TRIGONOMETRIC FUNCTIONS

Transcription

1 Chapter TRIGONOMETRIC FUNCTIONS.1 Introduction A mathematician knows how to solve a problem, he can not solve it. MILNE The word trigonometr is derived from the Greek words trigon and metron and it means measuring the sides of a triangle. The subject was originall developed to solve geometric problems involving triangles. It was studied b sea captains for navigation, surveor to map out the new lands, b engineers and others. Currentl, trigonometr is used in man areas such as the science of seismolog, designing electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analsing a musical tone and in man other areas. In earlier classes, we have studied the trigonometric ratios of acute angles as the ratio of the sides of a right Ara Bhatt ( B.C.) angled triangle. We have also studied the trigonometric identities and application of trigonometric ratios in solving the problems related to heights and distances. In this Chapter, we will generalise the concept of trigonometric ratios to trigonometric functions and stud their properties.. Angles Angle is a measure of rotation of a given ra about its initial point. The original ra is Vertex Fig.1

2 50 MATHEMATICS called the initial side and the final position of the ra after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative (Fig.1). The measure of an angle is the amount of rotation performed to get the terminal side from the initial side. There are several units for measuring angles. The definition of an angle Fig. suggests a unit, viz. one complete revolution from the position of the initial side as indicated in Fig.. This is often convenient for large angles. For example, we can sa that a rapidl spinning wheel is making an angle of sa 15 revolution per second. We shall describe two other units of measurement of an angle which are most commonl used, viz. degree measure and radian measure Degree measure If a rotation from the initial side to terminal side is 60 of a revolution, the angle is said to have a measure of one degree, written as 1. A degree is divided into 60 minutes, and a minute is divided into 60 seconds. One sixtieth of a degree is called a minute, written as 1, and one sixtieth of a minute is called a second, written as 1. Thus, 1 60, 1 60 Some of the angles whose measures are 60,180, 70, 40, 0, 40 are shown in Fig.. th Fig.

3 TRIGONOMETRIC FUNCTIONS 51.. Radian measure There is another unit for measurement of an angle, called the radian measure. Angle subtended at the centre b an arc of length 1 unit in a unit circle (circle of radius 1 unit) is said to have a measure of 1 radian. In the Fig.4(i) to (iv), OA is the initial side and OB is the terminal side. The figures show the angles whose measures are 1 radian, 1 radian, 1 1 radian and 1 1 radian. (i) (ii) (iii) Fig.4 (i) to (iv) We know that the circumference of a circle of radius 1 unit is. Thus, one complete revolution of the initial side subtends an angle of radian. More generall, in a circle of radius r, an arc of length r will subtend an angle of 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1 radian, an arc of length l will subtend an angle whose measure is l radian. Thus, if in r a circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have (iv) θ l r or l r θ.

4 5 MATHEMATICS.. Relation between radian and real numbers Consider the unit circle with centre O. Let A be an point on the circle. Consider OA as initial side of an angle. Then the length of an arc of the circle will give the radian measure of the angle which the arc will subtend at the centre of the circle. Consider the line PAQ which is tangent to the circle at A. Let the point A represent the real number zero, AP represents positive real number and AQ represents negative real numbers (Fig.5). If we rope the line AP in the anticlockwise direction along the circle, and AQ in the clockwise direction, then ever real number will correspond to a radian measure and conversel. Thus, radian measures and real numbers can be considered as one and the same. O Fig.5 1 A P Q..4 Relation between degree and radian Since a circle subtends at the centre an angle whose radian measure is and its degree measure is 60, it follows that radian 60 or radian 180 The above relation enables us to express a radian measure in terms of degree measure and a degree measure in terms of radian measure. Using approximate value of as 7, we have Also 1 1 radian approximatel. 180 radian radian approximatel. The relation between degree measures and radian measure of some common angles are given in the following table: Degree Radian 6 4

5 TRIGONOMETRIC FUNCTIONS 5 Notational Convention Since angles are measured either in degrees or in radians, we adopt the convention that whenever we write angle θ, we mean the angle whose degree measure is θ and whenever we write angle β, we mean the angle whose radian measure is β. Note that when an angle is expressed in radians, the word radian is frequentl omitted. Thus, 180 and 45 are written with the understanding that and 4 4 are radian measures. Thus, we can sa that Radian measure 180 Degree measure Degree measure 180 Radian measure Example 1 Convert 40 0 into radian measure. Solution We know that 180 radian. Hence degree radian 540 radian. Therefore radian. Example Convert 6 radians into degree measure. Solution We know that radian 180. Hence 6 radians degree degree 7 60 degree 4 + minute [as 1 60 ] minute [as 1 60 ] approximatel. Hence 6 radians approximatel. Example Find the radius of the circle in which a central angle of 60 intercepts an arc of length 7.4 cm (use ). 7

