivatives of inverse trigonometric functions:

(a) (arcsin x)' =V 1 ,1-x2

(b) (arccos~)' = - V 1 ,a a2-x2

(c) (arctan : )' = a2~x2'~ Let us prove, for instance, formula (a). By the definition of arc sine, for x E[-1, 1] there holds the identity

sin (arcsin x) = x,

Computing the derivatives of both sides, we get forIx 1<1:

cos (arcsin x) -(arcsin x)' = 1.

But cos (arcsin x) == V1- x2 , since if ex.:=; arcsin x, thensin a == x, -1f/2~a~'Jt/2, therefore cos a is nonnega-tive and cosa=V1-x2 • Consequently,

(arcsin x)' ==== 1/V1-x2 •

Equalities (b) and (c) are proved in a similar way byapplying Theorem 4.3, the rule (4.3), and identities(1.9), (1.10). ~

2. Applying the Derivative to Investigation of Trigonometric Functions. Let us consider some essentialthings needed for solving problems with the aid of thederivative.

Sufficient condition of monotonicity. I f a functionf (x) is differentiable on the interval (a, b) and f' (x) > 0tl' (x) < 0) on (a, b), then f (x) increases (decreases) on thisinterval.

Remember that the converse is not always true; forinstance, the function f (x) = x3 increases monotonicallyon the interval (-1, 1), and its derivative j'(x) = 2x2

is not positive everywhere, t' (0) = O.A point X o E D (/) is called the point of (local) maxi

mum (minimum) of the function j if for all x ED(f) fromsome neighbourhood of X o the inequality t (x) ~ f (xo)

(I (x) ~ f (xo» is fulfilled. The points of maximum and


minimum are also called the points of extremum of a function. One should remember that if a function f (x) isdefined on a certain set X, and X o is a point of maximumof I, X O EX, then the value t (xo) is not necessarily thegreatest value of the function f (x) on the set X. Consideran example. Figure 33 represents the graph of a functionhaving on the interval X === (-1, 3) three points of maximum Xl' x 2 , x a' however. none of them is the greatest

2

ff~ --------------"() a

Fig. 33

value of the function t on the set X. Moreover, the givenfunction does not attain the greatest value on X since allthe values of the function are less than the number 2, butno number a < 2 is the greatest value. Indeed, we canchoose a point X o such that a < f (xo) < 2 (see Fig. 33).The same note refers to the points of minimum and leastvalues.

Let us note (wi thout proof) that if a function f (x) is defined and continuous on an interval [a, b], then this function takes on the least and greatest values on this interval. This is one of the fundamental theorems in the courseof mathematical analysis. Also note that OIl an openinterval, half-open interval or throughout the numberline, the function may not take on the greatest or theleast value. This is exemplified by the function whosegraph is shown in Fig. 33.

A critical point of a function is a point from the domainof the function at which the derivative is zero or doesnot exist at all,


A necessary condition of an extremum, I f a functionf differentiable at a point X o has an extremum at thispoint, then I' (xo) === o.

A sufficient condition of an extremum. If, when passingthrough a critical point x o, the derivative of the junctionchanges sign [rom plus to minus, then Xo is a point of maximum; and if the derioatire changes sign from minus toplus, then Xo is a point of minimum of this function; if thederivative does not change sign, then X o is not a point ofextremum.

We often come across problems on finding the greatestor the least value of a continuous function on a closedinterval fa, b). When solving such problems, it is notsufficient to find the greatest local maximum of the function (the least minimum), we have also to compare thesenumbers with the values of the function attained at theend points x == a and x === b of the interval. The greatest(least) value is frequently attained at the end points.

In practice, to find the greatest or the least value of a(continuous) function on a given interval, one finds all thecritical points inside the interval and compares the va1-:ues of the function at the cri.tical points with each othe1and with the values at the end points of the interval,without further investigating the critical points.

Consider several examples.Example 4.3.6. Graph the function

f(x)= ~ sin2x+-2lcos2xl

and find its greatest and least values on the closed interval [0, nJ.~ In solving problems with functions involving a modulus sign, we have always to consider t\VO cases since

{a if a~O,

lal== -a if a<O.

The function cos 2x vanishes if 2x;= ~ + sik, k EZ~

that is, x = : +- ~k, k EZ, and if the number k is

even, then cos 2x changes sign from plus to minus, and ifk is odd, the sign changes from minus to plus. Conse-


quently, the inequality cos 2x ~ 0 is equivalent to that

and cos 2x < 0 is equivalent to that

n 3nT+nm<x< 4+nm, mEZ. (4.9)

Case 1: x satisfies the condition (4.8). Then cos 2x ~ 0,~ cos 2x I = cos 2x, and, by Theorem 2.2,

f(x)= ~ sin2x-I-2Icos2xl =oi-sin (2x+cp),

h ._- . . 4:t: n ( E' I 11 '1 r"!)were q>-arcslllT' T<CP<T see xamp C .L .1.

Case 2: x satisfies the condition (4.9). Then cos 2x < 0,I cos 2x I == -cos 2x and

f (x) = 4- sin 2x- 2 cos 2x ,,'- +sin (2x- rp},

"-here cp = arcsin +.Consequently, to construct the graph of a given func

tion, it is first necessary to graph the functions (seeSec. 4.1)

and

+sin (2x + cp)

52" sin (2x - rp)

1248

(4.10)

(4.11)

using dashed lines, and then for the values of x lying OIl

the intervals (4.8) to outline the graph of (4.10) usingcontinuous lines, and on the intervals (4.9) the graphof (4.11); as a result, we get the graph of the functionunder consideration (the continuous line in Fig. 34).

To find the critical points, note first that on the interval (0, rr) I' (x) = 5 cos (2x + cp) if 0 < x < rr./4 or311:/4 < x < rt, and I' (x) == 5 cos (2x - rp) if n/4 <x < 3n/4. At the points x = n/4 and x = 331/4 the functipn is not differentiable. This is proved exactly in the


same manner as in Example 4.3.1 by determining the derivative, and geometrically the absence of the derivativemeans that it is impossible to draw a tangent line to thegraph of the given function at these points.

