Chapter A mathematician knows how to solve a problem, he can not solve it. – MILNE • 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 Vertex Fig 3.1 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 Fig 3.2measuring 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.  1 th 3.2.1 Degree measure If a rotation from the initial side to terminal side is  of 360  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 360°,180°, 270°, 420°, – 30°, – 420° are shown in Fig 3.3. Fig 3.3 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 11angles whose measures are 1 radian, –1 radian, 1 radian and –1 radian.22(ii)(i) (iii) Fig 3.4 (i) to (iv) We know that the circumference of a circle of radius 1 unit is 2π. Thus, one complete revolution of the initial side subtends an angle of 2π 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 1lradian, an arc of length l will subtend an angle whose measure is radian. Thus, if inra circle of radius r, an arc of length l subtends an angle θ radian at the centre, we have l θ = or l = r θ. rcentre of the circle. Consider the line PAQ which is tangent to the circle at A. Let the point A represent the 1 radian = = 57° 16′ approximately.π πAlso 1° = radian = 0.01746 radian approximately.180The relation between degree measures and radian measure of some common angles are given in the following table: Degree 30° 45° 60° 90° 180° 270° 360° Radian ￀ 6 ￀ 4 ￀ 3 ￀ 2 π 3￀ 2 2π TRIGONOMETRIC FUNCTIONS 53 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 frequently omitted. Thus, π 180 and  π 45  are written with the understanding that π and π 44 are radian measures. Thus, we can say that πRadian measure = 180  Degree measure 180 Degree measure =  Radian measureπ Example 1 Convert 40° 20′ into radian measure. Solution We know that 180° = π radian. 1 π 121 121πHence 40° 20′ = 40 degree =  radian = radian.3 180 3540121πTherefore 40° 20′ = radian.540Example 2 Convert 6 radians into degree measure. Solution We know that π radian = 180°. 180 1080  7 Hence 6 radians =  6 degree = degreeπ 22 7 760  = 343 degree = 343° + minute [as 1° = 60′]1111 2 = 343° + 38′ + minute [as 1′ = 60″]11= 343° + 38′ + 10.9″ = 343°38′ 11″ approximately. Hence 6 radians = 343° 38′ 11″ approximately. Example 3 Find the radius of the circle in which a central angle of 60° intercepts an 22arc of length 37.4 cm (use π  ).7 54 MATHEMATICS 60ππSolution Here l = 37.4 cm and θ = 60° = radian = 180 3 lHence, by r = , we haveθ 37.4×3 37.4×3×7 r = = = 35.7 cmπ 22Example 4 The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14). Solution In 60 minutes, the minute hand of a watch completes one revolution. Therefore, 22in 40 minutes, the minute hand turns through of a revolution. Therefore, θ = × 360° 33 4π or radian. Hence, the required distance travelled is given by3 4πl = r θ = 1.5 cm = 2π cm = 2  3.14 cm = 6.28 cm. 3 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 π 13πθ = 65° = 65 = radian1180 36π 22πand θ2 = 110° = 110 = radian180 36 Let l be the length of each of the arc. Then l = r1θ1 = r2θ2, which gives 13π 22π r1 22  r1 =  r2 , i.e., = 36 36 r2 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° 2. Find the degree measures corresponding to the following radian measures (Use π  22 ).7 11 5π 7π(i) (ii) – 4 (iii) (iv)16 36 3. A wheel makes 360 revolutions in one minute. Through how many 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 by an arc of length 22 cm (Use π  22).7 5. In a circle of diameter 40 cm, the length of a chord is 20 