3.1 Introduction The word ‘trigonometry’ is derived from the Greek words ‘trigon’ and ‘metron’ and it means ‘measuring the sides of a triangle’. The subject was originally developed to solve geometric problems involving triangles. It was studied by sea captains for navigation, surveyor to map out the new lands, by engineers and others. Currently, trigonometry is used in many areas such as the science of seismology, designing electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analysing a musical tone and in many other areas. In earlier classes, we have studied the trigonometric ratios of acute angles as the ratio of the sides of a right 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 study their properties. 3.2 Angles Angle is a measure of rotation of a given ray about its initial point. The original ray is Chapter 3 TRIGONOMETRIC FUNCTIONS Arya Bhatt (476-550) Fig 3.1 Vertex 2022-23 50 MATHEMATICS called the initial side and the final position of the ray 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 3.1). 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 suggests a unit, viz. one complete revolution from the position of the initial side as indicated in Fig 3.2. This is often convenient for large angles. For example, we can say that a rapidly spinning wheel is making an angle of say 15 revolution per second. We shall describe two other units of measurement of an angle which are most commonly used, viz. degree measure and radian measure. 3.2.1 Degree measure If a rotation from the initial side to terminal side is th 1 360 æ ö ç ÷ è ø 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 1². Thus, 1° = 60¢, 1¢ = 60² Some of the angles whose measures are 360°,180°, 270°, 420°, – 30°, – 420° are shown in Fig 3.3. Fig 3.2 Fig 3.3 2022-23 TRIGONOMETRIC FUNCTIONS 51 3.2.2 Radian measure There is another unit for measurement of an angle, called the radian measure. Angle subtended at the centre by 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 3.4(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 2 radian and –1 1 2 radian. (i) (ii) (iii) Fig 3.4 (i) to (iv) (iv) We know that the circumference of a circle of radius 1 unit is 2p. Thus, one complete revolution of the initial side subtends an angle of 2p radian. More generally, 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 r radian. Thus, if in a circle of radius r, an arc of length l subtends an angle q radian at the centre, we have q = l r or l = r q. 2022-23 52 MATHEMATICS 3.2.3 Relation between radian and real numbers Consider the unit circle with centre O. Let A be any 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 3.5). If we rope the line AP in the anticlockwise direction along the circle, and AQ in the clockwise direction, then every real number will correspond to a radian measure and conversely. Thus, radian measures and real numbers can be considered as one and the same. 3.2.4 Relation between degree and radian Since a circle subtends at the centre an angle whose radian measure is 2p and its degree measure is 360°, it follows that 2p radian = 360° or p 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 p as 22 7 , we have 1 radian = 180
° = 57° 16¢ approximately. Also 1° =
180 radian = 0.01746 radian approximately. The relation between degree measures and radian measure of some common angles are given in the following table: A O 1 P 1 2 −1 −2 Q 0 Fig 3.5 Degree 30° 45° 60° 90° 180° 270° 360° Radian
6
4
3
2
3
2 2
2022-23 TRIGONOMETRIC FUNCTIONS 53 Notational Convention Since angles are measured either in degrees or in radians, we adopt the convention that whenever we write angle q°, we mean the angle whose degree measure is q and whenever we write angle b, we mean the angle whose radian measure is b. Note that when an angle is expressed in radians, the word ‘radian’ is frequently omitted. Thus,

180 and 45 4 = ° = ° are written with the understanding that p and
4 are radian measures. Thus, we can say that Radian measure =
180 × Degree measure Degree measure = 180
× Radian measure Example 1 Convert 40° 20¢ into radian measure. Solution We know that 180° = p radian. Hence 40° 20¢ = 40 1 3 degree =
180 × 121 3 radian = 121
540 radian. Therefore 40° 20¢ = 121
540 radian. Example 2 Convert 6 radians into degree measure. Solution We know that p radian = 180°. Hence 6 radians = 180
× 6 degree = 1080 7 22 × degree = 343 7 11 degree = 343° + 7 60 11 × minute [as 1° = 60¢] = 343° + 38¢ + 2 11 minute [as 1¢ = 60²] = 343° + 38¢ + 10.9² = 343°38¢ 11² approximately. Hence 6 radians = 343° 38¢ 11² approximately. Example 3 Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use 22
7 = ). 2022-23 54 MATHEMATICS Solution Here l = 37.4 cm and q = 60° = 60

radian = 180 3 Hence, by r =  l , we have r = 37.4×3 37.4×3×7

22 = 35.7 cm Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use p = 3.14). Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through 2 3 of a revolution. Therefore, 2  = × 360° 3 or 4
3 radian. Hence, the required distance travelled is given by l = r q = 1.5 × 4
3 cm = 2p cm = 2 × 3.14 cm = 6.28 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 r1 and r2 be the radii of the two circles. Given that q1 = 65° =
65 180 × = 13
36 radian and q2 = 110° =
110 180 × = 22
36 radian Let l be the length of each of the arc. Then l = r 1q1 = r 2q2, which gives 13
36 × r1 = 22
36 × r2 , i.e., 1 2 r r