6 54 MATHEMATICS Solution Here l 7.4 cm and θ radian 180 Hence, b r θ l, we have r cm Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use.14). Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through of a revolution. Therefore, θ 60 or 4 radian. Hence, the required distance travelled is given b l r θ cm cm.14 cm 6.8 cm. Example 5 If the arcs of the same lengths in two circles subtend angles 65 and 110 at the centre, find the ratio of their radii. Solution Let r 1 and r be the radii of the two circles. Given that θ radian and θ radian Let l be the length of each of the arc. Then l r 1 θ 1 r θ, which gives 1 6 r 1 6 r, i.e., r1 r 1 Hence r 1 : r : 1. EXERCISE.1 1. Find the radian measures corresponding to the following degree measures: (i) 5 (ii) 47 0 (iii) 40 (iv) 50

7 TRIGONOMETRIC FUNCTIONS 55. Find the degree measures corresponding to the following radian measurs (Use ) (i) (ii) 4 (iii) (iv) A wheel makes 60 revolutions in one minute. Through how man radians does it turn in one second? 4. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm b an arc of length cm (Use ) In a circle of diameter 40 cm, the length of a chord is 0 cm. Find the length of minor arc of the chord. 6. If in two circles, arcs of the same length subtend angles 60 and 75 at the centre, find the ratio of their radii. 7. Find the angle in radian through which a pendulum swings if its length is 75 cm and th e tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 1 cm. Trigonometric Functions In earlier classes, we have studied trigonometric ratios for acute angles as the ratio of sides of a right angled triangle. We will now extend the definition of trigonometric ratios to an angle in terms of radian measure and stud them as trigonometric functions. Consider a unit circle with centre at origin of the coordinate axes. Let P (a, b) be an point on the circle with angle AOP x radian, i.e., length of arc AP x (Fig.6). We define cos x a and sin x b Since OMP is a right triangle, we have OM + MP OP or a + b 1 Thus, for ever point on the unit circle, we have a + b 1 or cos x + sin x 1 Since one complete revolution subtends an angle of radian at the centre of the circle, AOB, Fig.6

8 56 MATHEMATICS AOC and AOD. All angles which are integral multiples of are called quadrantal angles. The coordinates of the points A, B, C and D are, respectivel, (1, 0), (0, 1), ( 1, 0) and (0, 1). Therefore, for quadrantal angles, we have cos 0 1 sin 0 0, cos 0 sin 1 cos 1 sin 0 cos 0 sin 1 cos 1 sin 0 Now, if we take one complete revolution from the point P, we again come back to same point P. Thus, we also observe that if x increases (or decreases) b an integral multiple of, the values of sine and cosine functions do not change. Thus, sin (n + x) sin x, n Z, cos (n + x) cosx, n Z Further, sin x 0, if x 0, ±, ±, ±,..., i.e., when x is an integral multiple of and cos x 0, if x ±, ± multiple of. Thus, ± 5,... i.e., cos x vanishes when x is an odd sin x 0 implies x n, where n is an integer cos x 0 implies x (n + 1), where n is an integer. We now define other trigonometric functions in terms of sine and cosine functions: cosec x 1 sin x, x n, where n is an integer. sec x 1 cos x, x (n + 1), where n is an integer. tan x sin x cos x, x (n +1), where n is an integer. cot x cos x, x n, where n is an integer. sin x

9 TRIGONOMETRIC FUNCTIONS 57 We have shown that for all real x, sin x + cos x 1 It follows that 1 + tan x sec x (wh?) 1 + cot x cosec x (wh?) In earlier classes, we have discussed the values of trigonometric ratios for 0, 0, 45, 60 and 90. The values of trigonometric functions for these angles are same as that of trigonometric ratios studied in earlier classes. Thus, we have the following table: 0 sin 0 cos 1 tan The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cos x and tan x, respectivel...1 Sign of trigonometric functions Let P (a, b) be a point on the unit circle with centre at the origin such that AOP x. If AOQ x, then the coordinates of the point Q will be (a, b) (Fig.7). Therefore cos ( x) cos x and sin ( x) sin x 1 not not defined defined Since for ever point P (a, b) on the unit circle, 1 a 1 and Fig.7

