INVERSE TRIGONOMETRIC FUNCTIONS

1 Chapter2 INVERSE INVERSE functionInverse of a function f exists, if the function is one-one and onto, , TRIGONOMETRIC FUNCTIONS are many-one over their domains, we restrict theirdomains and co-domains in order to make them one-one and onto and then findtheir INVERSE . The domains and ranges (principal value branches) of inversetrigonometric FUNCTIONS are given below:FunctionsDomainRange (Principal valuebranches)y = sin 1x[ 1,1] ,22 y = cos 1x[ 1,1][0, ]y = cosec 1xR ( 1,1) , {0}22 y = sec 1xR ( 1,1)[0, ] 2 y = tan 1xR ,22 y = cot 1xR(0, )Notes: (i)The symbol sin 1x should not be confused with (sinx) sin 1x is anangle, the value of whose sine is x, similarly for other TRIGONOMETRIC FUNCTIONS .(ii)The smallest numerical value, either positive or negative, of is called theprincipal value of the TRIGONOMETRIC FUNCTIONS 19(iii)Whenever no branch of an INVERSE TRIGONOMETRIC function is mentioned, we meanthe principal value branch.

2 The value of the INVERSE trigonometic function whichlies in the range of principal branch is its principal of an INVERSE TRIGONOMETRIC functionThe graph of an INVERSE TRIGONOMETRIC function can be obtained from the graph oforiginal function by interchanging x-axis and y-axis, , if (a, b) is a point on the graphof TRIGONOMETRIC function, then (b, a) becomes the corresponding point on the graph ofits INVERSE TRIGONOMETRIC can be shown that the graph of an INVERSE function can be obtained from thecorresponding graph of original function as a mirror image ( , reflection) along theline y = of INVERSE TRIGONOMETRIC 1 (sin x) = x : ,22x cos 1(cos x) = x :[0,]x tan 1(tan x) = x : ,22x cot 1(cot x) = x :()0, x sec 1(sec x) = x : [0, ] 2x cosec 1(cosec x) = x : , {0}22x (sin 1 x) = x :x [ 1,1]cos (cos 1 x) = x :x [ 1,1]tan (tan 1 x) = x :x Rcot (cot 1 x) = x :x Rsec (sec 1 x) = x :x R ( 1,1)cosec (cosec 1 x) = x :x R ( 1,1)3.

3 1 11sincosecxx = :x R ( 1,1) 1 11cossecxx = :x R ( 1,1)20/04/201820 MATHEMATICS 1 11tancotxx = :x > 0= + cot 1x :x < 1 ( x) = sin 1x :x [ 1,1]cos 1 ( x) = cos 1x :x [ 1,1]tan 1 ( x) = tan 1x :x Rcot 1 ( x) = cot 1x :x Rsec 1 ( x) = sec 1x :x R ( 1,1)cosec 1 ( x) = cosec 1x :x R ( 1,1) 1x + cos 1x = 2 :x [ 1,1]tan 1x + cot 1x = 2 :x Rsec 1x + cosec 1x = 2 :x R [ 1,1] 1x + tan 1y = tan 11 xyxy + :xy < 1tan 1x tan 1y = tan 1; 11xyxyxy > + 1x = sin 1221xx+ : 1 x 12tan 1x = cos 1221 1xx+ : x 02tan 1x = tan 1221 xx : 1 < x < Solved ExamplesShort Answer ( )Example 1 Find the principal value of cos 1x, for x = TRIGONOMETRIC FUNCTIONS 21 Solution If cos 132 = , then cos = we are considering principal branch, [0, ].

4 Also, since 32 > 0, being inthe first quadrant, hence cos 132 = 2 Evaluate tan 1 sin2 .Solution tan 1 sin2 = tan 1 sin2 = tan 1( 1) = 4 .Example 3 Find the value of cos 113 cos6 .Solution cos 113 cos6 = cos 1 cos(2)6 + = 1 coscos6 = 6 .Example 4 Find the value of tan 1 9 tan8 .Solution tan 1 9 tan8 = tan 1 tan 8 + = 1tantan8 = 8 Example 5 Evaluate tan (tan 1( 4)).Solution Since tan (tan 1x) = x, x R, tan (tan 1( 4) = 6 Evaluate: tan 13 sec 1 ( 2) .20/04/201822 MATHEMATICSS olution tan 13 sec 1 ( 2) = tan 13 [ sec 12] = 112cos32333 += += .Example 7 Evaluate: 1 13sincossin2 .Solution 1 1 13 sincossinsincos23 = = 11 sin26 =.)

