Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Inverse Trigonometric Functions Chapter – 9 Inverse Trigonometric Functions Introduction: To study inverse trigonometric functions with full rigour is a big challenge and it requires a purely analytical approach. Almost all results are true in a given interval only. These interval are discovered when we attempt to prove these results methodically. Unfortunately the results given in most of the books are not true for all the values within the domain. In any such crisis we may say that LHS is one of the values of RHS or vice versa. Let us now define inverse trigonometric functions. Suppose a≤1 and sin x = a then as we know there are infinite values of x satisfying this equation and which are which are given by x = n + (-1)n where is the numerically smallest angle (For instance if a = ½ then = /6) whose since is a. Then we write sin -1 a = n + (-1)n Thus SIN-1 a is multiple valued function taking infinite values for any a satisfying a≤1. We also note that is the numerically smallest angle whose sine is a i.e. a 1. 1 1 5 , and if a , not 2 6 2 6 6 Principal Values Of Inverse Functions: Recall basic properties of sin x we know that its domain is (, ) and range is [-1, 1], but to attain does not have to run from - to . We easily notice from the graph of sin x that to attain all the values in [-1, 1], x has to run either from to or from 2 2 3 3 5 to or from to and so on the most natural choice of which is to since 2 2 2 2 2 2 (i) sin x monotonically increasing in this interval (ii) Normal acute angles are covered. This particular is denoted by sin-1 a. Thus remember whose sine is a where a≤1.” , 2 2 “sin-1 a is the angle in To invert the equality y = sin x we sin-1 y = x provided y≤1 and x , otherwise we can 2 2 we proceed as , 2 2 not write it as sin-1 y = x. To invert the equality y = sin x when x follows We write sin x = sin x’ where x , which is always possible since any value in [-1, 1] is 2 2 , through the function sin x. For instance suppose we want to 2 2 3 invert y = sin x when x , then 2 2 attained by an angle in Text Book of Trigonometry 44 Inverse Trigonometric Functions 2 x 3 2 3 x 2 2 2 x 2 Also sin x = sin( - x). whence it , 2 2 Therefore the given equality can be written as y = sin( - x) where x 2. follows that sin-1 y = - x. A similar discussion follows for other inverse trigonometric functions. Domain And Range Of Inverse Trigonometric Functions: Function domain rangeReason for choosing range sin-1 x [-1, 1] 2 , 2 cos-1 x [-1, 1] [0, ] tan 1 x , Same as sin-1 x cot 1 x , , 2 2 (0, ) Same as cos-1 x sec 1 x , 1, 0, 2 2 , same as cos-1 x Discussed Since cos x is MD in [0, ] and [0, ] covers acute angles. (/2 is not attained) cosec-1 x , 1, 2 , 0 0, 2 same as sin-1 x (0 is not attained) (i) (ii) (iii) (iv) (v) (vi) The