22 13 Hence r1 : r2 = 22 : 13. EXERCISE 3.1

1. Find the radian measures corresponding to the following degree measures: (i) 25° (ii) – 47°30¢ (iii) 240° (iv) 520° 2022-23 TRIGONOMETRIC FUNCTIONS 55 2 . Find the degree measures corresponding to the following radian measures (Use 22
   7 = ). (i) 11 16 (ii) – 4 (iii) 5
   3 (iv) 7
   6
2. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
3. Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (Use 22
   7 = ).
4. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
5. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
6. 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) 21 cm 3.3 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 any angle in terms of radian measure and study them as trigonometric functions. Consider a unit circle with centre at origin of the coordinate axes. Let P (a, b) be any point on the circle with angle AOP = x radian, i.e., length of arc AP = x (Fig 3.6). We define cos x = a and sin x = b Since DOMP is a right triangle, we have OM2 + MP2 = OP2 or a2 + b2 = 1 Thus, for every point on the unit circle, we have a2 + b2 = 1 or cos2 x + sin2 x = 1 Since one complete revolution subtends an angle of 2p radian at the centre of the circle, ÐAOB =
   2 , Fig 3.6 2022-23 56 MATHEMATICS ÐAOC = p and ÐAOD = 3
   2 . All angles which are integral multiples of
   2 are called quadrantal angles. The coordinates of the points A, B, C and D are, respectively, (1, 0), (0, 1), (–1, 0) and (0, –1). Therefore, for quadrantal angles, we have cos 0° = 1 sin 0° = 0, cos
   2 = 0 sin
   2 = 1 cosp = − 1 sinp = 0 cos 3
   2 = 0 sin 3
   2 = –1 cos 2p = 1 sin 2p = 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) by any integral multiple of 2p, the values of sine and cosine functions do not change. Thus, sin (2np + x) = sin x, n Î Z , cos (2np + x) = cos x , n Î Z Further, sin x = 0, if x = 0, ± p, ± 2p , ± 3p, …, i.e., when x is an integral multiple of p and cos x = 0, if x = ±
   2 , ± 3
   2 , ± 5
   2 , … i.e., cos x vanishes when x is an odd multiple of
   2 . Thus sin x = 0 implies x = np, where n is any integer cos x = 0 implies x = (2n + 1)
   2 , where n is any integer. We now define other trigonometric functions in terms of sine and cosine functions: cosec x = 1 sin x , x ¹ np, where n is any integer. sec x = 1 cos x , x ¹ (2n + 1)
   2 , where n is any integer. tan x = sin cos x x , x ¹ (2n +1)
   2 , where n is any integer. cot x = cos sin x x , x ¹ n p, where n is any integer. 2022-23 TRIGONOMETRIC FUNCTIONS 57 not defined not defined We have shown that for all real x, sin2 x + cos2 x = 1 It follows that 1 + tan2 x = sec2 x (why?) 1 + cot2 x = cosec2 x (why?) In earlier classes, we have discussed the values of trigonometric ratios for 0°, 30°, 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°
   6
   4
   3
   2
   3
   2 2
   sin 0 1 2 1 2 3 2 1 0 – 1 0 cos 1 3 2 1 2 1 2 0 – 1 0 1 tan 0 1 3 1 3 0 0 The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cos x and tan x, respectively. 3.3.1 Sign of trigonometric functions Let P (a, b) be a point on the unit circle with centre at the origin such that ÐAOP = x. If ÐAOQ = – x, then the coordinates of the point Q will be (a, –b) (Fig 3.7). Therefore cos (– x) = cos x and sin (– x) = – sin x Since for every point P (a, b) on the unit circle, – 1 £ a £ 1 and Fig 3.7 2022-23 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 <
   2 ) a and b are both positive, in the second quadrant (
   2 < x <p) a is negative and b is positive, in the third quadrant (p < x < 3
   2 ) a and b are both negative and in the fourth quadrant ( 3
   2 < x < 2p) a is positive and b is negative. Therefore, sin x is positive for 0 < x < p, and negative for p < x < 2p. Similarly, cos x is positive for 0 < x <
   2 , negative for
   2 < x < 3
   2 and also positive for 3
   2 < x < 2p. 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 + – + – 3.3.2 Domain and range of trigonometric functions From the definition of sine and cosine functions, we observe that they 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 y = sin x and y = cos x is the set of all real numbers and range is the interval [–1, 1], i.e., – 1 £ y £ 1. 2022-23 TRIGONOMETRIC FUNCTIONS 59 Since cosec x = 1 sin x , the domain of y = cosec x is the set { x : x Î R and x ¹ n p, n Î Z} and range is the set {y : y Î R, y ³ 1 or y £ – 1}. Similarly, the domain of y = sec x is the set {x : x Î R and x ¹ (2n + 1)
   2 , n Î Z} and range is the set {y : y Î R, y £ – 1or y ³ 1}. The domain of y = tan x is the set {x : x Î R and x ¹ (2n + 1)
   2 , n Î Z} and range is the set of all real numbers. The domain of y = cot x is the set {x : x Î R and x ¹ n p, n Î 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
   2 , sin x increases from 0 to 1, as x increases from
   2 to p, sin x decreases from 1 to 0. In the third quadrant, as x increases from p to 3
   2 , sin x decreases from 0 to –1and finally, in the fourth quadrant, sin x increases from –1 to 0 as x increases from 3
   2 to 2p. Similarly, 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 ¥ (infinity) for 0 < x <
   2 simply means that tan x increases as x increases for 0 < x <
   2 and 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–¥ 2022-23 60 MATHEMATICS Fig 3.10 Fig 3.11 Fig 3.8 Fig 3.9 assumes arbitraily large positive values as x approaches to
   2 . Similarly, to say that cosec x decreases from –1 to – ¥ (minus infinity) in the fourth quadrant means that cosec x decreases for x Î ( 3
   2 , 2p) and assumes arbitrarily large negative values as x approaches to 2p. The symbols ¥ and – ¥ simply specify certain types of behaviour of functions and variables. We have already seen that values of sin x and cos x repeats after an interval of 2p. Hence, values of cosec x and sec x will also repeat after an interval of 2p. We 2022-23 TRIGONOMETRIC FUNCTIONS 61 shall see in the next section that tan (p + x) = tan x. Hence, values of tan x will repeat after an interval of p. Since cot x is reciprocal of tan x, its values will also repeat after an interval of p. 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 = – 3 5 , x lies in the third quadrant, find the values of other five trigonometric functions. Solution Since cos x = 3 5 − , we have sec x = 5 3 − Now sin2 x + cos2 x = 1, i.e., sin2 x = 1 – cos2 x or sin2 x = 1 – 9 25 = 16 25 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 Fig 3.12 Fig 3.13 2022-23 62 MATHEMATICS Further, we have tan x = sin cos x x = 4 3 and cot x = cos sin x x = 3 4 . Example 7 If cot x = – 5 12 , x lies in second quadrant, find the values of other five trigonometric functions. Solution Since cot x = – 5 12 , we have tan x = – 12 5 Now sec2 x = 1 + tan2 x = 1 + 144 25 = 169 25 Hence sec x = ± 13 5 Since x lies in second quadrant, sec x will be negative. Therefore sec x = – 13 5 , which also gives 5 cos 13 x = − Further, we have sin x = tan x cos x = (– 12 5 ) ×(– 5 13 ) = 12 13 and cosec x = 1 sin x = 13 12 . Example 8 Find the value of sin 31
   3 . Solution We know that values of sin x repeats after an interval of 2p. Therefore sin 31
   3 = sin (10p +
   3 ) = sin
   3 = 3 2 . 2022-23 TRIGONOMETRIC FUNCTIONS 63 Example 9 Find the value of cos (–1710°). Solution We know that values of cos x repeats after an interval of 2p or 360°. Therefore, cos (–1710°) = cos (–1710° + 5 × 360°) = cos (–1710° + 1800°) = cos 90° = 0. EXERCISE 3.2 Find the values of other five trigonometric functions in Exercises 1 to 5.
7. cos x = – 1 2 , x lies in third quadrant.
8. sin x = 3 5 , x lies in second quadrant.
9. cot x = 4 3 , x lies in third quadrant.
10. sec x = 13 5 , x lies in fourth quadrant.
11. tan x = – 5 12 , x lies in second quadrant. Find the values of the trigonometric functions in Exercises 6 to 10.
12. sin 765° 7. cosec (– 1410°)
13. tan 19
    3
14. sin (– 11
    3 )
15. cot (– 15
    4 ) 3.4 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
16. sin (– x) = – sin x
17. cos (– x) = cos x We shall now prove some more results: 2022-23 64 MATHEMATICS
18. cos (x + y) = cos x cos y – sin x sin y Consider the unit circle with centre at the origin. Let x be the angle P4OP1and y be the angle P1OP2. Then (x + y) is the angle P4OP2. Also let (– y) be the angle P4OP3. Therefore, P1, P2, P3 and P4 will have the coordinates P1(cos x, sin x), P2 [cos (x + y), sin (x + y)], P3 [cos (– y), sin (– y)] and P4 (1, 0) (Fig 3.14). Consider the triangles P1OP3 and P2OP4. They are congruent (Why?). Therefore, P1P3 and P2P4 are equal. By using distance formula, we get P1P3 2 = [cos x – cos (– y)]2 + [sin x – sin(–y]2 = (cos x – cos y)2 + (sin x + sin y)2 = cos2 x + cos2 y – 2 cos x cos y + sin2 x + sin2 y + 2sin x sin y = 2 – 2 (cos x cos y – sin x sin y) (Why?) Also, P2P4 2 = [1 – cos (x + y)] 2 + [0 – sin (x + y)]2 = 1 – 2cos (x + y) + cos2 (x + y) + sin2 (x + y) = 2 – 2 cos (x + y) Fig 3.14 2022-23 TRIGONOMETRIC FUNCTIONS 65 Since P1P3 = P2P4, we have P1P3 2 = P2P4 2. Therefore, 2 –2 (cos x cos y – sin x sin y) = 2 – 2 cos (x + y). Hence cos (x + y) = cos x cos y – sin x sin y 4 . cos (x – y) = cos x cos y + sin x sin y Replacing y by – y in identity 3, we get cos (x + (– y)) = cos x cos (– y) – sin x sin (– y) or cos (x – y) = cos x cos y + sin x sin y
19. cos ( x
    – 2 ) = sin x If we replace x by
    2 and y by x in Identity (4), we get cos (
    2 − x ) = cos
    2 cos x + sin
    2 sin x = sin x.
20. sin ( x
    – 2 ) = cos x Using the Identity 5, we have sin (
    2 − x ) = cos
    