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) 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 ΔOMP 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 2π radian at the πcentre of the circle, ∠AOB = Fig 3.6 2, 3∠AOC = π and ∠AOD =ππ . All angles which are integral multiples of are called2 2quadrantal 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, ππ= 0 sin = 1cos 2 2 cosπ = − 1 sinπ = 0 3π cos 2 3= 0 sinπ= –12 cos 2π = 1 sin 2π = 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 2π, the values of sine and cosine functions do not change. Thus, sin (2nπ + x) = sin x, n ∈ Z , cos (2nπ + x) = cos x, n ∈ Z Further, sin x = 0, if x = 0, ± π, ± 2π , ± 3π, ..., i.e., when x is an integral multiple of π π3π5πand cos x = 0, if x = ± , ± , ± , ... i.e., cos x vanishes when x is an odd2 22 πmultiple of . Thus2 sin x = 0 implies x = nπ, where n is any integer π 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: 1 cosec x = , x ≠ nπ, where n is any integer.sin x 1π , x ≠ (2n + 1) , where n is any integer.sec x = 2cos x sin xπ , x ≠ (2n +1), where n is any integer.tan x = 2cos x cos x cot x =, x ≠ n π, where n is any integer.sin x 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 π 4 ￀ 3 ￀ 2 π 3￀ 2 2π 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 not defined 0 not defined 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 ∠AOP = x. If ∠AOQ = – 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 – 1 ≤ b ≤ 1, we have – 1 ≤ cos x ≤ 1 and –1 ≤ sin x ≤ 1 for all x. We have learnt in πprevious classes that in the first quadrant (0 < x < ) a and b are both positive, in the2 πsecond quadrant ( < x <π) a is negative and b is positive, in the third quadrant23π3) a and b are both negative and in the fourth quadrant (π(π < x < < x < 2π) a is2 2positive and b is negative. Therefore, sin x is positive for 0 < x < π, and negative for ππ3ππ < x < 2π. Similarly, cos x is positive for 0 < x < , negative for < x < and also2 223positive forπ < x < 2π. Likewise, we can find the signs of other trigonometric2 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. TRIGONOMETRIC FUNCTIONS 59 1 Since cosec x = sin x , the domain of y = cosec x is the set { x : x ∈ R and x ≠ n π, n ∈ Z} and range is the set {y : y ∈ R, y ≥ 1 or y ≤ – 1}. Similarly, the domain 2, n ∈ Z} and range is the setπof y = sec x is the set {x : x ∈ R and x ≠ (2n + 1) {y : y ∈ R, y ≤ – 1or y ≥ 1}. The domain of y = tan x is the set {x : x ∈ R and 2, n ∈ Z} and range is the set of all real numbers. The domain ofπ x ≠ (2n + 1) y = cot x is the set {x : x ∈ R and x ≠ n π, n ∈ Z} and the range is the set of all real numbers. πWe further observe that in the first quadrant, as x increases from 0 to , sin x2 πincreases from 0 to 1, as x increases from to π, sin x decreases from 1 to 0. In the23third quadrant, as x increases from π to π, sin x decreases from 0 to –1and finally, in 2 3the fourth quadrant, sin x increases from –1 to 0 as x increases fromπto 2π.2Similarly, we can discuss the behaviour of other trigonometric functions. In fact, we have the following table: 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–∞ Remark In the above table, the statement tan x increases from 0 to ∞ (infinity) for ππ0 < x < simply means that tan x increases as x increases for 0 < x < and22πassumes 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 3π cosec x decreases for x ∈ ( , 2π) and assumes arbitrarily large negative values as2 x approaches to 2π. The symbols ∞ and – ∞ simply specify certain types of behaviour of functions and variables. Fig 3.8 shall see in the next section that tan (π + x) = tan x. Hence, values of tan x will repeat after an interval of π. Since cot x is reciprocal of tan x, its values will also repeat after an interval of π. Using this knowledge and behaviour of trigonometic functions, we can sketch the graph of these functions. The graph of these functions are given above: 3Example 6 If cos x = – , x lies in the third quadrant, find the values of other five5 trigonometric functions. 