following additional remarks on inverse functions will be useful. sin-1 x is positive or negative according as x is positive or negative. Same is true for tan-1 x. cos-1 x 0 for all x within its domain cot-1 x > 0 for all x sin-1 x, cos-1 x are continuous and differentiable in (-1, 1) tan-1 x, cot-1 x are continuous and differentiable every where. sin-1 x, tan-1 x are monotonically increasing while cos-1 x, cot-1 x are decreasing in their respective domains. Results 1. 1 1 , x 0, x 1 (ii) cos-1 x = sec 1 , x 0, x x 1 cot 1 if x 0 x 1 tan x cot 1 1 if x 0 x (i) sin-1 x = cos ec 1 (iii) Text Book of Trigonometry 45 x 1 Inverse Trigonometric Functions Proof: (i) if x 0,1 sin 1 x cos1 1 x2 if x 1,0 cos1 x 1 x2 (ii) if x 0,1 cos1 x sin 1 1 x2 if x 1,0 sin 1 1 x2 Indeed we can relate any inverse trigonometric function with any other inverse trigonometric function. Suppose we want to express tan-1 x in term of cos-1 x. 2 First of all let x > 0 then tan 1 x 0, and from tan = x we can conclude that cos 1 1 x2 0 and 0, . 2 1 x 1 1 But if x < 0 then , 0 and cos can not be inverted since is still 2 2 2 1 x 1 x positive and cos 1 positive 0, 2 To invert we write cos = cos (-) and note that 0, 2 1 1 Thus cos 1 Or tan 1 cos 1 1 x2 1 x2 1 1 if x 0 cos 1 x2 1 Consequently tan x 1 cos 1 if x 0 1 x2 1 cot If x > 0, put tan-1 x = , then 0, and tan = x Or x 2 1 1 since 0 and 0, this can be inverted and we can write cot 1 . x x 2 1 Again if x < 0 and we put tan-1 x = then , 0 .Now cot can not be x 2 which can be inverted since (iii) 1 2 1 0 and , 0 cot 1 negative , we try to write x 2 2 cot where , . Indeed cot cot and , 2 2 1 Therefore cot = x can be written as cot and we will invert it to get x inverted since Text Book of Trigonometry 46 cot = Inverse Trigonometric Functions cot 1 2. 3. 4. 1 x tan 1 1 1 cot 1 . x x -1 - sin x, x 1 (i) sin-1 (- x) = (ii) cos-1 (- x) = - cos-1 x, -1 -1 x 1 (iii) (i) tan (- x) sin(sin-1 x) = = - tan x for all x x, x 1 (ii) cos(cos-1 x) = x, x for all x -1 (iii) tan(tan x) = (i) sin-1(sin x) = 1 (ii) -1 cos (cos x) n x 1 x n , x n , n 2 2 if x 2n 1 , 2n = 2n - x if x 2n, 2n 1 = x – 2n , n 2 2 x 2 m 2 2 Proof (i): If n is even (Say n = 2m) then x n 2 m x 2 m 2 2 Also sin(x – 2m) = sin x for all x sin-1sin x = x – 2m = (-1)n (x - n) If n is odd (n is odd (n = 2m + 1) then x n , n 2m 1 x 2m 1 2 2 2 2 2m 1 x 2m 1 2 2 2m 1 x 2 2 n 1 sin sin x 2m 1 x 1 x n n 2m 1 (ii) If x [(2n – 1), 2n ] then 2n - x [0, ] and when x [2n, (2n + 1)], then x – 2n [0, ] whence the result (ii) follows. 19 3 ,3 6 2 2 19 19 1 We can calculate sin sin 6 directly also. Indeed sin 6 1 1 sin 3 and sin 1 . 