    2 2 x é æ öù ê − ç − ÷ú ë è øû = cos x.
21. sin (x + y) = sin x cos y + cos x sin y We know that sin (x + y) = cos
    ( ) 2 x y æ ö ç − + ÷ è ø = cos
    ( ) 2 x y æ ö ç − − ÷ è ø = cos (
    2 − x ) cos y + sin
    ( ) 2 − x sin y = sin x cos y + cos x sin y
22. sin (x – y) = sin x cos y – cos x sin y If we replace y by –y, in the Identity 7, we get the result.
23. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the following results: cos x
    ( + ) 2 = – sin x sin x
    ( + ) 2 = cos x cos (p – x) = – cos x sin (p – x) = sin x 2022-23 66 MATHEMATICS cos (p + x) = – cos x sin (p + x) = – sin x cos (2p – x) = cos x sin (2p – x) = – sin x Similar results for tan x, cot x, sec x and cosec x can be obtained from the results of sin x and cos x.
24. If none of the angles x, y and (x + y) is an odd multiple of
    2 , then tan (x + y) = x y x y tan + tan 1 – tan tan Since none of the x, y and (x + y) is an odd multiple of
    2 , it follows that cos x, cos y and cos (x + y) are non-zero. Now tan (x + y) = sin( ) cos( ) x y x y + + = sin cos cos sin cos cos sin sin x y x y x y x y + − . Dividing numerator and denominator by cos x cos y, we have tan (x + y) = x y x y x y x y x y x y x y x y cos cos sin sin cos cos cos cos cos cos cos sin cos cos sin cos − + = tan tan 1 – tan tan x y x y +
25. tan ( x – y) = x y x y tan – tan 1+ tan tan If we replace y by – y in Identity 10, we get tan (x – y) = tan [x + (– y)] = tan tan ( ) 1 tan tan ( ) x y x y