5 Example 8 Prove that tan(cot 1x) = cot (tan 1x). State with reason whether theequality is valid for all values of Let cot 1x = . Then cot = xor, tan = 2x 1 tan 2x=So 1 1 1 tan(cot)tan cot cotcotcot(tan)22xxx === = The equality is valid for all values of x since tan 1x and cot 1x are true for x 9 Find the value of sec 1tan2y .Solution Let 1tan= 2y, where ,22 . So, tan = 2y,which gives 24sec =2y+.Therefore, 2 14sectan=sec =22yy+ .Example 10 Find value of tan (cos 1x) and hence evaluate tan 18cos17 .Solution Let cos 1x = , then cos = x, where [0, ]20/04/2018 INVERSE TRIGONOMETRIC FUNCTIONS 23 Therefore, tan(cos 1x) = 221 cos 1 tan =.cos xx=Hence 2 181 17815tancos =817817 = .Example 11 Find the value of 1 5sin2cot12 Solution Let cot 1 512 = y . Then cot y = 512 .Now 1 5sin2cot12 = sin 2y= 2siny cosy = 12 521313 sincecot0,so, 2yy < 120169=Example 12 Evaluate 1 114cossinsec43 + Solution 1 114cossinsec43 + = 1 113cossincos44 + = 1 1 1 11313cossincoscos sinsinsincos4444 = 2231131 1 4444 = 31517315 7 444416=.

6 20/04/201824 MATHEMATICSLong Answer ( )Example 13 Prove that 2sin 135 tan 1 1731 = 4 Solution Let sin 1 35= , then sin = 35, where ,22 Thus tan = 34, which gives = tan , 2sin 135 tan 1 1731= 2 tan 1 1731 = 2 tan 134 tan 1 1731= 1 tan9311 16 = tan 1 12417tan731 = + = 4 Example 14 Prove thatcot 17 + cot 18 + cot 118 = cot 13 Solution We have cot 17 + cot 18 + cot 118= tan 117 + tan 118 + tan 1118 (since cot 1 x = tan 11x, if x > 0)= 1 111178tantan1118178 + + (since x . y = < 1)20/04/2018 INVERSE TRIGONOMETRIC FUNCTIONS 25= 1 131tantan1118+ = 1311118tan3111118 + (since xy < 1)= 165tan195= 11tan3 = cot 1 3 Example 15 Which is greater, tan 1 or tan 1 1?Solution From Fig. , we note that tan x is an increasing function in the interval,22 , since 1 > 4 tan 1 > tan 4.

7 This givestan 1 > 1 tan 1 > 1 > 4 tan 1 > 1 > tan 1 (1).Example 16 Find the value of 1 12sin2tancos(tan3)3 + .Solution Let tan 1 23 = x and tan 1 3 = y so that tan x = 23 and tan y = , 1 12sin2tancos(tan3)3 + = sin (2x) + cos y= 222tan11tan1tanxxy+++ = () +++= 1213713226+=. /2pp/4p/2 XtanxYO20/04/201826 MATHEMATICSE xample 17 Solve for x 1 111tantan,012xxxx => + Solution From given equation, we have 1 112tantan1xxx = + 1 1 12tan1tantanxx = 123tan4x = 1tan6x = 13x=Example 18 Find the values of x which satisfy the equationsin 1 x + sin 1 (1 x) = cos 1 From the given equation, we havesin (sin 1 x + sin 1 (1 x)) = sin (cos 1x) sin (sin 1 x) cos (sin 1 (1 x)) + cos (sin 1 x) sin (sin 1 (1 x) ) = sin (cos 1 x) 2221 (1 )(1)11xxxxx+ = 222 1(11)0xxxxx+ = ()222 10xxxx = x = 0or2x x2 = 1 x2 x = 0orx = 19 Solve the equation sin 16x + sin 1 63x = 2 Solution From the given equation, we have sin 1 6x = 1sin632x 20/04/2018 INVERSE TRIGONOMETRIC FUNCTIONS 27 sin (sin 1 6x) = sin 1sin632x 6x = cos (sin 1 63x) 6x = 21108x.