6 2 6 2 For example sin 1 sin 19 3 19 3 1 6 6 6 As another example to evaluate sin-1(sin 20) we note that 20 6 , 6 2 2 (i.e. n = 6) sin-1(sin 20) = (-1)6(20 - 6) = 20 - 6 Text Book of Trigonometry 47 Inverse Trigonometric Functions 5. x y , x, y 0, xy 1 1 xy x y (ii) tan-1 x + tan-1 y = tan 1 , x, y 0, xy 1 1 xy Proof: (i) Put tan-1 x = , tan-1 y = , then , 0, and since 2 tan tan x y tan 0 ** xy 1 1 tan tan 1 xy x y 0, and this can be inverted to get + = tan 1 1 xy 2 tan-1 x + tan-1 y = tan 1 (i) whence (i) follows. If xy > 1 then (**) can not be inverted since tan( + ) < 0 and inversion will yield + = tan 1 x y 1 xy ,0 2 angle in (0, ) = an angle in , since tan( + ) < 0 2 x y , 0 and tan( + - ) = 1 xy 2 x y which can be inverted and we get tan 1 1 xy x y Or tan-1 x + tan-1 y = tan 1 1 xy We note that 6. sin-1 x + cos-1 x = /2, x 1 tan-1 x + cot-1 x = /2, for all x sec-1 x + cosec-1 x = /2, x 1 7. sin 1 2 1 x 2 1 1 2sin x if x 1, 2 1 1 2sin 1 x if x , 2 2 1 2sin 1 x if x ,1 2 On putting x = sin , LHS = sin-1(sin 2) (*) 1 . Then 2 , 4 2 2 , 2 , 2 , 0 2 2 2 1 1 1 sin sin 2 sin sin 2 2sin x when x 1, Text Book of Trigonometry 48 Inverse Trigonometric Functions 1 then 4 , 4 2 2 1 1 sin (sin 2) 2 2sin x . Again if x 1 2 , 2 2 , ,1 then , 2 , 2 0, 2 2 4 2 2 1 1 1 2sin x. sin (sin 2) sin sin ( 2) 2 1 Finally when x sin 1 ( x 1 y 2 y 1 x 2 ) , if x, y 0, x 2 y 2 1 8. sin 1 x sin 1 y sin 1 ( x 1 y 2 y 1 x 2 ) if x 0, y 0, x 2 y 2 1 1 2 2 if x , y [0,1] cos ( xy 1 x 1 y ), Proof : (i) If we put sin 1 x ,sin 1 y then , 0, ( x , y 0) 2 and x 2 y 2 1 sin 2 sin 2 1 cos 2 cos 2 0 cos( + ) cos ( - ) 0 cos( ) 0 ( cos ( ) 0 ) 0 , 2 the equality sin ( ) = sin cos cos sin = x 1 y 2 y 1 x2 (*) can be inverted and we can get sin 1 ( x 1 y 2 y 1 x 2 ) (ii) If x 2 y 2 1 then following the same steps we will arrive at cos( ) 0 (iii) , 2 ( ) , 2 The equality (*) can be inverted and we get ( ) Or sin 1 [ x 1 y 2 y 1 x 2 ] sin 1 x sin 1 y [ sin 1 x 1 y 2 y 1 x 2 ] Put cos 1 x , cos 1 y then , 0, since x , y 0,1 2 Now cos (+ ) = cos cos - sin sin = xy - 1 x 2 ( x , y (0,1) , x 1 x 2 , y 1 y 2 ) The equality can be inverted since + , 2 Thus we get , cos 1 ( xy 1 x 2 Or 1 y2 0 cos 1 x cos 1 y Text Book of Trigonometry 1 y2 ) cos 1 ( xy 1 x 2 49 1 y2 ) Inverse Trigonometric Functions 9. 1 2 x 1 x 1 sin 1 x 2 2x 1 2 tan x sin 1 if x 1 1 x2 2x 1 sin 1 x 2 if x 1 Put x = tan on RHS. If -1 x 1, then 4 4 2 2 2 1 2 tan 1 1 sin sin sin 2 2 2 tan x 2 1 tan Again if x > 1 then 2 , 2 0, 4 2 2 2 2x 2 tan 1 x 2 tan 1 x sin-1 (sin 2) = sin-1 (sin( - 2)) = sin 1 2 1 x Finally 2 0 sin -1 (sin 2) = sin -1 (sin(-- 2)) = - - 2 tan -1 x 2 2x sin 1 2 tan 1 x 2 tan 1 x LHS. 2 1 x 2x 2 1 x Now sin 1 Solved Examples 1. Prove the following numerical equalities 1 1 (ii) cot cos 1 2 2 3 1 1 2 6 (iii) (iv) cos 2 cos 1 sin 1 5 5 5 1 14 (v) (vi) sin 2 tan 1 cos tan 1 2 2 3 15 1 Solution: (i) Take cos 1 t , then t , and cos t 3 2 1 3 cos t 1 cot t sin t 1 cos2 t 2 2 (i) (ii) Put 1 3 2 sin cot 1 5 4 2 33 