• − − − = tan tan 1 tan tan x y x y − +

12. If none of the angles x, y and (x + y) is a multiple of p, then cot ( x + y) = x y y x cot cot – 1 cot +cot 2022-23 TRIGONOMETRIC FUNCTIONS 67 Since, none of the x, y and (x + y) is multiple of p, we find that sin x sin y and sin (x + y) are non-zero. Now, cot ( x + y)= cos ( ) cos cos – sin sin sin ( ) sin cos cos sin x y x y x y x y x y x y + =

• + Dividing numerator and denominator by sin x sin y, we have cot (x + y) = cot cot –1 cot cot x y y + x

13. cot (x – y)= x y y x cot cot +1 cot – cot if none of angles x, y and x–y is a multiple of p If we replace y by –y in identity 12, we get the result
14. cos 2x = cos2x – sin2 x = 2 cos2 x – 1 = 1 – 2 sin2 x = x x 2 2 1 – tan 1 + tan We know that cos (x + y) = cos x cos y – sin x sin y Replacing y by x, we get cos 2x = cos2x – sin2 x = cos2 x – (1 – cos2 x) = 2 cos2x – 1 Again, cos 2x = cos2 x – sin2 x = 1 – sin2 x – sin2 x = 1 – 2 sin2 x. We have cos 2x = cos2 x – sin 2 x = 2 2 2 2 cos sin cos sin x x x x − + Dividing numerator and denominator by cos2 x, we get cos 2x = 2 2 1 – tan 1+ tan x x ,
    
    2 x ¹ n + , where n is an integer
15. sin 2x = 2 sinx cos x = x x 2 2tan 1 + tan
    
    2 x ¹ n + , where n is an integer We have sin (x + y) = sin x cos y + cos x sin y Replacing y by x, we get sin 2x = 2 sin x cos x. Again sin 2x = 2 2 2sin cos cos sin x x x+ x 2022-23 68 MATHEMATICS Dividing each term by cos2 x, we get sin 2x = 2 2tan 1 tan x

• x

16. tan 2x = x x 2 2tan 1 – tan if
    2
    2 x ¹ n + , where n is an integer We know that tan (x + y) = tan tan 1 tan tan x y – x y + Replacing y by x , we get 2 2 tan tan 2 1 tan x x x = −
17. sin 3x = 3 sin x – 4 sin3 x We have, sin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x = 2 sin x cos x cos x + (1 – 2sin2 x) sin x = 2 sin x (1 – sin2 x) + sin x – 2 sin3 x = 2 sin x – 2 sin3 x + sin x – 2 sin3 x = 3 sin x – 4 sin3 x
18. cos 3x= 4 cos3 x – 3 cos x We have, cos 3x = cos (2x +x) = cos 2x cos x – sin 2x sin x = (2cos2 x – 1) cos x – 2sin x cos x sin x = (2cos2 x – 1) cos x – 2cos x (1 – cos2 x) = 2cos3 x – cos x – 2cos x + 2 cos3 x = 4cos3 x – 3cos x.
19. = x x x x 3 2 3 tan – tan tan3 1– 3tan if
    3
    2 x ¹ n + , where n is an integer We have tan 3x =tan (2x + x) = tan 2 tan 1 tan 2 tan x x – x x

• 2 2 2tan tan 1 tan 2tan tan 1 1 tan x x – x x . x – – x + = 2022-23 TRIGONOMETRIC FUNCTIONS 69 3 3 2 2 2 2tan tan tan 3 tan tan 1 tan 2tan 1 3tan x x – x x – x – x – x – x + = =