35Solution Since cos x =  , we have sec x =  53 Now sin2 x + cos2 x = 1, i.e., sin2 x = 1 – cos2 x 9 16 or sin2 x = 1 – = 2525 4Hence sin x = ± 5 Since x lies in third quadrant, sin x is negative. Therefore 4sin x = – 5 which also gives 5 cosec x = – 4 Further, we have sin x 4 cos x 3 tan x= = and cot x= = 4. cos x3sin x5Example 7 If cot x= – 12, xlies in second quadrant, find the values of other five trigonometric functions. 5 12Solution Since cot x= – , we have tan x = –12 5 144 169 Now sec2 x= 1 + tan2 x= 1 + = 2525 13Hence sec x=± 5 Since xlies in second quadrant, sec xwill be negative. Therefore 13 sec x=– 5, which also gives 5 cos x 13 Further, we have 12 512sin x= tan xcos x= (– ) (– ) =5 13 13 1 13and cosec x== sin x 12 . 31πExample 8 Find the value of sin .3 Solution We know that values of sin xrepeats after an interval of 2π. Therefore 31π ππsin = sin (10π + ) = sin = .3332 TRIGONOMETRIC FUNCTIONS 63 Example 9 Find the value of cos (–1710°). Solution We know that values of cos x repeats after an interval of 2π 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. 1 1. cos x = – 2, x lies in third quadrant. 3 2. sin x = 5, x lies in second quadrant. 3. cot x = , x lies in third quadrant.34 13 4. sec x = 5, x lies in fourth quadrant. 5. tan x = – 12, x lies in second quadrant. 5 Find the values of the trigonometric functions in Exercises 6 to 10. 6. sin 765° 7. cosec (– 1410°) 19π 11π8. tan 9. sin (– )33 15π10. cot (– )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 1. sin (– x) = – sin x 2. cos (– x) = cos x We shall now prove some more results: 3. 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, P, P, P and P will have the coordinates P(cos x, sin x),12341P2 [cos (x + y), sin (x + y)], P3 [cos (– y), sin (– y)] and P4 (1, 0) (Fig 3.14). 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 P1P32 = [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, P2P42 = [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) Since P1P3 = P2P4, we have P1P32 = P2P42. 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 π5. cos( – x ) = sin x2 πIf we replace x by and y by x in Identity (4), we get2x π π) = cos 26.sin () = cos– x x2 Using the Identity 5, we have ππcos (cos x + sin sin x = sin x.2 2ππ 2 x  π x ) = cossin ( = cos x.22 7. sin (x + y) = sin x cos y + cos x sin y We know thatπ2 π 2   (x ) (  x)sin (x + y) = cos y y= cos   ππ 2 x ) cos y + sin ( x) sin y= cos (2 = sin x cos y + cos x sin y 8. 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. 