10 58 MATHEMATICS 1 b 1, we have 1 cos x 1 and 1 sin x 1 for all x. We have learnt in previous classes that in the first quadrant (0 < x < ) a and b are both positive, in the second quadrant ( < x <) a is negative and b is positive, in the third quadrant ( < x < ) a and b are both negative and in the fourth quadrant ( < x < ) a is positive and b is negative. Therefore, sin x is positive for 0 < x <, and negative for < x <. Similarl, cos x is positive for 0 < x <, negative for < x < and also positive for < x <. Likewise, we can find the signs of other trigonometric functions in different quadrants. In fact, we have the following table. I II III IV sin x + + cos x + + tan x + + cosec x + + sec x + + cot x Domain and range of trigonometric functions From the definition of sine and cosine functions, we observe that the are defined for all real numbers. Further, we observe that for each real number x, 1 sin x 1 and 1 cos x 1 Thus, domain of sin x and cos x is the set of all real numbers and range is the interval [ 1, 1], i.e., 1 1.

11 TRIGONOMETRIC FUNCTIONS 59 1 Since cosec x, the domain of cosec x is the set { x : x R and sin x x n, n Z} and range is the set { : R, 1 or 1}. Similarl, the domain of sec x is the set {x : x R and x (n + 1), n Z} and range is the set { : R, 1or 1}. The domain of tan x is the set {x : x R and x (n + 1), n Z} and range is the set of all real numbers. The domain of cot x is the set {x : x R and x n, n Z} and the range is the set of all real numbers. We further observe that in the first quadrant, as x increases from 0 to, sin x increases from 0 to 1, as x increases from to, sin x decreases from 1 to 0. In the third quadrant, as x increases from to, sin x decreases from 0 to 1and finall, in the fourth quadrant, sin x increases from 1 to 0 as x increases from to. I quadrant II quadrant III quadrant IV quadrant sin increases from 0 to 1 decreases from 1 to 0 decreases from 0 to 1 increases from 1 to 0 cos decreases from 1 to 0 decreases from 0 to 1 increases from 1 to 0 increases from 0 to 1 tan increases from 0 to increases from to 0 increases from 0 to increases from to 0 cot decreases from to 0 decreases from 0 to decreases from to 0 decreases from 0to sec increases from 1 to increases from to 1 decreases from 1to decreases from to 1 cosec decreases from to 1 increases from 1 to increases from to 1 decreases from 1to Similarl, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table: Remark In the above table, the statement tan x increases from 0 to (infinit) for 0 < x < simpl means that tan x increases as x increases for 0 < x < and

12 60 MATHEMATICS assumes arbitrail large positive values as xapproaches to. Similarl, to sa that cosec x decreases from 1 to (minus infinit) in the fourth quadrant means that cosec x decreases for x (, ) and assumes arbitraril large negative values as x approaches to. The smbols and simpl specif certain tpes of behaviour of functions and variables. We have alread seen that values of sin x and cos x repeats after an interval of. Hence, values of cosec x and sec x will also repeat after an interval of. We Fig.8 Fig.9 Fig.10 Fig.11

13 TRIGONOMETRIC FUNCTIONS 61 Fig.1 Fig.1 shall see in the next section that tan ( + x) tan x. Hence, values of tan x will repeat after an interval of. Since cot x is reciprocal of tan x, its values will also repeat after an interval of. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given above: Example 6 If cos x, x lies in the third quadrant, find the values of other five 5 trigonometric functions. 5 Solution Since cos x, we have sec x 5 Now sin x + cos x 1, i.e., sin x 1 cos x or sin x Hence sin x ± 4 5 Since x lies in third quadrant, sin x is negative. Therefore sin x 4 5 which also gives cosec x 5 4

14 6 MATHEMATICS Further, we have tan x sin x cos x 4 and cos x cot x sin x 4. Example 7 If cot x 5, x lies in second quadrant, find the values of other five 1 trigonometric functions. Solution Since cot x 5 1, we have tan x 1 5 Now sec x 1 + tan x Hence sec x ± 1 5 Since x lies in second quadrant, sec x will be negative. Therefore which also gives Further, we have sec x 1 5, 5 cos x 1 and cosec x sin x tan x cos x ( 1 5 ) ( 5 1 ) sin x 1 1. Example 8 Find the value of sin 1. Solution We know that values of sin x repeats after an interval of. Therefore sin 1 sin (10 + ) sin.