20. (i) cos x + cos y = x + y x – y 2cos cos 2 2 (ii) cos x – cos y = – x + y x – y 2sin sin 2 2 (iii) sin x + sin y = x + y x – y 2sin cos 2 2 (iv) sin x – sin y = x + y x – y 2cos sin 2 2 We know that cos (x + y) = cos x cos y – sin x sin y … (1) and cos (x – y) = cos x cos y + sin x sin y … (2) Adding and subtracting (1) and (2), we get cos (x + y) + cos(x – y) = 2 cos x cos y … (3) and cos (x + y) – cos (x – y) = – 2 sin x sin y … (4) Further sin (x + y) = sin x cos y + cos x sin y … (5) and sin (x – y) = sin x cos y – cos x sin y … (6) Adding and subtracting (5) and (6), we get sin (x + y) + sin (x – y) = 2 sin x cos y … (7) sin (x + y) – sin (x – y) = 2cos x sin y … (8) Let x + y = q and x – y = f. Therefore   and 2 2 x y æ +f ö æ −f ö =ç ÷ =ç ÷ è ø è ø Substituting the values of x and y in (3), (4), (7) and (8), we get cos q + cos f = 2 cos   cos 2 2 æ +f ö æ −f ö ç ÷ ç ÷ è ø è ø cos q – cos f = – 2 sin   sin 2 æ + f ö æ – f ö ç ÷ ç ÷ è ø è 2 ø sin q + sin f = 2 sin   cos 2 2 æ +f ö æ −f ö ç ÷ ç ÷ è ø è ø 2022-23 70 MATHEMATICS sin q – sin f = 2 cos   sin 2 2 æ +f ö æ −f ö ç ÷ ç ÷ è ø è ø Since q and f can take any real values, we can replace q by x and f by y. Thus, we get cos x + cos y = 2 cos cos 2 2 x + y x− y ; cos x – cos y = – 2 sin sin 2 2 x + y x− y , sin x + sin y = 2 sin cos 2 2 x + y x− y ; sin x – sin y = 2 cos sin 2 2 x + y x − y . Remark As a part of identities given in 20, we can prove the following results:
21. (i) 2 cos x cos y = cos (x + y) + cos (x – y) (ii) –2 sin x sin y = cos (x + y) – cos (x – y) (iii) 2 sin x cos y = sin (x + y) + sin (x – y) (iv) 2 cos x sin y = sin (x + y) – sin (x – y). Example 10 Prove that 5 3sin sec 4sin cot 1 6 3 6 4 p p p p − = Solution We have L.H.S. = 5 3sin sec 4sin cot 6 3 6 4 p p p p − = 3 × 1 2 × 2 – 4 sin 6 æ p ö çp− ÷ è ø × 1 = 3 – 4 sin 6 p = 3 – 4 × 1 2 = 1 = R.H.S. Example 11 Find the value of sin 15°. Solution We have sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° = 1 3 1 1 3 1 2 2 2 2 2 2 – × − × = . Example 12 Find the value of tan 13 12 p . 2022-23 TRIGONOMETRIC FUNCTIONS 71 Solution We have tan 13 12 p = tan 12 æ p ö çp + ÷ è ø = tan tan 12 4 6 p æ p p ö = ç − ÷ è ø = tan tan 4 6 1 tan tan 4 6 p p − p p + = 1 1 3 3 1 2 3 1 3 1 1 3 − − = = −

• + Example 13 Prove that sin ( ) tan tan sin ( ) tan tan x y x y x y x y
• + = − − . Solution We have L.H.S. sin ( ) sin cos cos sin sin ( ) sin cos cos sin x y x y x y x y x y x y
• + = = − − Dividing the numerator and denominator by cos x cos y, we get sin ( ) tan tan sin ( ) tan tan x y x y x y x y
• + = − − . Example 14 Show that tan 3 x tan 2 x tan x = tan 3x – tan 2 x – tan x Solution We know that 3x = 2x + x Therefore, tan 3x = tan (2x + x) or tan 2 tan tan3 1– tan 2 tan x x x x x + = or tan 3x – tan 3x tan 2x tan x = tan 2x + tan x or tan 3x – tan 2x – tan x = tan 3x tan 2x tan x or tan 3x tan 2x tan x = tan 3x – tan 2x – tan x. Example 15 Prove that cos cos 2 cos 4 4 x x x æ p ö æ p ö ç + ÷+ ç − ÷= è ø è ø Solution Using the Identity 20(i), we have 2022-23 72 MATHEMATICS L.H.S. cos cos 4 4 x x æ p ö æ p ö = ç + ÷ + ç − ÷ è ø è ø ( ) 2cos 4 4 cos 4 4 2 2 x x x – x æ p p ö æ p p ö ç + + − ÷ ç + − ÷ = ç ÷ ç ÷ çç ÷÷ çç ÷÷ è ø è ø = 2 cos 4 p cos x = 2 × 1 2 cos x = 2 cos x = R.H.S. Example 16 Prove that cos 7 cos 5 cot sin 7 – sin 5 x x x x x + = Solution Using the Identities 20 (i) and 20 (iv), we get L.H.S. = 7 5 7 5 2cos cos 2 2 7 5 7 5 2cos sin 2 2 x x x x x x x x
• −
• − = cos sin cot x x = x = R.H.S. Example 17 Prove that sin 5 2sin 3 sin tan cos5 cos x x x x x x − + = = − Solution We have L.H.S. sin 5 2sin3 sin cos5 cos x x x x x − + = − sin 5 sin 2sin 3 cos5 cos x x x x x
• − = − 2sin3 cos 2 2sin3 – 2sin3 sin 2 x x x x x − = sin 3 (cos2 1) sin 3 sin 2 x x – x x − = 2 1 cos 2 2sin sin 2 2sin cos x x x x x − = = = tan x = R.H.S. 2022-23 TRIGONOMETRIC FUNCTIONS 73 EXERCISE 3.3 Prove that:

1. sin2
   6

• cos2 3 p – tan2 1 – 4 2 p = 2. 2sin2 6 p
• cosec2 7 2 3 cos 6 3 2 p p = 3. 2 5 2 cot cosec 3tan 6 6 6 6 p p p
• • = 4. 2 3 2 2 2sin 2cos 2sec 10 4 4 3 p p p
• • =