9. By taking suitable values of x and y in the identities 3, 4, 7 and 8, we get the following results: π πcos (+ x) = – sin x sin (+ x) = cos x2 2cos (π– x) = – cos x sin (π– x) = sin x cos (π+ x) = – cos x sin (π+ x) = – sin x cos (2π– x) = cos x sin (2π– x) = – sin x Similar results for tan x, cot x, sec xand cosec xcan be obtained from the results of sin xand cos x. π10. If none of the angles x, y and (x + y) is an odd multiple of , then2 tan x + tan ytan (x + y) = 1– tan x tan y πSince none of the x, yand (x+ y) is an odd multiple of , it follows that cos x,2 cos yand cos (x+ y) are non-zero. Nowsin(x y) sin xcos y cos xsin ytan (x+ y)= = .cos(xy) cos xcos y sin sinxy Dividing numerator and denominator by cos xcos y, we havesin xcos y cos xsin y cos xcos ycos xcos ytan (x+ y)= cos xcos y sin xsin y cos xcos ycos xcos y tan x tan y = 1– tan xtan y tan x – tan y11. tan ( x – y)= 1+ tan x tan y If we replace yby – yin Identity 10, we get tan (x– y) = tan [x+ (– y)] tan x tan(y) tan x tan y= = 1tan xtan (y)  xtan y 1tan 12. If none of the angles x, y and (x + y) is a multiple of π, thencot x cot y –1 cot ( x + y) = cot y +cot x Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and sin (x + y) are non-zero. Now, cos ( x  y) cos x cos y – sin x sin ycot ( x + y)=  sin ( x  y) sin x cos y  cos x sin y Dividing numerator and denominator by sin x sin y, we have cot x cot y –1cot (x + y)= cot y  cot x cot x cot y +1 13. cot (x – y)= if none of angles x, y and x–y is a multiple of π cot y – cotx If we replace y by –y in identity 12, we get the result 1– tan 2 x14. cos 2x = cos2x – sin2 x = 2 cos2 x – 1 = 1 – 2 sin2 x = 1+ tan 2 x 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. cos 2 x sin 2 xWe have cos 2x =cos2 x – sin 2 x = 2 2cos x sin x Dividing numerator and denominator by cos2 x, we get 1– tan 2 x πcos 2x = 2, xn π , where n is an integer1+ tan x 2 2tan x π15. sin 2x = 2 sinx cos x = 2 xn π , where n is an integer1+ tan x 2 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. 2sin xcos x Again sin 2x = 2 2cos x sin x Dividing each term by cos2 x, we get 2tan x sin 2x= 21tan x 2tan x π xn16. tan 2x =2 if 2 π ,where n is an integer1– tanx 2 We know that tan x tan ytan (x+ y) = 1tan xtan –y 2 tan xReplacing yby x, we get tan 2 x 1tan 2 x 17. sin 3x = 3 sin x – 4 sin3 x We have, sin 3x = sin (2x+ x) = sin 2xcos x+ cos 2xsin x = 2 sin xcos xcos 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 2xcos x– sin 2xsin x = (2cos2 x– 1) cos x– 2sin xcos xsin x = (2cos2 x– 1) cos x– 2cos x(1 – cos2 x) = 2cos3 x– cos x– 2cos x+ 2 cos3 x = 4cos3 x– 3cos x. 3 tan x – tan 3 x π xnπ19. tan 3 x  2 if 3 2 ,where n is an integer1– 3tanx We have tan 3x=tan (2x+ x) 2tan x 2  tan x tan 2 x tan x 1tan x– =  1 –tan 2 xtan x 2tan x.tan x1 – 1tan 2 x– 2tan x tan x–tan 3 x 3 tan x– tan 3 x22 21 –tan x– 2tan x 1 –3tan x x + yx – y20. (i) cos x + cos y = 2cos cos 22 x + yx – y(ii) cos x – cos y = – 2sin sin 22 x + yx – y(iii) sin x + sin y = 2sin 2cos 2 x + yx – y(iv) sin x – sin y = 2cos 2sin 2 We know that cos (x+ y) = cos xcos y– sin xsin y ... (1) and cos (x– y) = cos xcos y+ sin xsin y ... (2) Adding and subtracting (1) and (2), we get cos (x+ y) + cos(x– y) = 2 cos xcos y ... (3) and cos (x+ y) – cos (x– y) = –2 sin xsin y ... (4) Further sin (x+ y) = sin xcos y+ cos xsin y ... (5) and sin (x– y) = sin xcos y– cos xsin y ... (6) Adding and subtracting (5) and (6), we get sin (x+ y) + sin (x– y) = 2 sin xcos y ... (7) sin (x+ y) – sin (x– y) = 2cos xsin y ... (8) Let x+ y= θ and x– y= φ. Therefore θ θ  x and y  2  2  Substituting the values of xand yin (3), (4), (7) and (8), we get θ θ  cos cos θ + cos φ = 2 cos    2  2  θ θ – cos θ – cos φ = – 2 sin sin   2   θ θ sin θ + sin φ = 2 sin cos  2  2  70 MATHEMATICS θ θ sin sin θ – sin φ = 2 cos    2  2  Since θ and φ can take any real values, we can replace θ by x and φ by y. Thus, we get x yx yx yx ycos x + cos y = 2 cos cos ; cos x – cos y = – 2 sin sin ,22 22 x yx yx yx ysin x + sin y = 2 sin cos ; sin x – sin y = 2 cos