Thus, f' (x) = 0 (on (0, :It)) at the points Xi =' ~ - ! 'X2 == : + ~ (both of them are points of maximumsince at these points the derivative changes sign from

\ x\\,

IIIII

I, I',.I

y=~ Sin2X+:':\j\

I \I \I \I \

III

JI

III

I~ig. 34

plus to minus). The derivative is not existent at thepoints Xs == rrJ4 and X 4 === 331/4 (both of them are points ofminimum since the derivative changes sign from minus toplus when passing through these points). Evaluating thefunction at the critical points

j(xt ) = f ( : - r) = -} , f (x 2) = f ( : -+- i ) = -} ,

f (x3) = t ( : ) = 4-, f (x4) = t (3: )= - -} ,

and also the values f (0) === f (n) ~ 2 at the ond points of theinterval [0, nl, we see that the greatest value of the func-

142 4. Investigating Trigonometric Function,

!/y=x-cos2x

,

//

// /1= f,(x)

/'/

Fig. 35

tion is equal to 5/2, and the least value to -3/2. Notethat, by virtue of the periodicity of the function (itsperiod being equal to a), these values are the greatest andleast values on the entire number; line.~

Example 4.3.7. Graph the function

y = x - cos 2x

and find its points of extremum.~To construct the graph of the given function, we applythe method of "adding the graphs" of two functions. Letus construct the graphs of the functions /1 (x) == x andf 2 (x) == -cos 2x, on one drawing using dashed lines(Fig. 35). Now, the ordinate of any point of the graphof the function x - cos 2x is equal to the sum of the ordinates of the points on the graphs of auxiliary functions(for an arbitrary value of x). By adding the ordinates of


points, it is possible to construct a sufficient number ofpoints belonging to the graph of the function x - cos 2x,and then to join them with a continuous line. Prior tofinding the critical points, let us note that the givenfunction is differentiable everywhere and that y' ==

1 -t- 2 sin 2x. The derivative vanishes at the points ,vhere

1 -t- 2 sin 2x = 0, that is, at the points x ~ (_1)n+l~+

,n; (n EZ). Note that if n == 2m -+- 1, m EZ, then x ==

~~ + scm, At these points, the derivative changes sign

from plus to minus, and therefore x = ~~ + nni (m EZ)

are points of maximum. If n == 2m, mE Z, then x =- 1~ + stni (m EZ), and, when passing through these

points, the derivative changes sign from minus to plus,therefore these points are points of minimum. ~

Example 4.3.8. Find the greatest and least values ofthe function f (x) == x - cos 2x on the closed interval[ - 1'£/2, n/12].<lllIIII \Ve use the investigation of the critical pointsof the given function (from Example 4.3.7), and seethat the interval (- 11/2, n/12) contains both the pointof maximum x == - 51[/12 and the point of minimumx === - 1'£/12. Consequently , to find the greatest valueof the function on the closed interval [-n/2, 11/12],We have to compare the value f (-51'£/12) attained bythe function at the point of maximum with the valuesreached at the end points /(-11/2) and f(n/12).

(5", ) 5It (511 ) Jig 5n

We have f -12 ~ -12-cOS -If ==2-12'

t ( - ; ) = - ~ + 1, f ( 1~ ) 1~ - ~3, and the

following inequalities hold:

va 5.rt It--->1--2 12 2 '

which are derived from the estimate 3 < rr< 3.2.Therefore the greatest value of the function on the


given interval is equal to f ( - ~) ==_ V3 - ~ Simi-12 2J 12·

larly, we find that the least value of the function on

th · . t I· I ( 3t ) n V3e given In erva IS -12::'= -12-2. ~

Example 4.3.9. Find all the values of the parameter afor each of which the function

f (x) = sin 2x - 8 (a + 1) sin x + (4a2 -t.- 8a - 14) x

increases throughout the number line and has no criticalpoints.~ For any fixed a the given function is differentiable atevery point of the number line. If the function t (x) increases, then the inequality t' (x) > 0 holds at everypoint. If, in addition, f (x) has no critical points, the relationship If (x) =t= 0 is true for any x, and, consequently,I' (x) > O. On the other hand, if for all x's the inequality/' (x) > 0 holds, then, obviously, the function has nocritical points and increases.

In view of the fact that

t' (x) = 2 cos 2x - 8 (a + 1) cos x + (4a2 + 8a - 14),

the problem can now be reformulated as follows: find allthe values of the parameter a for each of which the inequality

cos 2x - 4 (a + 1) cos x -t- (2a2 + 4a - 7) > 0

holds for any x E R. Since cos 2x = 2 cos" X - 1, bysetting cos x = t, we reformulate the problem as follows:find all the values of the parameter a for each of whichthe least value of the function

2t2- 1 - 4 (a + 1) t + (2a2 + 4a - 7),

or the function

g (t) = t2- 2 (a + 1) t + a2 + 2a - 4,

on the closed interval [-1, 1] is positive. The derivativeg' (t) = 2t - 2 (a + 1) vanishes at the point to = a + 1.Therefore the least value m of the quadratic function

4.H. Derivatives 01 Trigonometric Functions 145

g (t) on the closed interval [-1, 1] is equal to

{

g ( - 1) = a2 + 4a - 1 if a + 1~ - 1,

m=-= g(a+1)==-5 if -1<a+1<1,g (1) == a2.- 5 if a + 1~ 1.

Since the least value of the function g (t) on [-1, 1] mustbe positive, the values of the parameter a satisfying theconditions of the problem lie in two intervals: a ~ -2and a ~ O. If a ~ -2, then the least value of g (t) onthe closed interval [-1, 1] is equal to a2 -f-- 4a - 1, andthe desired values of the parameter a satisfy the inequality a2 + 4a - 1 > 0. If a ~ 0, then the least value ofg (t) on [-1, 1] equals a2 - 5, and the sought-for valuesof the parameter satisfy the inequality a2 - 5 > O. Thus,the set of the sought-for values of a is the union of solutions of two systems of inequalities

{a~-2

a;:4a-1 >0,

The set of solutions of the first system is the intervala < - 2 - ·V5 and the set of solutions of the secondsystem is a > VS. Hence the required set of valuesofais (-00, -2-VS)U(V5, +00). ~

Example 4.3.10. Construct the graph of the function

f (x) = arcsin (sin x)

and find all of its critical points.~ The given function is defined throughout the numberline R. By virtue of the periodicity of the function sin z,the function t (x) is also periodic with period 231, and itsuffices to analyze it, say, on the closed interval [-11:/2,331/2]. By the definition of arc sine, on the closed interval[-n/2, n/2] we have arcsin (sin x) = x (see Sec. 1.4,Item 1), therefore for these values of x there holds the

~ equality f (x) == x. If x E [n/2, 331/2], then 3t - x E[-rr./2, n/2], and the equality sin (n - x) = sin x implies that arcsin (sin x) = n - x, The final graph is represented in Fig. 36. In order to find the critical points ofthe given function, it suffices to investigate only thoseto-01644

l46 4. Investigating Trigonometric Functions

a:

-n/2

Fig. 36

values of x for which I sin x I = 1, that is, x = ~ +stk, If, EZ. Using the definition of the derivative, we canshow that the derivative is nonexistent at these points,

the points ofl the form x === ~ + 2nm, m EZ, being points

of maximum, and the points of the form x == ~ +n (2m - 1) == -~ + 2nm, m EZ, being points of

minimum.