5. Find the value of: (i) sin 75° (ii) tan 15° Prove the following:
6. cos cos sin sin sin ( ) 4 4 4 4 x y x y x y æ p ö æ p ö æ p ö æ p ö ç − ÷ ç − ÷ − ç − ÷ ç − ÷ = + è ø è ø è ø è ø 7. 2
   tan 4 1 tan
   1 tan tan 4 x x x x æ ö ç + ÷ è ø æ + ö = ç ÷ æ ö è − ø ç − ÷ è ø 8. cos ( ) cos ( ) 2 cot sin ( ) cos 2 x x x x x p + − = æ p ö p − ç + ÷ è ø 9. 3
   3
   cos cos (2
   ) cot cot (2
   ) 1 2 2 x x x x æ ö é æ ö ù ç + ÷ + ê ç − ÷ + + ú = è ø ë è ø û
7. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x 11. 3 3 cos cos 2 sin 4 4 x x x æ p ö æ p ö ç + ÷ − ç − ÷ = − è ø è ø
8. sin2 6x – sin2 4x = sin 2x sin 10x 13. cos2 2x – cos2 6x = sin 4x sin 8x
9. sin2 x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x
10. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x) 16. cos cos sin sin sin cos 9 5 17 3 2 10 x x x x x x − − = − 17. sin sin cos cos tan 5 3 5 3 4 x x x x x + + = 18. sin sin cos cos tan x y x y − x y + = − 2 19. sin sin cos cos tan x x x x x + + = 3 3 2 20. sin sin sin cos sin x x x x x − − = 3 2 2 2 21. cos cos cos sin sin sin cot 4 3 2 4 3 2 3 x x x x x x x

• +
• + = 2022-23 74 MATHEMATICS

22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1 23. 2 2 4 4tan (1 tan ) tan 4 1 6 tan tan x x x x x − = − +
23. cos 4x = 1 – 8sin2 x cos2 x
24. cos 6x = 32 cos6 x – 48cos4 x + 18 cos2 x – 1 3.5 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 already learnt that the values of sin x and cos x repeat after an interval of 2p and the values of tan x repeat after an interval of p. The solutions of a trigonometric equation for which 0 £ x < 2p 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 = 3 2 . Solution We know that,
    3 sin 3 2 = and 2
    
    
    3 sin sin
    sin 3 3 3 2 æ ö = ç − ÷ = = è ø . Therefore, principal solutions are
    3 x = and 2
    3 . Example 19 Find the principal solutions of the equation tan x = − 1 3 . Solution We know that,
    1 tan 6 3 = . Thus,
    
    1 tan
    – = – tan = – 6 6 3 æ ö ç ÷ è ø and
    
    1 tan 2
    tan 6 6 3 æ ö ç − ÷ = − = − è ø Thus 5
    11
    1 tan tan 6 6 3 = = − . Therefore, principal solutions are 5
    6 and 11
    6 . We will now find the general solutions of trigonometric equations. We have already 2022-23 TRIGONOMETRIC FUNCTIONS 75 seen that: sin x =0 gives x = np, where n Î Z cos x =0 gives x = (2n + 1)
    2 , where n Î Z. We shall now prove the following results: Theorem 1 For any real numbers x and y, sin x = sin y implies x = np + (–1)n y, where n Î Z Proof If sin x = sin y, then sin x – sin y = 0 or 2cos x + y x − y 2 2 sin = 0 which gives cos x + y 2 = 0 or sin x − y 2 = 0 Therefore x + y 2 = (2n + 1) p 2 or x − y 2 = np, where n Î Z i.e. x = (2n + 1) p – y or x = 2np + y, where nÎZ Hence x = (2n + 1)p + (–1)2n + 1 y or x = 2np +(–1)2n y, where n Î Z. Combining these two results, we get x = np + (–1)n y, where n Î Z. Theorem 2 For any real numbers x and y, cos x = cos y, implies x = 2np ± y, where n Î Z Proof If cos x = cos y, then cos x – cos y = 0 i.e., –2 sin x + y 2 sin x − y 2 = 0 Thus sin x + y 2 = 0 or sin x − y 2 = 0 Therefore x + y 2 = np or x − y 2 = np, where n Î Z i.e. x = 2np – y or x = 2np + y, where n Î Z Hence x = 2np ± y, where n Î Z Theorem 3 Prove that if x and y are not odd mulitple of
    2 , then tan x = tan y implies x = np + y, where n Î Z 2022-23 76 MATHEMATICS Proof If tan x = tan y, then tan x – tan y = 0 or sin cos cos sin cos cos x y x y x y − = 0 which gives sin (x – y) = 0 (Why?) Therefore x – y = np, i.e., x = np + y, where n Î Z Example 20 Find the solution of sin x = – 3 2 . Solution We have sin x = – 3 2 =
    
    4
    sin sin
    sin 3 3 3 æ ö − = ç + ÷ = è ø Hence sin x = 4
    sin 3 , which gives 4
    
    ( 1) 3 n x = n + − , where n Î Z. A Note 4
    3 is one such value of x for which 3 sin 2 x = − . One may take any other value of x for which sin x = − 3 2 . The solutions obtained will be the same although these may apparently look different. Example 21 Solve cos x = 1 2 . Solution We have, 1
    cos cos 2 3 x = = Therefore
    2
    3 x = n ± , where n Î Z. Example 22 Solve
    tan 2 cot 3 x x æ ö = − ç + ÷ è ø . Solution We have,
    tan 2 cot 3 x x æ ö = − ç + ÷ è ø =
    
    tan 2 3 x æ ö ç + + ÷ è ø 2022-23 TRIGONOMETRIC FUNCTIONS 77 or 5
    tan2 tan 6 x x æ ö = ç + ÷ è ø Therefore 5
    2
    6 x = n + x + , where nÎZ or 5
    