sin .22 22 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 63 64 Solution We have  5L.H.S. = 3sin sec 4sin cot 63 64 1  =3 × × 2 – 4 sin  × 1 = 3 – 4 sin2 6 6 1=3 – 4 × = 1 = R.H.S.2Example 11 Find the value of sin 15°. Solution We have sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° 1 311 3 –1 =   .22 2222 13Example 12 Find the value of tan .12 TRIGONOMETRIC FUNCTIONS 71 Solution We have 13    tan = tan   = tan tan 12  12 12 46   1 tan tan 1 4 6 331 2  3==  1 311tan tan  146 3 Example 13 Prove that sin ( x y) tan x tan y .sin ( x y) tan x tan y Solution We havesin ( x y) sin x cos y cos xsin yL.H.S. sin ( x y) sin xcos y cos xsin y Dividing the numerator and denominator by cos x cos y, we get sin ( x y) tan x tan y .sin ( x y) tan x tan y Example 14 Show thattan 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) tan 2 x tan x or tan3 x  1–tan 2 x tan 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 x cos x  2 cos x4 4  Solution Using the Identity 20(i), we have  cos x cos xL.H.S.  4 4     x x– ( x)x 44 442cos cos  22    1 = 2 cos cos x = 2× cos x = 2 cos x = R.H.S.42cos 7x cos 5x cot xExample 16 Prove that sin 7x – sin 5x Solution Using the Identities 20 (i) and 20 (iv), we get 7x 5x 7x 5x2cos cos cos x22L.H.S. = = cot x = R.H.S.7x 5x 7x 5x sin x2cos sin22 sin5 x 2sin3x sin xExample 17 Prove that tan x cos5x cos x Solution We have sin5 x 2sin3x sin x sin 5 x sin x 2sin 3xL.H.S.  cos5x cos x cos5x cos x 2sin 3 x cos2x 2sin3x sin 3 x (cos 2x 1)– –2sin3 sin2xx xsin3 sin2x 1cos 2x 2sin2 x= tan x = R.H.S.sin 2x 2sin x cos x EXERCISE 3.3 Prove that: π 1  7 2 31. sin2 + cos2 – tan2 – 2. 2sin2 + cosec2 cos  6 342 6 632 2  5 2  23 2  2 3. cot cosec 3tan 6 4. 2sin 2cos 2sec 10 666 443 5. Find the value of: (i) sin 75° (ii) tan 15° Prove the following:  6. cos x cos y sinx sin y sin ( x y) 4 4 4 4  tan  π x  24 1tan  cos ( x) cos ( x)7.    x  8.  cot 2 x π  1tan  x   tan x sin ( x) cos x 4  2  3π  3π  cos x cos (2πx)cot x cot (2πx) 19.   2   2  10. sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x 3 3 cos x cos x  2 sin x11.   4  4  12. sin2 6x – sin2 4x = sin 2x sin 10x 13. cos2 2x – cos2 6x = sin 4x sin 8x 14. sin2 x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x 15. cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x) cos 9x cos 5x sin 2x sin 5x sin 3x 16.  17. tan 4xsin17x sin 3x cos 10x cos 5x cos 3x sin x sin yx  y sin x sin 3x 18.  tan 19.  tan 2x cos x cos y 2 cos x cos 3x sin x sin 3x cos 4x cos 3x cos 2x 20. 22  2 sin x 21.  cot 3x sin x cos x sin 4x sin 3x sin 2x 22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1 4tan x (1 tan 2 x)23. tan 4 x  24 24. cos 4x = 1 – 8sin2 x cos2 x 16 tan x tan  x 25. 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 2π and the values of tan x repeat after an interval of π. The solutions of a trigonometric equation for which 0 ≤ x < 2π 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: 3Example 18 Find the principal solutions of the equation sin x  . 2 π 32π  π π 3 Solution We know that, sin  and sin sin π sin  .32 3  3  32 Therefore, principal solutions are x π and 2π .33 1 Example 19 Find the principal solutions of the equation tan x  π 1  π π 1Solution We know that, tan  . Thus, tan π – =–tan =– 63  6  63  π  π 1and tan 2π   tan   6  63 5π 11π 1Thus tan tan  66 5π 11πTherefore, principal solutions are and .66 We will now find the general solutions of trigonometric equations. We have already TRIGONOMETRIC FUNCTIONS 75 seen that: sin x =0 gives x = nπ, where n ∈ Z π cos x =0 