PROBLEMS

In Problems 4.1 to 4.10, graph the given functions.4.1. Y= Isin2xl+Y3cos2x.4.2. Y === I sin 2x I + y3 I cos 2x I·4.3. Y == arcsin (cos x). 4.4. y == arccos (cos x).4.5. Y == sin x-x. 4.6. Y == I x I - cos 2x.

{

sin _1_, x =1= 0,4.7. Y= x

0, x==O.

.. {" x sin _1_, x ::;6 O}4.8. u> x

0, X =-==0.

4: 9 . 2x• • y:::::= arCSIn 1+x2 •

4.10. y == 4 sin" x + 4 cos" x.In Problems 4.11 to 4.14, compute the indicated Iimi ts.

. . 4 . (x2+3x-1) tan z4.11. l1m(x+1)sln -. 4.12. 11m 2+2x .

x~oo x X~O x

Problem;

J. 13 1· (x2-4x+3) sin (x-i)q. t. im ( -1)2 .

x~1 x

~ 14 1- sin x+sin 5xj. • Hfl r.:: •

1t COS x -1- cos ;)xx~- 4

t47

In Problems 4.15 to 4.20, find the derivatives ot thegiven Iunct.ions,

4.15. y=,=~+~~:;~. 4.16. y=(sin2 x + 1) ex•

14.17. y==tan2x-cot2x. 4.18. y==x2cos- ..r

4.19. y === x -+- sin x cos x.4.20. (a) y === tan sin x; (b) y == tan" x.In Problems 4.21 to 4.24, find the critical points and

compute the least and greatest values of the given functions.

4.21. Y==lsin2xl+V3cos2x.

4.22. y === [sin 2xl-+- V3 [cos 2xl.4.23. y == are-sin (cos z). 4.24. y = arccos (cos x).In Problems 4.25 anti 4.26, find the intervals of increase

and decrease of the given functions.

{

sin _1_, x=¥=O,4.25. Y=-==-lxl-cos2x. ~.26. Y== x

0, x=O.4.27. Find all the values of x for each of which the

tangent lines to the graphs of the functions

y (x) = 3 cos 5x and y (x) == 5 cos 3x -t- 2

at points wi th abscissa x are parallel.4.28. Find all the values of the parameter b for each

of which the function

f (x) = sin 2x - 8 (b + 2) cos x - (4b2 -t 1Gb + 6) x

decreases throughout the number line and has no criticalpoints.

4.29. Find the greatest value of the expression

• 2 ( 15rt ) ( 177Jt

_ 4x) 0 /8SIn -s--tfX -sin2 for ~X~Jl.

to*

for O<x<n/8.

t48 4. Investigating Trigonometric Functions

4.30. Find the least value of the expression

cot 2x- tan 2x

1+sin (5; -8x)In Problems 4.31 to 4.35, find the greatest and least

values of the given functions on the indicated intervals.4.31. f (x) = x + cos" X, x E [0, n/2l.4.32. f (x) = tan x + cot 2x, x E [rt/6, n/3l.

4.33. f(x) =i- cos 2x + sin x, x E[0, 31/2].

4.34. f (x) = ~ - -{- sin 2x++ cos3 x-cos x,

xE[-n/2, n/2].4.35. f(x)=cos2x+sinx, xErO, rt/41.

Chapter 5

Trigonometric Inequalities

5.1. Proving Inequalities Involving TrigonometricFunctions

Problems on proving trigonometric inequalities fall intoseveral types. Some problems require to prove a numerical inequality which is satisfied by some value of a trigonometric function or an expression composed of valuesof trigonometric functions; while other problems requireto prove that an inequality is satisfied for all values ofthe arguments of a given trigonometric expression orfor the permissible values of the arguments that satisfyan additional constraint of the hypothesis. However, inany case the solution is reduced to investigating the values of a trigonometric function on some interval fromits domain of definition or on the entire domain. In simple cases, we succeed in transforming a trigonometricexpression in similar problems so that it is then possibleto directly apply an inequality of the form I sin x I~ 1or I cos x I~ 1. In other cases, we can transform thetrigonomet.ric expression under consideration so that, asa result, it takes the form ·of a function F (z) in which theargument z is represented by some other, simpler trigonometric function or trigonometric expression. In sucha case, the solution of the problem is reduced to investigating the function F (z), bearing in mind that z canmeet the additional conditions connected with propertiesof trigonometric functions. In this case the function F (z)can be analyzed using either the derivative or elementaryconsiderations such as, for instance, the inequality between an arithmetic and a geometric mean of two nonnegative numbers a and b:

lj- a+b.' ab~-2- ,

150 5. Trigonometric Inequalities

in which the equality sign occurs only for a ~ b. Animportant class of problems comprise inequalities whoseproof is reduced to comparison of the size of an acuteangle, its sine and tangent (see equation (4.1)). We beginto consider some examples wi th this inequality.

Example 5.1.1. Prove that for 0 < t < n/2 the following inequalities hold:

sin t < t < tan tsin t

cos t «: -t- < 1.

A

Fig. 37

~ Consider the trigonometric circle and mark the pointP t in the first quadrantcorresponding to a realnumber t. Then LAOP t~ t(radians). The first inequality is obtained from thecomparison of the areas ofthe triangle OAP t , the sector OAEP t, and the trian-

.z gle OAWt, where W t is apoint on the line of tangents(that is, on the tangen t tothe trigonometric circle atthe point A) correspondingto the point P t (sec Fig. 37and Sec. 1.2, Item 3). It is

known from geometry that the following equalities holdtrue:

Sf; OAPt = {- IOAI·IOPtl sin t = +sin t ,

1 18 6 UAW t ==T IOAI·IA1Vtl ~T tan t,

SOAEPt.~ t/2

(the last formula follows from the fact that the area of acomplete circle of unit radius, equal to Jtr 2 == 1(, may beconsidered as the area of a sector of 2n radians, and,consequently, the area of a sector of t radians is foundfrom an obvious proportion and equals t/2).

5.1. Proving Inequalities 151

Since ~ OAPt is contained entirely in the sectorOAEPt , and the latter is contained in 60AWt , wehave S~OAPt<SOA.EPt<S~OAWt'whence

sin t < t < tan t. (5.2)

Dividing the terms of (5.2) by sin t > 0, we get 1 <t 1 . sin t

'"'-.-t < -----t ' or cos t < -t- < 1. ~SIn cos .Example 5.1.2. Prove that sin 1 > n/4.