15 TRIGONOMETRIC FUNCTIONS 6 Example 9 Find the value of cos ( 1710 ). Solution We know that values of cos x repeats after an interval of or 60. Therefore, cos ( 1710 ) cos ( ) cos ( ) cos EXERCISE. Find the values of other five trigonometric functions in Exercises 1 to cos x 1, x lies in third quadrant.. sin x, x lies in second quadrant. 5. cot x 4, x lies in third quadrant. 4. sec x 1 5, x lies in fourth quadrant. 5. tan x 5, x lies in second quadrant. 1 Find the values of the trigonometric functions in Exercises 6 to sin cosec ( 1410 ) 8. tan cot ( 15 4 ) 9. sin ( 11 ).4 Trigonometric Functions of Sum and Difference of Two Angles In this Section, we shall derive expressions for trigonometric functions of the sum and difference of two numbers (angles) and related expressions. The basic results in this connection are called trigonometric identities. We have seen that 1. sin ( x) sin x. cos ( x) cos x We shall now prove some more results:

16 64 MATHEMATICS. cos (x + ) cos x cos sin x sin Consider the unit circle with centre at the origin. Let x be the angle P 4 OP 1 and be the angle P 1 OP. Then (x + ) is the angle P 4 OP. Also let ( ) be the angle P 4 OP. Therefore, P 1, P, P and P 4 will have the coordinates P 1 (cos x, sin x), P [cos (x + ), sin (x + )], P [cos ( ), sin ( )] and P 4 (1, 0) (Fig.14). Consider the triangles P 1 OP and P OP 4. The are congruent (Wh?). Therefore, P 1 P and P P 4 are equal. B using distance formula, we get P 1 P [cos x cos ( )] + [sin x sin( ] (cos x cos ) + (sin x + sin ) cos + cos cos x cos + sin x + sin + sin x sin (cos x cos sin x sin ) (Wh?) Also, P P 4 [1 cos (x + )] + [0 sin (x + )] 1 cos (x + ) + cos (x + ) + sin (x + ) cos (x + ) Fig.14

17 TRIGONOMETRIC FUNCTIONS 65 Since P 1 P P P 4, we have P 1 P P P 4. Therefore, (cos x cos sin x sin ) cos (x + ). Hence cos (x + ) cos x cos sin x sin 4. cos (x ) cos x cos + sin x sin Replacing b in identit, we get cos (x + ( )) cos x cos ( ) sin x sin ( ) or cos (x ) cos x cos + sin x sin 5. cos ( x ) sin x If we replace x b and b x in Identit (4), we get cos ( x ) cos cos x + sin sin x sin x. 6. sin ( x ) cos x Using the Identit 5, we have sin ( x ) cos x cos x. 7. sin (x + ) sin x cos + cos x sin We know that sin (x + ) cos ( x + ) cos ( x) cos ( x ) cos + sin ( x) sin sin x cos + cos x sin 8. sin (x ) sin x cos cos x sin If we replace b, in the Identit 7, we get the result. 9. B taking suitable values of x and in the identities, 4, 7 and 8, we get the following results: cos ( + x) sin x sin ( + x) cos x cos ( x) cos x sin ( x) sin x

18 66 MATHEMATICS cos ( + x) cos x sin ( + x) sin x cos ( x) cos x sin ( x) sin x Similar results for tan x, cot x, sec x and cosec x can be obtianed from the results of sin x and cos x. 10. If none of the angles x, and (x + ) is an odd multiple of, then tan (x + ) tan x+ tan 1 tan xtan Since none of the x, and (x + ) is an odd multiple of, it follows that cos x, cos and cos (x + ) are non-zero. Now tan (x + ) sin( x + ) sin x cos + cos x sin. cos( x+ ) cos xcos sin xsin Dividing numerator and denominator b cos x cos, we have tan (x + ) sin x cos cos xsin + cos x cos cos x cos cos x cos sin x sin cos x cos cos x cos tan x + tan 1 tanxtan tan x tan 11. tan ( x ) 1+tan x tan If we replace b in Identit 10, we get tan (x ) tan [x + ( )] tan x + tan ( ) tan x tan 1 tanx tan( ) 1+ tanx tan 1. If none of the angles x, and (x + ) is a multiple of, then cot ( x + ) cot xcot 1 cot +cot x