    6 x = n + , where nÎZ. Example 23 Solve sin 2x – sin 4x + sin 6x = 0. Solution The equation can be written as sin 6x + sin 2x − sin 4x = 0 or 2 sin 4x cos2x − sin 4x = 0 i.e. sin 4x(2 cos2x − 1) = 0 Therefore sin 4x = 0 or 1 cos 2 2 x = i.e.
    sin4 0 or cos 2 cos 3 x = x = Hence
    4
    or 2 2
    3 x =n x = n ± , where nÎZ i.e.
    
    or
    4 6 n x = x = n ± , where nÎZ. Example 24 Solve 2 cos2 x + 3 sin x = 0 Solution The equation can be written as 2(1 ) 3 0 2 − sin x + sin x = or 2 3 2 0 2 sin x − sin x − = or (2sinx +1) (sinx − 2) =0 Hence sin x = 1 2 − or sin x = 2 But sin x = 2 is not possible (Why?) Therefore sin x = 1 2 − = 7
    sin 6 . 2022-23 78 MATHEMATICS Hence, the solution is given by 7
    
    ( 1) 6 x = n + − n , where n Î Z. EXERCISE 3.4 Find the principal and general solutions of the following equations:
25. tan x = 3 2. sec x = 2
26. cot x = − 3 4. cosec x = – 2 Find the general solution for each of the following equations:
27. cos 4 x = cos 2 x 6. cos 3x + cos x – cos 2x = 0
28. sin 2x + cos x = 0 8. sec2 2x = 1– tan 2x
29. sin x + sin 3x + sin 5x = 0 Miscellaneous Examples Example 25 If sin x = 3 5 , cos y = − 12 13 , where x and y both lie in second quadrant, find the value of sin (x + y). Solution We know that sin (x + y) = sin x cos y + cos x sin y … (1) Now cos2 x = 1 – sin2 x = 1 – 9 25 = 16 25 Therefore cos x = ± 4 5 . Since x lies in second quadrant, cos x is negative. Hence cos x = − 4 5 Now sin2y = 1 – cos2y = 1 – 144 169 25 169 = i.e. sin y = ± 5 13 . Since y lies in second quadrant, hence sin y is positive. Therefore, sin y = 5 13 . Substituting the values of sin x, sin y, cos x and cos y in (1), we get 2022-23 TRIGONOMETRIC FUNCTIONS 79 sin(x + y) = × − æ è ç ö ø ÷

• − æ è ç ö ø ÷ × 3 5 12 13 4 5 5 13 = 36 20 56 65 65 65 − − = − . Example 26 Prove that 9 5 cos 2 cos cos 3 cos sin 5 sin 2 2 2 x x x x − x = x . Solution We have L.H.S. = 1 9 2cos 2 cos 2cos cos 3 2 2 2 x x x x é ù ê − ú ë û = 1 9 9 cos 2 cos 2 cos 3 cos 3 2 2 2 2 2 x x x x x x x x é æ ö æ ö æ ö æ öù ê ç + ÷ + ç − ÷ − ç + ÷ − ç − ÷ú ë è ø è ø è ø è øû = 1 2 5 2 3 2 15 2 3 2 cos cos cos cos x x x x
• − − é ë ê ù û ú = 1 2 5 2 15 2 cos cos x x − é ë ê ù û ú = 5 15 5 15 1 2 2 2 2 2sin sin 2 2 2 é ì x x ü ì x x üù ê ï + ï ï − ïú ê− í ý í ýú ê ï ï ï ïú êë î þ î þúû = − − æ è ç ö ø ÷ sin5 sin = sin sin 5 2 5 5 2 x x x x = R.H.S. Example 27 Find the value of tan
  8 . Solution Let
  8 x = . Then
  2 4 x = . Now tan tan tan 2 2 1 2 x x x = − or 2
  2tan
  8 tan 4
  1 tan 8 = − Let y = tan
  8 . Then 1 = 2 1 2 y y − 2022-23 80 MATHEMATICS or y2 + 2y – 1 = 0 Therefore y = − ± = − ± 2 2 2 2 1 2 Since
  8 lies in the first quadrant, y = tan
  8 is positve. Hence
  tan 2 1 8 = − . Example 28 If 3 3
  tan = ,
  < < 4 2 x x , find the value of sin x 2 , cos x 2 and tan x 2 . Solution Since 3
  
  2 < x < , cos x is negative. Also
  3
  2 2 4 x < < . Therefore, sin x 2 is positive and cos x 2 is negative. Now sec2 x = 1 + tan2 x = 1 9 16 25 16
• = Therefore cos2 x = 16 25 or cos x = 4 5 – (Why?) Now 2 2 2 sin x = 1 – cos x = 1 4 5 9 5
• = . Therefore sin2 x 2 = 9 10 or sin x 2 = 3 10 (Why?) Again 2cos2 x 2 = 1+ cos x = 1 4 5 1 5 − = Therefore cos2 x 2 = 1 10 2022-23 TRIGONOMETRIC FUNCTIONS 81 or cos x 2 = − 1 10 (Why?) Hence tan x 2 = sin cos x x 2 2 3 10 10 1 = × æ − è ç ö ø ÷ = – 3. Example 29 Prove that cos2 x + cos2 2
  