~ Using the reduction formula (2.7), we pass to cosineand make use of formula (2.39) for cosine of double argument:

sin 1 = cos ( ~ - 1) = 1- 2 sin? ( ~ -+) .We now use inequality (5.2), which implies thatsin (~__1_) <~__1_ therefore

4 2 4 2'

1-2 sin> (~__1_) > 1-2 (~_~)2. 4 2 4 2 '

and, to prove the inequality, it suffices to ascertain

(rt 1 ) 2 Jt n 2 n 1

that 1-2 4 - 2 >4 or 8---' 4-2<0 or

n 2 - 2Jt- 4 < O. Note that all the solutions of theinequality x2 - 2.'C - 4 < 0 are specified by the condi-tion 1- 11 5< x < 1+ ·V5. Using the estimatesVS> 2.2 and Jt < 3.2, we get rt < 3.2 < 1 +V5,Jt> 0> 1-V5. Consequently, together with the inequality n2 - 2n - 4 < 0, the original inequality is alsotrue. ~

Note that inequality (5.2) enables us to calculate approximately the number sr. Indeed, it follows from (5.2)that for n > 2 the following ineqnalities hold true:

nsin ~<J1<ntan~.n n

Note that the number 2nsin~ is the perimeter of an

regular n-goll j nscriherl in the trigonometric ci rc le ,

while the number 2n tan~ is the perimeter of n rcgun


lar n-gon circumscribed about this circle (see Fig. 38,where n == 5). It follows from Theorem 4.2 that

. nSID-

lim n (sin~ -tan"'::') =; lim n ( 1 )n--+oo n n 11--+00 ~ 1- cos~

n n

=n·1 (1-1)==:0,

which makes it possible to approximate the number n

by numbers of the form n sin rr n tan~n' n

II

a:

t'ig. 38

consequently,3t 1(-tan 8== V 2- 1,

with an arbitrarily high accuracy. For instance, if

n = 2R., then the numbers sin ; , tan 2~ can be found

from formulas (2.47) and (2.49) for half-argument func-j-

tions. Thus, if k = 2, then sin : = t2

2, tan ~ "'--' 1,

1[- n V? __ "r2therefore 2 V 2 < rr< 4. If k == 3, then sin 8:'='=- .... 2 * ,

n 'V2+72cos 8 =.- 2


---I., " ~_ V2-V2+V'2 . nFor k - 1 we get SIn 16 - 2 ,COS 16 ===

~/2+V2+V2 :rt Vy 1/-2 ' tan 16 ~ 2 2 -t- 1 - y 2 - 1, there-

fore 8 V2 - V2+Y2 < n < 16 (2 Y11 2 + 1- 112- 1).In addition, the following estimates hold true:

8 V 2- V 2+V 2> 3.1,

16 (2 VV2+1-112-1) < 3.2,

which are proved by squaring both sides several times.Hence, in particular, we get the inequality 3.1 < rr < 3.2,which we used repeatedly when solving problems.

Example 5.1.3. Prove that for 0 < t < 'It/2 the in-

equality t- ~ < sin t holds.

<lIlIIII Let us use inequality (5.2) which implies that

t tT<tan z · (5.3)

Mul ti pl ying both sides of (5.a) by tile positive numhert t t t2 cos- - we get t cos- - < 2 tan - cos2 _ -2 ' 2 2 2

2 · t t . tSID 2 cos 2: :.=::. SIn , or

t ( 1- sin- -;- ) < sin t.

S· . ttL I· . tl I It 1 I . IIDee SIn 2 < 2 ' JJy rep acing In . re {} - ianr SIt o

of (5.4) sin -} by i 'we got t ( 1 - ~ ) < sin i, that is,

t3 •t-T<slnt. ~

Example 5.1.4. Prove the inequality

sin a sin 2a sin 3a < 3/4.

154 5. Trigonometric lnequalittes

~We have

sin ex sin 2a sin 3ex = sin 2a cos 2a - cos 4a2

2 sin 2(1. cos 2a - 2 sin 2(1. cos 4a4

=== sin 4a+sin 2a-sin 6a ~~4 ~ 4 '

since I sin t I~ 1. But an equality would occur only ifsin 4a == sin 2a == -sin 6a == 1. Let us show that thisis impossible. To this end, it suffices to show that sin 4aand sin 2a are not equal to 1 simultaneously. Indeed, letsin 2a == 1, then cos 2a == 0, therefore sin 4a ==2 sin 2a cos 2a == 2·0 == 0 =1= 1. ~

Example 5.1.5. Prove that for 0 < x < n/4 the ine-o cos x

qual ity . 2 ( .) > 8 takes place.SIn x cos X-SIn x

~ Since cos .» =I=- 0, by dividing the numerator anddenominator of the left-hand side by cos" x, we get

sin2 x (COS x sin X )

cos2 X Cos X - cos x

1+ tan2 X

tan2 x (1- tan x)

(1- tan x)2 + 2 tan .T

tan 2 x (1- tan x)

i-tan x 2:=:.; tan2 x + tan x (1- tan x) •

Since tanx+(1-tanx)==1 (a constant quantity), andthe numbers a==tanx, b==.:1-tanx are nonnegativeby hypothesis, inequality (5.1) implies that fortan x == 1/2 the expression tan x (1- tan x) takes OIl thegreatest value. Consequently, the least value of the

expression t (12

t ) is equal to 8. But since foranx - an x

O 11: th . 1- tan z > 0 h< x < T e expression tau2 x ,we ave

1-tanx __ I 2 8 ........tan2 x r tan x (i-tan x) > 0 ....



sin" ex + cos" ex ~ 1/8.

For what values of a does the equality occur?~We have sin" a + cos- a = 1,

(sirr'« -t- cos2a)2 == sin" ex + cos' o: + 2 sin2a., cos'' o: == 1,but since, according to (5.1), sirr' ex, + cos! a ~2 sin? ex cos" «, we have sin! ex, + cos! ex ~ 1/2, the equality occurring when sin ex == +cos a. Further,

(sin" ex + cos" a)2

== sins a., + cos" ex + 2 sin" ex cos'' CG ~ 1/4,

and since, according to (5.1), sins o: + cos" a ~

2 sirr' a cos" ex, we have sins ex + cos" a ~ 1/8. As before,the equality occurs when sin a = +cos a." that is, for

11: 1tk 1ex = 7; + 2' tc EZ. ~


(x -f- y) (x -t- y -t- 2 cos x) + 2 ~ 2 sin2 x,

For what values of x and y does the equality occur?~ Let us rewrite the given inequality as follows:

(x -f- y)2 -t- 2 (x -i-- y) cos x + 2 (1 - sin" x) ;?; 0,,-

or (x + y)2 + 2 (x + y) cos x + cos- X + cos" X ~ 0,

that is,((x + y)2 + cos X)2 + cos" X~ o.

In other words, the inequality has been reduced to anobvious one, since both terms are nonnegative. For theequality to occur, it is necessary and sufficient that

{

X + y -r- cos x=:::0, { x + y==-O tor

cos- x === 0, cos x:=:: o.

Consequently, the equality holds when x = ~ (2k+ 1)

y~::~-~ (2k-+-1), kEZ. For any other values of x,

y we have a strict inequality. ~


Example 5.1.8. Prove that there occurs the in

equality -4~Y~2+, where y=cos2x-I--3sinx.