19 TRIGONOMETRIC FUNCTIONS 67 Since, none of the x, and (x + ) is multiple of, we find that sin x sin and sin (x + ) are non-zero. Now, cos ( x + ) cos xcos sin xsin cot ( x + ) sin ( x + ) sin xcos + cos xsin Dividing numerator and denominator b sin x sin, we have cot (x + ) cot x cot 1 cot + cot x cot xcot cot (x ) cot cot x If we replace b in identit 1, we get the result 14. cos x cos x sin x cos x 1 1 sin x We know that cos (x + ) cos x cos sin x sin Replacing b x, we get cos x cos x sin x cos x 1 cos x (1 cos x) cos x 1 Again, cos x cos x sin x 1 sin x sin x 1 sin x. We have cos x cos x sin x Dividing each term b cos x, we get 1 tan x cos x 1+tan x cos cos x sin x + sin x x 1 tan 1+tan x x tan x 15. sin x sinx cos x 1+tan x We have sin (x + ) sin x cos + cos x sin Replacing b x, we get sin x sin x cos x. sin xcos x Again sin x cos x+ sin x

20 68 MATHEMATICS Dividing each term b cos x, we get tan x sin x 1+ tan x tan x 16. tan x 1 tan x We know that tan (x + ) tan x + tan 1 tanx tan tan x Replacing b x, we get tan x 1 tan x 17. sin x sin x 4 sin x We have, sin x sin (x + x) sin x cos x + cos x sin x sin x cos x cos x + (1 sin x) sin x sin x (1 sin x) + sin x sin x sin x sin x + sin x sin x sin x 4 sin x 18. cos x 4 cos x cos x We have, cos x cos (x +x) cos x cos x sin x sin x (cos x 1) cos x sin x cos x sin x (cos x 1) cos x cos x (1 cos x) cos x cos x cos x + cos x 4cos x cos x. tan x tan x 19. tan x 1 tan x We have tan x tan (x + x) tan x + tan x tan x + tan x 1 tan x 1 tan xtan x tan x. tan x 1 1 tan x

21 TRIGONOMETRIC FUNCTIONS 69 tan x + tan x tan x tan x tan x 1 tan x tan x 1 tan x 0. (i) cos x + cos (ii) cos x cos (iii) sin x + sin x + x cos cos x + x sin sin x + x sin cos (iv) x + x sin x sin cos sin We know that cos (x + ) cos x cos sin x sin... (1) and cos (x ) cos x cos + sin x sin... () Adding and subtracting (1) and (), we get cos (x + ) + cos(x ) cos x cos... () and cos (x + ) cos (x ) sin x sin... (4) Further sin (x + ) sin x cos + cos x sin... (5) and sin (x ) sin x cos cos x sin... (6) Adding and subtracting (5) and (6), we get sin (x + ) + sin (x ) sin x cos... (7) sin (x + ) sin (x ) cos x sin... (8) Let x + θ and x φ. Therefore θ θ x +φ and φ Substituting the values of x and in (), (4), (7) and (8), we get cos θ + cos φ cos θ +φ θ cos φ cos θ cos φ sin θ+ φ θ φ sin sin θ + sin φ sin θ +φ θ cos φ

22 70 MATHEMATICS θ θ sin θ sin φ cos +φ sin φ Since θ and φ can take an real values, we can replace θ b x and φ b. Thus, we get cos x + cos cos sin x + sin sin x + x x + x cos ; cos x cos sin sin, x + x x + x cos ; sin x sin cos sin. Remarks As a part of identities given in 0, we can prove the following results: 1. (i) cos x cos cos (x + ) + cos (x ) (ii) sin x sin cos (x + ) cos (x ) (iii) sin x cos sin (x + ) + sin (x ) (iv) cos x sin sin (x + ) sin (x ). Example 10 Prove that 5 sin sec 4sin cot Solution We have 5 L.H.S. sin sec 4sin cot sin sin R.H.S. Example 11 Find the value of sin 15. Solution We have sin 15 sin (45 0 ) sin 45 cos 0 cos 45 sin Example 1 Find the value of tan 1 1.