  3 cos 3 3 2 x x æ ö æ ö ç + ÷ + ç − ÷ = è ø è ø . Solution We have L.H.S. = 2
  2
  1 cos 2 1 cos 2 1 cos 2 3 3 2 2 2 x x x æ ö æ ö
• ç + ÷ + ç − ÷ + è ø è ø + + . = 1 2
  2
  3 cos 2 cos 2 cos 2 2 3 3 x x x é æ ö æ öù ê + + ç + ÷ + ç − ÷ú ë è ø è øû = 1 2
  3 cos 2 2cos 2 cos 2 3 x x é ù ê + + ú ë û = 1
  3 cos 2 2cos 2 cos
  2 3 x x é æ öù ê + + ç − ÷ú ë è øû = 1
  3 cos 2 2cos 2 cos 2 3 x x é ù ê + − ú ë û = 1 3 3 cos 2 cos 2 2 2
• x − x = = R.H.S. Miscellaneous Exercise on Chapter 3 Prove that:

1. 0 13 5
   cos 13 3
   cos 13 9
   cos 13
   2cos + + =
2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 2022-23 82 MATHEMATICS
3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2 2 x + y
4. (cos x – cos y)2 + (sin x – sin y)2 = 4 sin2 2 x − y
5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x 6. (sin7 sin5 ) (sin9 sin3 ) tan6 (cos7 cos5 ) (cos9 cos3 ) x x x x x x x x x

• • + =
• • +

7. sin 3x + sin 2x – sin x = 4sin x cos x 2 cos 3 2 x Find sin x 2 , cos x 2 and tan x 2 in each of the following :
8. tan x = − 4 3 , x in quadrant II 9. cos x = − 1 3 , x in quadrant III
9. sin x = 4 1 , x in quadrant II Summary ®If in a circle of radius r, an arc of length l subtends an angle of q radians, then l = r q ®Radian measure =
   180 × Degree measure ®Degree measure = 180
   × Radian measure ®cos2 x + sin2 x = 1 ®1 + tan2 x = sec2 x ®1 + cot2 x = cosec2 x ®cos (2np + x) = cos x ®sin (2np + x) = sin x ®sin (– x) = – sin x ®cos (– x) = cos x 2022-23 TRIGONOMETRIC FUNCTIONS 83 ®cos (x + y) = cos x cos y – sin x sin y ®cos (x – y) = cos x cos y + sin x sin y ®cos (
   2 − x ) = sin x ®sin (
   2 − x ) = cos x ®sin (x + y) = sin x cos y + cos x sin y ®sin (x – y) = sin x cos y – cos x sin y ®cos
   + 2 x
        = – sin x sin
   + 2 x
        = cos x cos (p – x) = – cos x sin (p – x) = sin x cos (p + x) = – cos x sin (p + x) = – sin x cos (2p – x) = cos x sin (2p – x) = – sin x ®If none of the angles x, y and (x ± y) is an odd multiple of
   2 , then tan (x + y) = tan tan tan tan x y x y + 1− ®tan (x – y) = tan tan tan tan x y x y − 1+ ®If none of the angles x, y and (x ± y) is a multiple of p, then cot (x + y) = cot cot 1 cot cot x y y x − + ®cot (x – y) = cot cot 1 cot cot x y y − x + ®cos 2x = cos2 x – sin2 x = 2cos2 x – 1 = 1 – 2 sin2 x 2 2 1 tan 1 tan – x x = + 2022-23 84 MATHEMATICS ®sin 2x = 2 sin x cos x 2 2 tan 1 tan x x = + ®tan 2x = 2 2tan 1 tan x − x ®sin 3x = 3sinx – 4sin3 x ®cos 3x = 4cos3x – 3cos x ®tan 3x = 3 2 3tan tan 1 3tan x x x − − ® (i) cos x + cos y = 2cos cos 2 2 x + y x − y (ii) cos x – cos y = – 2sin sin 2 2 x + y x− y (iii) sin x + sin y = 2 sin cos 2 2 x + y x − y (iv) sin x – sin y = 2cos sin 2 2 x + y x− y ® (i) 2cos x cos y = cos ( x + y) + cos ( x – y) (ii) – 2sin x sin y = cos (x + y) – cos (x – y) (iii) 2sin x cos y = sin (x + y) + sin (x – y) (iv) 2 cos x sin y = sin (x + y) – sin (x – y). ®sin x = 0 gives x = np, where n Î Z. ®cos x = 0 gives x = (2n + 1)
   2 , where n Î Z. ® sin x = sin y implies x = np + (– 1)n y, where n Î Z. ®cos x = cos y, implies x = 2np ± y, where n Î Z. ®tan x = tan y implies x = np + y, where n Î Z. 2022-23 TRIGONOMETRIC FUNCTIONS 85 Historical Note The study of trigonometry was first started in India. The ancient Indian Mathematicians, Aryabhatta (476), Brahmagupta (598), Bhaskara I (600) and Bhaskara II (1114) got important results. All this knowledge first went from India to middle-east and from there to Europe. The Greeks had also started the study of trigonometry but their approach was so clumsy that when the Indian approach became known, it was immediately 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 history of mathematics. Bhaskara I (about 600) gave formulae to find the values of sine functions for angles more than 90°. A sixteenth century Malayalam work Yuktibhasa (period) contains a proof for the expansion of sin (A + B). Exact expression for sines or cosines of 18°, 36°, 54°, 72°, etc., are given by Bhaskara II. The symbols sin–1 x, cos–1 x, etc., for arc sin x, arc cos x, etc., were suggested by the astronomer Sir John F.W. Hersehel (1813) The names of Thales (about 600 B.C.) is invariably associated with height and distance problems. He is credited with the determination of the height of a great pyramid in Egypt by measuring shadows of the pyramid and an auxiliary staff (or gnomon) of known height, and comparing the ratios: H S h s = = tan (sun’s altitude) Thales is also said to have calculated the distance of a ship at sea through the proportionality of sides of similar triangles. Problems on height and distance using the similarity property are also found in ancient Indian works.