gives x = (2n + 1) , where n ∈ Z.2 We shall now prove the following results: Theorem 1 For any real numbers x and y, sin x = sin y implies x = nπ + (–1)ny, where n ∈ Z Proof If sin x = sin y, then x  yx  ysin x – sin y = 0 or 2cos sin = 022 x  yx  ywhich gives cos = 0 or sin = 022x  y  x  yTherefore = (2n + 1) or = nπ, where n ∈ Z222 i.e. x = (2n + 1) π – y or x = 2nπ + y, where n∈Z Hence x = (2n + 1)π + (–1)2n + 1 y or x = 2nπ +(–1)2ny, where n ∈ Z. Combining these two results, we get x = nπ + (–1)ny, where n ∈ Z. Theorem 2 For any real numbers x and y, cos x = cos y, implies x = 2nπ± y, where n ∈ Z Proof If cos x = cos y, then x  yx  y cos x – cos y = 0 i.e., –2 sin sin = 022x  yx  yThus sin = 0 or sin = 022x  yx  yTherefore = nπ or = nπ, where n ∈ Z22i.e. x =2nπ – y or x = 2nπ + y, where n ∈ Z Hence x =2nπ± y, where n ∈ Z πTheorem 3 Prove that if x and y are not odd mulitple of 2 , then tan x = tan y implies x = nπ + y, where n ∈ Z Proof If tan x= tan y, then tan x– tan y= 0 sin xcos ycos xsin y0or cos xcos y which gives sin (x– y) = 0 (Why?) Therefore x– y= nπ, i.e., x= nπ + y, where n∈ Z 3Example 20 Find the solution of sin x= – . 2 3 π  π 4π Solution We have sin x= – = sin sin π sin 23  3  3 4πHence sin x= sin , which gives3 n4π  xnπ (1) , where n∈ Z.3 4π 3is one such value of xfor which sin x . One may take any32 3other value of xfor which sin x . The solutions obtained will be the same 2 although these may apparently look different. 1 Example 21 Solve cos x .2 1 πSolution We have, cos xcos 23 πTherefore x2nπ  , where n∈ Z.3  πExample 22 Solve tan 2 xcot x.  3   πππtan Solution We have, tan 2 xcot x =  x  3 23   5πtan2 x tan xor   6  5π xn xTherefore 2  π , where n∈Z6 5π xnor  π , where n∈Z.6 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 2sin 4xcos 2x sin 4x 0 i.e. sin 42x( cos 2x 1)  0 1Therefore sin 4x= 0 or cos2 x 2 πi.e. sin4 x 0 or cos 2 x cos 3 πHence 4xnπ or 2x 2nπ , where n∈Z  3 nπ πi.e. x or  π ∈Z.xn , where n4 6 Example 24 Solve 2 cos2 x+ 3 sin x= 0 Solution The equation can be written as  sin x 3sin x 021 2 or 2sin2 x 3sin x 2  0 or (2sin x1) (sin x 2) 0 1Hence sin x=  or sin x= 22 But sin x= 2 is not possible (Why?) 17πTherefore sin x=  = sin .26 Hence, the solution is given by n7π  xnπ (1) , where n∈ Z.6 EXERCISE 3.4 Find the principal and general solutions of the following equations: 2. sec x= 21. tan x4. cosec x= – 2 Find the general solution for each of the following equations: 3. cot x5. cos 4 x= cos 2 x 6. cos 3x+ cos x– cos 2x= 0 7. sin 2x+ cos x= 0 8. sec2 2x= 1– tan 2x 9. sin x+ sin 3x+ sin 5x= 0 Miscellaneous Examples 3 12 Example 25 If sin x= , cos y= , where xand yboth lie in second quadrant,5 13 find the value of sin (x+ y). Solution We know that sin (x+ y) = sin xcos y+ cos xsin y ... (1) 9 16 Now cos2 x= 1 – sin2 x= 1 – = 2525 4 Therefore cos x= .5 Since xlies in second quadrant, cos xis negative. 4 Hence cos x= 5 144 25 Now sin2y= 1 – cos2y= 1 – 169 169 5 i.e. sin y= .135 Since ylies in second quadrant, hence sin yis positive. Therefore, sin y= . Substituting13the values of sin x, sin y, cos xand cos yin (1), we get 3  12  4  5 36 20 56   sin( xy)   =  .5 135 13 6565 65 Example 26 Prove that x 9x 5xcos 2 xcos cos 3 xcos sin 5 xsin .22 2 Solution We have 1  x 9x 2cos 2 xcos 2cos cos 3 xL.H.S. = 2  22  1  x x9x 9x  cos 2xcos 2xcos 3xcos 3x=   2  2  2 2 2  1  5x 3x 15x 3x 1  5x 15x cos cos cos cos cos cos = = 2 22 2 2 2 22 5x15 x5x15 x1   22 222sin sin   = 22 2       5x 5x= sin 5xsin  sin 5xsin = R.H.S.2 2 πExample 27 Find the value of tan 8. ππSolution Let x . Then 2x .84 2 tan x Now tan 2x 1 tan2 x π2tan π 8tan  or π421tan  8 π 2yLet y= tan . Then 1 =81 y2 or y2 + 2y – 1 = 0  222Therefore y =  21 2 ππSince lies in the first quadrant, y = tan is positve. Hence88πtan  2 1.8 33π xx x xExample 28 If tan x =, π << , find the value of sin , cos and tan .4 2 222 Solution Since π x3π , cos x is negative.2 Also π 22 x 3π 4 . 