23 TRIGONOMETRIC FUNCTIONS 71 Solution We have tan 1 tan tan tan tan tan tan tan Example 1 Prove that sin ( x + ) tan x+ tan sin ( x ) tan x tan. Solution We have L.H.S. sin ( x+ ) sin xcos + cos xsin sin ( x ) sin xcos cos xsin Dividing the numerator and denominator b cos x cos, we get sin ( x + ) tan x+ tan sin ( x ) tan x tan. Example 14 Show that tan x tan x tan x tan x tan x tan x Solution We know that x x + x Therefore, tan x tan (x + x) or tan x+ tan x tanx 1 tan x tan x or tan x tan x tan x tan x tan x + tan x or tan x tan x tan x tan x tan x tan x or tan x tan x tan x tan x tan x tan x. Example 15 Prove that cos + x + cos x cos x 4 4 Solution Using the Identit 0(i), we have

24 7 MATHEMATICS L.H.S. cos + x + cos x 4 4 x x x ( + + x) cos cos 4 4 cos 4 cos x 1 cos x cos x R.H.S. Example 16 Prove that cos 7 x + cos 5 x cot x sin 7 x sin 5x Solution Using the Identities 0 (i) and 0 (iv), we get L.H.S. 7x + 5x 7x 5x cos cos 7x + 5x 7x 5x cos sin cos x sin x cot x R.H.S. Example 17 Prove that sin 5 x sin x+ sin x tan x cos5x cos x Solution We have L.H.S. sin 5 x sinx+ sin x cos5x cos x sin 5 x+ sin x sin x cos5x cos x sin x cos x sinx sinxsin x sin x (cosx 1) sin xsin x 1 cosx sin x tan x R.H.S. sin x sin xcos x

25 TRIGONOMETRIC FUNCTIONS 7 Prove that: EXERCISE. 1. sin cos tan. sin 4 + cosec 6. 5 cot + cosec + tan Find the value of: (i) sin 75 (ii) tan Prove the following: 7 cos 6 sin + cos + sec cos x cos sin x sin sin ( x+ ) tan tan + x 4 1+ tan x 1 tan x x 4 8. cos ( + x) cos ( x) sin ( x) cos + x cot x 9. cos + x cos ( + x) cot x + cot ( + x) sin (n + 1)x sin (n + )x + cos (n + 1)x cos (n + )x cos x 11. cos + x cos x 4 4 sin x 1. sin 6x sin 4x sin x sin 10x 1. cos x cos 6x sin 4x sin 8x 14. sin x + sin 4x + sin 6x 4 cos x sin 4x 15. cot 4x (sin 5x + sin x) cot x (sin 5x sin x) cos9x cos5x sin x 17. sin17x sin x cos10x sin x sin x tan 19. cos x+ cos sin x sin x sin x 1. sin x cos x sin5x + sin x tan 4x cos5x + cosx sin x + sin x tan x cos x + cosx cos4x + cosx + cosx cot x sin 4x + sin x + sin x

26 74 MATHEMATICS. cot x cot x cot x cot x cot x cot x 1. 4tan x (1 tan x) tan 4x 4. cos 4x 1 8sin x cos x 4 1 6tan x + tan x 5. cos 6x cos 6 x 48cos 4 x + 18 cos x 1.5 Trigonometric Equations Equations involving trigonometric functions of a variable are called trigonometric equations. In this Section, we shall find the solutions of such equations. We have alread learnt that the values of sin x and cos x repeat after an interval of and the values of tan x repeat after an interval of. The solutions of a trigonometric equation for which 0 x < are called principal solutions. The expression involving integer n which gives all solutions of a trigonometric equation is called the general solution. We shall use Z to denote the set of integers. The following examples will be helpful in solving trigonometric equations: Example 18 Find the principal solutions of the equation sin x Solution We know that, Therefore, principal solutions are sin and x and.. sin sin sin. Example 19 Find the principal solutions of the equation tan x 1. 1 Solution We know that, tan. Thus, 6 1 tan tan 6 6 and Thus 1 tan tan tan tan. 6 6 Therefore, principal solutions are 5 11 and 6 6. We will now find the general solutions of trigonometric equations. We have alread

27 TRIGONOMETRIC FUNCTIONS 75 seen that: sin x 0 gives x n, where n Z cos x 0 gives x (n + 1), where n Z. We shall now prove the following results: Theorem 1 For an real numbers x and, sin x sin implies x n + ( 1) n, where n Z Proof If sin x sin, then sin x sin 0 or cos x + sin x 0 which gives cos x + 0 or sin x x + Therefore (n + 1) or x n, where n Z i.e. x (n + 1) or x n +, where n Z Hence x (n + 1) + ( 1) n + 1 or x n +( 1) n, where n Z. Combining these two results, we get x n + ( 1) n, where n Z. Theorem For an real numbers x and, cos x cos, implies x n ±, where n Z Proof If cos x cos, then cos x cos 0 i.e., sin x sin x 0 0 Thus sin x + 0 or sin x 0 Therefore x + n or x n, where n Z i.e. x n or x n +, where n Z Hence x n ±, where n Z Theorem Prove that if x and are not odd mulitple of, then tan x tan implies x n +, where n Z