~We have y=cos2x+3sinx=~-2sin2 x -t-3 sill x -t·· 1.

Let sinx=z, where -1~z~1, then Y= -2Z2+3z~-1.3 • 3

For z= 2~ 2) = T the function y (z) takes on the

greatest value equal to -2· 1~ +3· f+ 1 = 2+.

To find the least value of y (z) on the closed interval[-1, 1], it suffices (by virtue of the properties of thequadratic function) to compare its values at the endpoints of the interval. We have: Y (-1) = -2 - 3 +1 = -4, y (1) = -2 + 3 -t- '1 = 2. 'rhus, for the numbers z belonging to the closed interval [-1, 1] the least

value of y is -4, the greatest value is 2~...


o< sins x -·f- COS14 X ~ 1.

~Since sin" x ~ 1 and cos" x ~ 1 for any x E R, wehave: sins x ~ sin" x and COS14 x ~ cos" x. Adding theseinequali ties termwise, we get sins x + COS14 X ~ 1. Sincesins x ~ 0 and COS14 x ~ 0, we have sins x + COS14 X > 0,an equality being satisfied only if sins x and COS14 x areboth equal to zero which is impossible, that is, sin" x -t-COS

14 X > o. ~

5.2. Solving Trigonometric Inequalities

For some time trigonometric inequalities have notbeen set at entrance exams, although some problemsinvolve the comparison of values of trigonometric functions. For instance, when sol ving equations and systemsof equations containing, along with trigonometric functions, logarithms or radicals, the domain of permissiblevalues of unknowns is given by conditions having theform of trigonometric i nequal i tics. In such problems,however, the only thing required is, from the set of rootsobtained, to choose those which belong to the domain ofpermissible values without finding this domain itself.

5.2. Solving Inequalities 157

Q

y

Fig. 39

But the ability to solve simple trigonometric inequalitiesmay turn out to be useful, for instance, when it is required to find the intervals of increase and decrease of afunction using its derivative, and the function and itsderivative are given with the aid of trigonometric expressions. In this section, weconsider some examples onsolving trigonometric inequalities.

The technique of solvingsimple trigonometric inequali ties is, in many re":spects, the same as that ofsolving corresponding trigonometric equations. Forinstance, let there be required to solve the inequality tan t ~ a. The number tan t is the ordi na te ofthe point W t on the lineof tangents correspondingto the point P t (see Sec.1.4, Item 3 and Fig. 39).Therefore in order to solvethis inequality, we havefirst to find all points P t

on the trigonometric circlesuch that the ordinates of the corresponding pointson the line of tangents are less than, or equal to, a.The set of such points is shown in Fig. 39; in thegiven case, it consists of two parts, one part beingobtained from the other when rotated about the point 0through an angle of 11: (radians). Then we have to passfrom the points on the circle to the corresponding realnumbers. Since the function tan t is periodic with periodn, it suffices to find all the solutions of the inequality inquestion belonging' to a definite interval of length n;since all the remaining solutions wi ll differ from the foundones by a shift to the right or left by numbers multipleof rr. To get the shortest possible answer, it is desiredthat the initial interval of length n be chosen so that thesolutions belonging to that interval, in turn, constitute

15S $. Trigonometric lnequalities

a continuous interval. In our case, we may take, for instance, (-'Jt/2, 'Jt/2) as such an interval. If t E (-'Jt/2,n/2), then the found set of points contains on the circlepoints P t such that -n/2 < t ~ arctan a (in Fig. 39the points P t1 and P t2' for example). 1'herefore, the setof solutions of the inequality tan x ~ a is a union of aninfinite number of intervals

-i -t- nn < x ~ arctan a -t- nn, n EZ.

Other simplest trigonometric inequalities are solved ina similar way. For the sake of convenience, we give alist of solutions of simple trigonometric inequalities:

(1) sin x ~ a. If I a I~ 1, then

-arcsin a + n (2n - 1) ~ x ~ arcsin a --1- 23tn, n EZ.

If a ~ 1, then x is an arbitrary real number. If a < -1,then there is no solution.

(2) sin x~a. If la 1~1, then

arcsin a +. 2nn ~ x ~ -arcsin a + (2n + 1) 31, n EZ.

If a > 1, then there is no solution. If a. ~ -1" then xis an arbitrary real number.

(3) cos x ~ a. If I a I~ 1, then

arccos a -f- 2nn ~ x ~ -arccos a + 2n (n + 1), n EZ.

If a ~ 1, then x is an arbitrary real number. If a < -1,then there is no solution.

(4) cos x ~ a. If I a I~ 1, then

-arccos a + 2nn ~ x ~ arccos a + 2 nn, n EZ.

If a > 1, then there is no solution. If a ~ -1, then x isan arbitrary real number.

(5) tan x~a. - ~ -1- nn < x~arctan a + nn, n EZ.

(6) tanx~a. arctan a-t-rtn~x<T+nn, nEZ.

(7) cot x ~ a. arccot a + sin ~ x < rr (n + 1), n EZ.(8) cot x ~ a. nn < x ~ arccot a -~ nn, n EZ.

5.~. Solving Inequaltttes

In most cases, solution of trigonometric inequali tiescan be reduced to solution of one or several simple inequalities using identical trigonometric transformationsand an auxiliary unknown. Consider some examples.

Example 5.2.1. Solve the inequality

• 5 . 2 1. 22 2TS1n x+Tsln x>cos x.

~ Using the half-angle formula for sine, we rewri te ourinequality in the form 5 (1 - cos 2x) + 2 (1 - cos'' 2x) >8 cos 2x, or

2 cos" 2x + 13 cos 2x - 7 < O.Setting y === cos 2x, we get the quadratic inequality2y2+13y--7 < 0 whose solution is the interval-7 < y < 1/2. Thus, the problem has been reduced tosolving the inequality - 7 < cos 2x < 1/2. The inequality -7<cos2.x is satisfied for any x. Solving the

inequality cos 2x < 1/2, we get ~ --f- 2:rrn < 2x < ~3t-+

2nn, nEZ, that is, ~ +nn<x< 5: +.Tln, nEZ.•


5 -t- 2 cos 2x ~ 3 I 2 sin x - 1 I.-4I11III Using the double-argument formula for cosine, wereduce the given inequality to the form

7 - 4 sin" x ~ 3 12 sin x - 1 I.Setting y = sin x, ,ve get

7 - 4y2 ~ 3 I 2y - 1 I.(a) Let y ~ 1/2, then 7 - 4y2 ~ 3 (2y - 1) or 2y2 +

3y - 5 ~ O. Solving this inequality, we get y ~ 1 andy ~ -5/2, but from the condition y ~ 1/2 we ha vey ~ 1.