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 sin2 2 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 x 1 or cos =  (Why?)210 xsin x 3  102 Hence tan =   = – 3. x2cos 10  1  2  π 2  π 3.Example 29 Prove that cos2 x + cos2 x  cos x   3  3  2 Solution We have 2π 2π1cos 2x  1 cos 2x   1cos 2 x 3 3L.H.S. =   . 22 2 1  2π 2π3  cos 2 x  cos 2x  cos 2x =  2  3  3  1  2π3  cos 2 x  2cos 2 x cos= 2  3  1  π3  cos 2 x  2cos 2 x cos π=  2  3  1  π3  cos 2 x  2cos 2 x cos = 2  3  13 = 3  cos 2 x  cos 2 x = R.H.S.22 Miscellaneous Exercise on Chapter 3 Prove that: π 9π 3π 5π 1. 2cos cos  cos  cos 0 1313 13 13 2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 MATHEMATICS x  y3. (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2 2 x  y4. (cos x – cos y)2 + (sin x – sin y)2 = 4 sin2 5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x 2 (sin 7x  sin 5x)  (sin 9x  sin 3x)6. tan 6x (cos 7x  cos 5x)  (cos 9x  cos 3x ) x 3x 7. sin 3x + sin 2x – sin x = 4sin x cos cos22 xx x Find sin , cos and tan in each of the following :22241 8. tan x =  , x in quadrant II 9. cos x =  , x in quadrant III33110. sin x = , x in quadrant II4 Summary If in a circle of radius r, an arc of length l subtends an angle of θradians, then l = r θ Radian measure = π  Degree measureDegree measure = 180  Radian measure180 π cos2 x + sin2 x = 1 1 + tan2 x = sec2 x 1 + cot2 x = cosec2 x cos (2nπ+ x) = cos x sin (2nπ+ x) = sin x sin (– x) = – sin x cos (– x) = cos x cos (x + y) = cos x cos y – sin x sin y cos (x – y) = cos x cos y + sin x sin y cos ( π  x ) = sin xsin ( π  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 2 2  π cos  2 cos (πcos (π+ x  = – sin x – x) = – cos x + x) = – cos x cos (2π – x) = cos x  π + xsin  = cos x 2 sin (π – x) = sin x sin (π + x) = – sin x sin (2π – x) = – sin x π If none of the angles x, y and (x  y) is an odd multiple of , then2 tan x  tan ytan (x + y) = 1 tan x tan y tan x  tan ytan (x – y) = 1 tan x tan y If none of the angles x, y and (x  y) is a multiple of π, then cotx cot y 1 cot (x + y) = cot y  cot x cot x cot y  1 cot (x – y) = cot y  cot x 1tan 2 x– cos 2x = cos2 x – sin2 x = 2cos2 x – 1 = 1 – 2 sin2 x  21tan x MATHEMATICS 2tan x sin 2x = 2 sin x cos x  21tan x 2tan x tan 2x = 21tan x sin 3x = 3sin x – 4sin3 x cos 3x = 4cos3 x – 3cos x 3tan x  tan 3 x tan 3x = 13tan 2 x • (i) (ii) (iii) (iv) • (i) (ii) (iii) (iv) sin x x  yx  ycos x + cos y = 2cos cos 22 x  yx  ycos x – cos y = – 2sin sin 22 x  yx  ysin x + sin y = 2 sin cos 22 x  yx  ysin x – sin y = 2cos sin 22 2cos x cos y = cos (x + y) + cos ( x – y) – 2sin x sin y = cos (x + y) – cos (x – y) 2sin x cos y = sin (x + y) + sin (x – y) 2 cos x sin y = sin (x + y) – sin (x – y). = 0 gives x = nπ, where n ∈ Z. π cos x = 0 gives x = (2n + 1) , where n ∈ Z.sin x = sin y implies x = nπ + (– 1)ny, where n ∈ Z. cos x = cos y, implies x = 2nπ± y, where n ∈ Z. tan x = tan y implies x = nπ + y, where n ∈ Z. 2 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  h = tan (sun’s altitude)S s 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. — • —

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