8 Squaring, we get36x2 = 1 108x2 144x2 = 1 x = 112 Note that x = 112 is the only root of the equation as x = 112 does not satisfy 20 Show that2 tan 1 = + Solution = 1 122since2tantan1xxx = = 1221tan22tan21tan2tan1tan21tan21tan2 + + = 2 + 20/04/201828 MATHEMATICS=2 12222tan1tan22tan1tan1tan2tan1tan2222 + ++ =222 12222tan1tan221tan1tan22tan1tan2tan221ta n1tan22 ++ + ++= 1sincostancossin + = type questionsChoose the correct answer from the given four options in each of the Examples 21 to 21 Which of the following corresponds to the principal value branch of tan 1?(A),22 (B),22 (C),22 {0}(D)(0, )Solution (A) is the correct 22 The principal value branch of sec 1 is(A){},022 (B)[]0,2 (C)(0, )(D),22 20/04/2018 INVERSE TRIGONOMETRIC FUNCTIONS 29 Solution (B) is the correct 23 One branch of cos 1 other than the principal value branch corresponds to(A)3,22 (B)[]3,22 (C)(0, )(D)[2 , 3 ]Solution (D) is the correct 24 The value of 143sincos5 is(A)35 (B)75 (C)10 (D) 10 Solution (D) is the correct answer.

9 1 14033sincossincos855 + = + = 1 133sincossinsin525 = = 1sinsin1010 = .Example 25 The principal value of the expression cos 1 [cos ( 680 )] is(A)29 (B)29 (C)349 (D)9 Solution (A) is the correct answer. cos 1 (cos (680 )) = cos 1 [cos (720 40 )]= cos 1 [cos ( 40 )] = cos 1 [cos (40 )] = 40 = 29 .Example 26 The value of cot (sin 1x) is(A)21xx+(B)21xx+20/04/201830 MATHEMATICS(C)1x(D)21xx .Solution (D) is the correct answer. Let sin 1 x = , then sin = x cosec = 1x cosec2 = 21x 1 + cot2 = 21x cot = 21xx .Example 27 If tan 1x = 10 for some x R, then the value of cot 1x is(A)5 (B)25 (C)35 (D)45 Solution (B) is the correct answer. We know tan 1x + cot 1x = 2 . Thereforecot 1x = 2 10 cot 1x = 2 10 = 25 .Example 28 The domain of sin 1 2x is(A)[0, 1](B)[ 1, 1](C)11,22 (D)[ 2, 2]Solution (C) is the correct answer.

10 Let sin 12x = so that 2x = sin .Now 1 sin 1, , 1 2x 1 which gives 1122x .Example 29 The principal value of sin 1 32 is20/04/2018 INVERSE TRIGONOMETRIC FUNCTIONS 31(A)23 (B)3 (C)43 (D)53 .Solution (B) is the correct answer. 1 1 13sinsin sin sinsin 2333 === .Example 30 The greatest and least values of (sin 1x)2 + (cos 1x)2 are respectively(A)225and48 (B)and22 (C)22and44 (D)2and04 .Solution (A) is the correct answer. We have(sin 1x)2 + (cos 1x)2 = (sin 1x + cos 1x)2 2 sin 1x cos 1 x= 2 1 12sinsin42xx = ()22 1 1sin2sin4xx += ()22 1 12sinsin28xx + = 22 12sin416x + .Thus, the least value is and the Greatest value is 2222416 + , 254 .Example 31 Let = sin 1 (sin ( 600 ), then value of is20/04/201832 MATHEMATICS(A)3 (B)2 (C)23 (D)23 .Solution (A) is the correct answer. 1 110sinsin600sinsin1803 = = 12sinsin43 = 12sinsin3 = 1 1sinsinsinsin333 ==.)