(b) Let y < 1/2. Then the inequality is rewritten asIol lows: 7 - 4y2 ~ -3 (2y - 1) or 2y2 - 3y - 2 ~ u.Solving the last inequali ty, we get y ~ 2 and y ~ -1/2Of, by hypothesis, y ~ -1/2.

Thus, all x's satisfying the inequalities sin x ~ 1 andsin x ~ -1/2 are solutions of the original inequality.The first inequality holds true only for x's satisfying the


equation sin x = 1, that is,

Solving the second inequality, \ve get5n n ,

-6+2nn~x~-6+2J[n, nEZ.

n 5nThus, x:--::T-l-2nn, nEZ, and -G+2nn~x~

- ~ +2nn, nEZ. ~


sin" x + cos" X > 13/16.

~ Transforming the Ieft-hand side, we havesin" x --t- cos" X

:::= (sin2 .7: -t·· cos2 x) (sin! x - sin? x cos2 x + cos! x)

=== (sin" x + cos- X)2 - 3 sin2 x cos- x

== 1-~ sin2 2x= 1- ~.1-cos4x==~ __1_ 3 cos4x4 · II: 2 8 8·

Hence, the problem has been reduced to solving the. 1· 5 3 cos 4x ...... 13 !J 1/2 Hinequa lty R -+- 8 .> 16 or cos JX> · ence,

-- ~ -r- 2nn < 4.1; < ~ + 2nn, nEZ, whence

3t + rtn n I 1tn-12 'T<x<12 T - 2- '


sin 2x -1- tan x ~ 2.

nEZ. ~

~The left-hand side is defined for x =1= ~ + sin, nEZ,

. 2 tan xand for the same x's the substitution SIn 2x -~ 1+ tan2 x

leads to the equivalent inequa l ity

2 tan xt+tan2 x + tan x-2~O.

5.2. Solving Inequalities 161

Setting y=tanx, we have 1~y2+y-2~O, or (since

y2 + 1 > 0) 2y + (y - 2) (y2 + 1) ~ O. Removing theparentheses, we get y3 - 2y2 + 3y - 2 > 0 which canbe rewritten as follows:

ory2 (y _ 1) - y (y - 1) + 2 (y - 1) ~ 0,

(y2 _ Y + 2) (y - 1) ~ o.The quadratic function y2 - Y + 2 is positive for any y(since the discriminant is negative), therefore the lastinequality is equivalent to the inequality y - 1 ~ 0, ortan x - 1 ~ 0, tan x ~ 1. Hence

~+rtn~x<~+nn, nEZ.~

Example 5.2.5. Solve the inequality cos" x < 3/4.~The given inequality is equivalent to the inequalities- V3/2 < cos x < Y3/2 or to the system of ineqnalities

{cosx> - y3/2,cos x < Y3/2.

The set representing the solution of this system is theintersection of the sets which are the solutions of twosimple trigonometric inequalities. In order to find thisintersection, it is convenient to consider the closed interval [-rt, n] and mark on it separately the solutions ofthe first and second inequalities. Then we get two subintervals

5n n n 5n--<x<-- and - <x<-6 6 6 6 ·

Noting that one interval can be obtained from the otherby shifting the latter by n and taking into account thatcos x is a periodic function with period 2n, we may writethe answer in the following form:

n 5nT+ nn<x<6+ nn, nEZ. ~

11-01644


PROBLEMS

In Problems 5.1 to 5.18, solve the given inequalities.5.1. cos x ~ 1/2. 5.2. sin 2x < Y2/2.5.3. sin ~ ~ -1/2. 5.4. tan (3x - 1) < 1tV3.5.5. r tan x 1~ Va. 5.6. 1 cos x I < 1/2.5.7. sin" x < 1/4. 5.8. tan" x > 1/3.5.9. 2 sin" x ~ sin x. 5.10. 2 cos" x + cos x < 1.5.11. Isinx 1< l cos z I· 5.12. l sin z I> l cos z I·5.13. l sin z Icosx>1.4.5.14. I sin x I + I cos x I > 1.5.15. 4 (sin 2 x - I cos x I) < 1.5.16. s~nx+cosx > Y3.

SIn x- cos x

5- 4 (sin 2 x +cos x)5.17. ~O.

COS x

5.18. Isi.n x+.cos x 'I'~ 1.SIn x cos x

5.19. Which is greater: tan 1 or arctan 1?~ 15.20. Which is smaller: (a) T or arctan T+

5 n . 2 2arctan S ? (b) T or arcsin T+arccos 3 1

5.21. Which is greater: sin (tan 1) or tan (sin 1)?In Problems 5.22 to 5.25, prove the indicated inequali

ties.

5.22. -+s;sinxsin(~ -x)sin( ~ +x)~+.5.23. 0 ~ cos" a + cos'' (ex, + ~)

- 2 cos ex cos ~ cos (ct + B) ~ 1.5.24. (cot" ex - 1) (3 cot" ex - 1) (cot 3ct tan 2ex - 1)~ -1.

5.25. s~nx-1 +~~ 2-s~nx •SID x-2 2 ~ 3-SID X

Answers"

CHAPTER 1

1.1. (a) In the fourth quadrant, (b) in the second quadrant,

(c) in the third quadrant. 1.2. (a) Ms, (b) A?2' (c) A:'Ao.1.3. iJ-;B4. 1.4. N=2·11·13=28G. 1.6. The number (Ct-~)/Jt must

be rational. 1.7. (a) sin 1 > sin ( 1+ 2; ), (b) cos ( 1+ 2; ) >

cos ( 1+ 4; ) . 1.8. (a) The plus sign, (b) the plus. 1.9. (a) No,

(b) yes. 1.10. The minus. 1.11. (a) 1/2, (b) - V2/2, (c) Y2/2,(d) lfS/2. 1.12. The minus. 1.13. (a) -1, (b) - ya. 1.14. Hint.

~

Consider the coordinates of the sum of vectors OP 23th anda+ N

prove that this sum is zero since it remains unchanged underthe rotation through an angle of 2n/N. 1.15. 91831. 1.16. No.1.19.2Jl. 1.21. (a) 6, (b) 30n. 1.22. n/3. 1.27. Y==!O+fl'where fo==(sin (x+1) sins (2x-3)+sin (x-1)sin3(2x+3»/2, /1=(sin (x+ 1) sin 3 (2x-3) -sin (x-1) sin'' (2x+3))/2. t.28. y= 10+11,

h f rt .:Tt 2 f . 1(;. +'v ere o==cosgcosx-sln12cos x, l=-sln g slll x

cos ;2 sin2x. 1.29. (a) a==O, b is arbitrary, (b) b==O, a is arbi

trary. 1.30. (a) The plus sign, (b) the minus. 1.31. (a) The minus,(b) the plus, (c) the plus, (d) the plus. 1.32. (a) The plus, (b) theminus, (c) the minus, (d) the minus. t.33. Increasing. 1.34. Increasing. 1.36. Decreasing. 1.37. (a) cos t == - 3/5, tan t = -4/3,cot t= -3/4, (b) cos t= -12/13, tan t=5/12, cot t=12/5,(c) cos t=4/5, tan t= -3/4, cot t= -4/3. 1.38. (a) sin t=24/25,tant=24/7, cott=7/24, (b) sint=-7/25, tant=7/24, cott=24/7, (c) sin t = -8/17, tan t == -8/15, cot t = -15/8.1.39•. (a) sin t = 3/5, cos t = 4/5, cot t = 4/3, (b) sin t = 3/5,cos t == -4/5, cot t = - 4/3. 1.40. (a) sin t = - 5/13, cos t =-12/13, tant=5/12, (b) sint=-12/13, cost==5/13, tant=

•* The letters k, l, m, n symbolize any integers if otherwise

is not stateo.