28 76 MATHEMATICS Proof If tan x tan, then tan x tan 0 or sin x cos cos x sin 0 cos x cos which gives sin (x ) 0 (Wh?) Therefore x n, i.e., x n +, where n Z Example 0 Find the solution of sin x. Solution We have sin x Hence sin x 4 sin, which gives 4 sin sin + sin n 4 x n + ( 1), where n Z. Note 4 is one such value of x for which sin x. One ma take an other value of x for which sin x. The solutions obtained will be the same although these ma apparentl look different. Example 1 Solve cos x 1. Solution We have, Therefore Example Solve 1 cos x cos x n ±, where n Z. tan x cot x+. Solution We have, tan x cot x+ tan + x +

29 TRIGONOMETRIC FUNCTIONS 77 or Therefore 5 tanx tan x+ 6 5 x n + x+, where n Z 6 5 or x n +, where n Z. 6 Example Solve sin x sin4 x + sin 6x 0. Solution The equation can be written as sin 6x + sin x sin 4x 0 or sin 4x cosx sin 4x 0 i.e. sin 4x( cos x 1) 0 Therefore sin 4x 0 or i.e. Hence i.e. 1 cos x sin4x 0 or cos x cos 4x n or x n ±, where n Z n x or x n ±, where n Z. 4 6 Example 4 Solve cos x + sin x 0 Solution The equation can be written as ( sin x) 1 + sin x 0 or sin x sin x 0 or (sinx+ 1) (sinx ) 0 Hence 1 sin x or sin x But sin x is not possible (Wh?) Therefore sin x 1 7 sin 6.

30 78 MATHEMATICS Hence, the solution is given b n 7 x n + ( 1), where n Z. 6 EXERCISE.4 Find the principal and general solutions of the following equations: 1. tan x. sec x. cot x 4. cosec x Find the general solution for each of the following equations: 5. cos 4 x cos x 6. cos x + cos x cos x 0 7. sin x + cosx 0 8. sec x 1 tan x 9. sin x + sin x + sin 5x 0 Miscellaneous Examples Example 5 If sin x 1, cos, where x and both lie in second quadrant, 5 1 find the value of sin (x + ). Solution We know that sin (x + ) sin x cos + cos x sin... (1) Now cos x 1 sin x Therefore cos x ± 4 5. Since x lies in second quadrant, cos x is negative. Hence cos x 4 5 Now sin 1 cos i.e. sin ± Since lies in second quadrant, hence sin is positive. Therefore, sin 5 1. Substituting the values of sin x, sin, cos x and cos in (1), we get

31 TRIGONOMETRIC FUNCTIONS 79 sin( x+ ) Example 6 Prove that x 9x 5x cos x cos cos x cos sin 5x sin. Solution We have L.H.S. 1 x cos cos cos 9 x x cos x x x cos cos cos 9 x cos 9 x x x x x x x 15x x cos + cos cos cos 1 5x 15x cos cos 5x 15x 5x 15x 1 + sin sin 5x 5x sin5x sin sin5x sin R.H.S. Example 7 Find the value of tan 8. Solution Let x. Then x. 8 4 tan x Now tan x 1 tan x tan tan 8 or 4 1 tan 8 Let tan 8. Then 1 1

32 80 MATHEMATICS or Therefore ± 1± Since 8 lies in the first quadrant, tan 8 is positve. Hence tan 1 8. Example 8 If Solution Since Also tan x, < x <, find the value of sin x 4, cos x and tan x. < x <, cos x is negative. x < <. 4 Therefore, sin x is positive and cos x is negative. 9 5 Now sec x 1 + tan x Therefore cos x 16 5 or cos x 4 (Wh?) 5 Now sin x 1 cos x Therefore sin x 9 10 or sin x (Wh?) 10 Again cos x 1+ cos x Therefore cos x 1 10

33 TRIGONOMETRIC FUNCTIONS 81 or cos x 1 10 (Wh?) x Hence tan x sin 10. x 10 1 cos Example 9 Prove that cos x + cos x+ + cos x. Solution We have L.H.S. 1+ cos x+ 1+ cos x 1+ cos x cos x cos x cos x cos cos cos x x cos x cos x cos cos cos cos x x R.H.S. [ cos x cos x] Miscellaneous Exercise on Chapter Prove that: cos cos + cos + cos (sin x + sin x) sin x + (cos x cos x) cos x 0

34 8 MATHEMATICS. (cos x + cos ) + (sin x sin ) 4 cos x + 4. x (cos x cos ) + (sin x sin ) 4 sin 5. sin x + sin x + sin 5x + sin 7x 4 cos x cos x sin 4x 6. (sin 7x + sin 5x) + (sin 9x + sin x) tan 6x (cos 7x + cos 5x) + (cos 9x + cosx) 7. sin x + sin x sin x 4sin x cos x cos x Find sin x, cos x and tan x in each of the following : 8. tan x 4, x in quadrant II 9. cos x 1, x in quadrant III 10. sin x 4 1, x in quadrant II Summar If in a circle of radius r, an arc of length l subtends and angle of θ radians, then l r θ Radian measure 180 Degree measure 180 Degree measure Radian measure cos x + sin x tan x sec x 1 + cot x cosec x cos (n + x) cos x sin (n + x) sin x sin ( x) sin x cos ( x) cos x

35 TRIGONOMETRIC FUNCTIONS 8 cos (x + ) cos x cos sin x sin cos (x ) cos x cos + sin x sin cos ( x ) sin x sin ( x ) cos x sin (x + ) sin x cos + cos x sin sin (x ) sin x cos cos x sin cos + x sin x sin + x cos x cos ( x) cos x sin ( x) sin x cos ( + x) cos x sin ( + x) sin x cos ( x) cos x sin ( x) sin x If none of the angles x, and (x ± ) is an odd multiple of, then tan (x + ) tan x + tan 1 tan xtan tan x tan tan (x ) 1+ tan xtan If none of the angles x, and (x ± ) is a multiple of, then cot xcot 1 cot (x + ) cot + cot x cot (x ) cot x cot + 1 cot cot x cos x 1 tan x cos x sin x cos x 1 1 sin x 1 + tan x

36 84 MATHEMATICS sin x sinx cos x tanx 1 + tan x tan x tanx 1 tan x sin x sinx 4sin x cos x 4cos x cosx tan x tanx tan x 1 tan x (i) cos x + cos cos x + x cos x + x (ii) cos x cos sin sin x + x (iii) sin x + sin sin cos x + x (iv) sin x sin cos sin (i) cos x cos cos ( x + ) + cos ( x ) (ii) sin x sin cos (x + ) cos (x ) (iii) sin x cos sin (x + ) + sin (x ) (iv) cos x sin sin (x + ) sin (x ). sin x 0 gives x n, where n Z. cos x 0 gives x (n + 1), where n Z. sin x sin implies x n + ( 1)n, where n Z. cos x cos, implies x n ±, where n Z. tan x tan implies x n +, where n Z.

37 TRIGONOMETRIC FUNCTIONS 85 Historical Note The stud of trigonometr was first started in India. The ancient Indian Mathematicians, Arabhatta (476A.D.), Brahmagupta (598 A.D.), Bhaskara I (600 A.D.) and Bhaskara II (1114 A.D.) got important results. All this knowledge first went from India to middle-east and from there to Europe. The Greeks had also started the stud of trigonometr but their approach was so clums that when the Indian approach became known, it was immediatel adopted throughout the world. In India, the predecessor of the modern trigonometric functions, known as the sine of an angle, and the introduction of the sine function represents the main contribution of the siddhantas (Sanskrit astronomical works) to the histor of mathematics. Bhaskara I (about 600 A.D.) gave formulae to find the values of sine functions for angles more than 90. A sixteenth centur Malaalam work Yuktibhasa (period) contains a proof for the expansion of sin (A + B). Exact expressin for sines or cosines of 18, 6, 54, 7, etc., are given b Bhaskara II. The smbols sin 1 x, cos 1 x, etc., for arc sin x, arc cos x, etc., were suggested b the astronomer Sir John F.W. Hersehel (181 A.D.) The names of Thales (about 600 B.C.) is invariabl associated with height and distance problems. He is credited with the determination of the height of a great pramid in Egpt b measuring shadows of the pramid and an auxiliar staff (or gnomon) of known height, and comparing the ratios: H S h tan (sun s altitude) s Thales is also said to have calculated the distance of a ship at sea through the proportionalit of sides of similar triangles. Problems on height and distance using the similarit propert are also found in ancient Indian works.