11'"

164 Answers

rt 2rt 1-12/5. 1.41. (a) ± T+ 2Jlk , (b) ± T+2nk, ± arccos 3+ 2nk .

1.42. (b) -4n/3, -2:t/3, 21[/3, 4n/3, 8n/3, 10n/3, 14rt/3, 16n/3.n nk n

1.43. (c) 6+2' (e) 2" +nk, (f) nk. 1.44. (a) n/2, (b) rt/2,

4 3 4 . 5(c) 4n/3. 1.48. (a) arccos "5' arctan "4' arccot"3' (b) arcsin 13 '

5 12 . 5 12 12arctan 12' arccot 5 (c) arcsin 13' arccos 13' arccot 5 '

. 4 3 4 2 ~/2(d) arcsin "5 ' arccos 5"' arctan "3' 1.49. (a) n-arcsin -3- ,

y--n-arctan2Y2, arccot(. __2_) n-arccos-1-

4 ' 3 '. ( 7) 24 24 7(b) arcsin - 25 ' - arccos 25 ' - arccot T' arctan 24 '

. 24 (7 ) 24 7(c) n-arcSID25 , arccos -25 ,n-arctan 7 , n-arccos 25 ,7 2 '/-5

n- arccot 24' 1.50. --5-

CHAPTER 2

2.18. sin t sin 4s. 2.19. -sin 2t sin 4s. 2.20. 1/4. 2.21. sin 4t.

2.22. i sin 8t. 2.23. -cos2 2t. 2.24. 2 sin ( 6t - ~ ) .

2.25. 8 sin ( t - ~ ) sin ( t + ~ ) sin ( t - ~ ) sin ( t +i )ICOS4 t,

2.26. -tan t tan s, 2.27. -2 sin 2t sin s cos (2t-s). 2.28. Bcos 16t Xcos3 2t . 2.33. 'It/4. 2.34. 1/2. 2.35. -2 J/-s/s. 2.36. 1. 2.37. 1 or-1/6.

CHAPTER 3

Jt . nk3.1. n (2k+1)/10, (-1)k 12+3' 3.2. 1t (4k-1)/12.

n rt nk3.3. n (2k+ 1)/2, (-1)k (3 +rtk. 3.4. rt (2k+ 1)/4, (-1)k 12+2.

3.5. n (2k+1)/4, n (2k+1)/14. 3.6. n (2k+1)/16, (-1)k+111+ ~k •

rt3.7. nk/2, n (6k ± 1)/12. 3.8. (-1)k+1 T+nk. 3.9. rr (4k+1)/4,

n 3arctan 5+n.k. 3.10. n (4k+3)/4. 3.11. -T+nk, arctan "4 +nk.

1t3.t2. rtk/5, n (4k-1)/2, n (4k+1)/10. 3.13. -T+nk, arctan 3+

Answers 165

2n :rt 1stk, 3.14. n(2k+1)/12, ±g +2rrk. 3.15. T+Jtk, arctan "3+ n k •

3.16. nk/3, jt (2k-f-1)/7. 3.17. 2 arctan 3 +2Jtk, -2 arctan 7+2n:k.3.18. n (2k+ 1)/6, n (4k-1)/4. 3.19. n (2k+ 1)/4. 3.20. nk,

rt nk(-1)k (3 +-2-' 3.21. 11 (4k-1)/4. 3.22. nk, rr; (4k+ 1)/4. 3.23. nk,

nk-arctan3. 3.24. n(4k-1)/4, rc(2k+1)/2. 3.25. n(2k+1)/6,1 V73-7

k =1= 31+1, :rk/5, k ~ 5l. 3.26. ± 2" are-cos 12 +nk,

n 1±3+Jtk, ±2arccos(-1/3)+nk. 3.27. ;r(4k+3)/32. 3.28. xk,

n (2k+ 1)/6. 3.29. Jl (3k ± 1)/3. 3.30. rr (2k+ 1)/8, 1£ (6k ± 1)/12.

3.31. nkj2. 3.32. ~ +2nk. 3.33. (-1)k arcsin ~ +sik, 3.34. stk,

3.35. rr (4k+1)/4. 3.36. ~ +Jtk. 3.37. Jt/2. 3.38. ~~ +Jtk,

17rr; Ts: 2n 4n~+nk, 24+nk. 3.39. sik, 3.40. g+4nk, 3+ 4nk,

1~Jt +4Jtk. 3.42. (2, ± 23"1: -2+2Jtk). 3.43. {(2, -1), (-2,

-1)}. 3.44. (rt (2k+1)j4, :t (6l+1)/6). 3.45. (n (4k+1)/4,-n (12k+ 1)j12), (rr (12k-1)j12, n (1-4k)j4). 3.46. (st (21c+3)/2,

Jt (6k-1)/6). 3.47. (~ +Jt (k-l), ; +Jt (k +l)), (- ~ -+Jt (k-l), 2

3Jt+ rt (k+ l) ). 3.48. «6k-1)j6), (6k+ 1)/6).

3.49. (arctan~ +sck, arctan ~ - Jtk). 3.50. (± ~ +sik,

± ~ + Jtl) , l-k=2m. 3.51. -2+12k, 2+12k.

CHAPTER 4

4.1. See Fig. 40. 4.2. See Fig. 41. 4.3. See Fig. 42. 4.4. SeeFig. 43. 4.5. See Fig. 44. 4.6. See Fig. 45. 4.7. See Fig. 46.4.8. See Fig. 47. 4.9. See Fig. 48. Hint.

{

-n-2 arctan x,y = 2 arctan a,

n-2 arctan x,

xE(-oo, -1],x E[-1, 1J,x E[1, + 00 ).

4.10. See Fig. 49. flint. y==3+cos4x. 4.11. 4. 4.12. -0.5.4.13. -2. 4.14. 1. 4.15. 2 (tan x+tan3 x). 4.16. (sin- z -l-sin 2x+

1) eX. 4.17. 8/sin24x. 4.18. 2x cos ...!-+sin .i.. 4.19. 2 cos- x,X x

4.20. (a) cos xjcos2 sin x, (b) 3 (tan2 x+ tan- x). 4.21. Critical points:x=rt (12k ± 1)/12 (y' =0 points of maximum), x=nk/2 (y' is

y2

Fig. 40

.Ii' Jl7i 2

Fig. 41

l~ig. 43

Answers 167

y

IIr-ff

I

-,-,

-,-,

-,-,

-,-,

-.Jr

-,-,

-,<,

-,-,

<,-,

Fig. 44

Fig. 45

.68 Answers

-I

Fig. 46

y, /, /-, I //

~---------~------ ------r----------, /~ .I.:

Fig. 47

Fig. 48

-.JilL; 0

Answers

Fig. 49

eX

169

nonexistent, points of minimum), max y (x) = 2, min y (x) = - V3.xER xER

4.22. Critical points: x=31(6k ± 1)/12 (y' =0, points of maximum),x = nk/4 (y' is nonexistent, points of minimum), max y (x) = 2,

xERmin y (x) = 1. 4.23. Critical points: x= nk (the derivative is nonxER

existent), max y (x) = ~ , min y (x) = - ~ . 4.24. Critical points:xER xER

x= tik (the derivative is nonexistent) max y (x) = 31, min y (x) = o.xER xER

4.25. Intervals of increase [ 0, ~~ J, rn (2~+ 1) ~,

n (2k+1)+-'::'J k 0 [Jt (2l+1) +..::. ; (2l+3) Jt ]4 6' <, 4 6' 4 6'

. [ t« J [3t (2l+1) 3tl > 0, Intervals of decrease -12' 0, 4 +6 '3t (2l +3) 3t ] l 0 [J"t (2k+ 1) n 3t (2k+ 1) , ~J

4 6' <, 4 6 ' 4 -r 6 '

k> O. 4.26. Intervals of inc~ease [n (4;+3) , n(4;+1J ' inter-

vals of decrease (-00, -; J, [; , +00), [n(4;+1)':It (4;-1)J, k =/= O. 4.27. nk, n (2k+1)j8. 4.28. b <-3- Y3,

b > -1+V3. 4.29. 1/ Y2. 4.30. 2. 4.31. Ymax = rt/2, Ymln = 1.4.32. Ymax=2 v3/3, Yroln=1. 4.33. Ynlax=3/4, Ymln=O.5.4.34. Ymax=n/4, Ymln= -rt/4. 4.35. Ymax=1.25, Ymln=1.

f70 Answers

CHAPTER 5

[rt 3t ] (Sn n )5.1. -3+2nn, g+2nn · 5.2. -g+nn, 8+'111, .

[5n . 35n ] ( 1 n nn 1

5.3. -T+10rtn, -6-+ 10nn . 5.4. 3-"6+3' '3+n nn ) (n 1t ] [11: 11: )18+3 · 5.5. -y+nn, -3+ '1n , .3+nn' T + nn •

(1t 2n) (n 11:)5.6. 3+ nn, g+nn · 5.7. -T+'1n, If+nn .

5.8. (- ~ +nn, _0 ~ +nnJ ' [~ +rrn, ~ +nn) .5.9. [2M,11: ] [51(; -'J ( 1t I6"+2nn , T+ 2rrn, n+2nn . 5.10. -n+2n:n, -3-1-

2nn ), (~ +2nn, rt+2JTn ). 5.11. (--~ +nn. ~ +nn ) .

(n 3rt) ( n 531 )5.12. 4+ nn, T+ ttn · 5.13. "12+ 2nn, 12+ 2nn ,

(511: 1t) nn (n-12+2nn, -12+ 2nn · 5.14. x 4= 2 . 5.15. -T+nn,

1£ ) ( n 5n) n ( rr3+ nn . 5.16. 4+:rn, -f2+ 1[11, • 5.17. ± T+ 2nn, 2+3n ) r n ]2nn, 2+ 2nn · 5.18. Lrtn, 2+ nn . 5.19. tan 1. 5.20. (a) n/4,

(b) n/4. 5.21. tan (sin 1).

Subject Index

Basic properties of trigonometric functions, 23

Basic trigonometric functions,18

Addition formulas, 41Arc cosine, 33Arc cotangent, 35Arc sine, 32Arc tangent, 35

Evaluation of trigonometricexpressions, 63-70

Even function, 26Extremum,

a necessary condition of, 139a sufficient condition of, 139

Computing limits, 126-131Coordinate circle, 10Cosine, 18

of sum (difference) ofreal numbers, 42-45

Cotangent, 20of a difference, 50of a sum, 50

Degree(s), 9Derivative(s), 132

two

reduction, 45of sum and difference of liketrigonometric functions,51for transforming a product oftrigonometric functionsinto a sum, 53of triples argument, 62

Function(s),continuous at a point, 126decreasing, 27differentiable, 132discontinuous, 129even, 26increasing, 27nondecreasing, 27nonincreasing, 27odd, 23periodic, 23

Fundamental trigonometricidentity, 18

Graph of harmonic oscillations,120-126

Graphs of basic trigonometricfunctions, 113

Half-angle (argument) formulas, 59

Harmonic oscillations, 120

Formula(s),addition of, 41of double argument, 55of half argument, 59

Intervals of monotonicity, 27Inverse trigonometric functions,

31Investigating trigonometric

functions with the aid ofderivative, 132

172 Subject Index

Limit(s), 126first remarkable, 129of a function, 126

Line,of cotangents, 21of tangents, 21

Monotonicity, 27interval of, 27sufficient condition of, 137

Nondecreasing function, 27Nonincreasing function, 27

Odd function, 26

Periodicity, 23Period(s)

fundamental, 23Point(s),

accumulation, 126critical, 138of extremum, 138limit, 126of maximum, 137of minimum, 137

Principal methods of solvingtrigonometric equations,87-90

Properties and graph of thefunctions:f (x) = cos I, 116-118f (x) .= sin I, 113-115f (x) = tan I, 118-120

Quadrantts), 17

Radian, 9Reduction formulas, 45

Relation bet,veen trigonometric functions, 30

Root(s),ari thmetic square, 87

Simplifying trigonometric expressions, 70

Sine. 18of a difference J 47of a sum, 47

Sine line l 115Sinusoid, 115Solving

the equation cos t = m; 33simplest trigonometric Iunctions, 31trigonometric equations andsystems of equations in several unknowns, 101-109

Tangent, 20or a difference, 48of a sum, 48

Transformingthe expression a sin t + b,54sums and products of trigonometric expressions, 74

Trigonometriccircle, 10equations and systems of equations, 80inequalities, 149-160

Unit of measurement, 9Universal substitution formu

las, 56

Vertices of a regular N-gon, 13

\V ina ing the real axis on thetrigonometric circle, 11

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