sin 1 cos 10 5 4 5 16 sin 1 sin 1 sin 1 5 13 65 2 1 and 3 3 cot 1 t then t , , 4 2 t 1 cos t 3 , But cot t and 4 2 2 3 t 2 cos t t , whence sin 5 2 5 2 and LHS = sin Text Book of Trigonometry 50 Inverse Trigonometric Functions 1 1 1 1 sin 1 cos 1 cos cos 1 5 5 5 5 2 (iii) LHS = cos cos 1 (iv) 24 2 6 1 1 sin cos 1 1 5 5 5 5 33 3 3 3 We have cos cos 3 cos sin 10 10 10 2 10 sin sin 5 5 2 whence sin 1 cos (v) 33 1 sin sin 10 5 5 1 , tan 1 2 2 then 3 2 tan 1 LHS = sin 2 cos 2 1 tan 1 tan 2 Put tan 1 sin ce 0, 2 we have taken sign with radical 14 1 on substituting tan , tan 2 2 15 3 (vi) First method of proving this equality could be applying formulae set (9) several We should prefer proceeding independently. Put sin 1 4 x, 5 sin 1 5 y, 13 sin 1 16 z 65 We have to show x + y + z = /2 It is sufficient to show that x + y = /2 – z We have sin(x + y) = sin x cos y + cos x sin y = 4 12 3 5 63 5 13 5 13 65 ( x, y are acute angles) 16 z cos z cos sin 1 65 2 and sin 2 16 1 65 sin x y sin z 2 65 16 2 65 2 2 81 49 65 2 x + y + z = /2 x + y – z = /2 (A) (B) If n = 2 we get x+y+z= (C) Text Book of Trigonometry 63 65 n x y n 1 z 2 If n = 0 we get If n = 1 we get If n = 3 we get 5 2 5 x+y–z= 2 (D) 51 times. Inverse Trigonometric Functions 3 , (C) and (D) are not possible. For n 4 and n < 0 2 3 relations obtained will again be impossible since 0 < x + y + z < 2 Since max(x + y + z) = the (B) is not possible either (why ?) therefore it can be concluded that x+y+z= 3 essentially 2 NOTE: A superior solution will be to convert every term in tan-1 . Indeed 4 5 1 4 1 5 1 63 1 3 12 tan 1 16 LHS = tan tan tan tan 4 5 3 12 16 63 1 3 12 63 16 63 63 tan 1 tan 1 tan 1 cot 1 16 63 16 16 2 2. Solve the following equations x 1 x 1 tan 1 x2 x2 4 (i) tan 1 (iii) 4 sin-1 x + cos-1 x = Solution: (i) sin-1 x + sin-1 2x = (iv) 1 sin cos 1 x 1 5 On taking tan of both sides we get x 1 tan tan 1 tan 1 x2 We easily get x are positive. 1 2 x 1 x 1 x 1 x 2 x 2 1 1 x2 x 1 x 1 1 x 2 x 2 for x 1 2 both expressions The equation is satisfied for x 1 2 x 1 x 1 and x2 x2 . 1 is possible. Thus solutions are . 4 2 The equation can be written as sin 1 2 x sin 1 x 3 sin sin 1 2 x sin sin 1 x 3 3 1 2x 1 x2 x 2x = sin cos sin 1 x cos sin sin 1 x 2 2 3 3 tan 1 tan 1 (ii) 3 (ii) 5x 3 3 1 x2 x 2 2 28 3 Now for x = . LHS = sin-1 (-ve) + sin-1 (-ve) = negative 28 Text Book of Trigonometry 52 Inverse Trigonometric Functions x (iii) 3 is not a solution. Thus the only solution is 28 x 3 (check !!) 28 The equation is sin-1 x + cos-1 x + 3 sin-1 x = 3sin 1 x 2 sin 1 x 1 x 6 2 Thus x = ½ is a solution. (iv) We must have 1 cos 1 x 5 2 cos 1 x which is impossible since max cos-1 x = NOTE:- The student may conclude that x 5 2 5 0 which is not correct since it 2 satisfy the original equation. 3. If cos-1 x + cos-1 y + cos-1 z = , prove that x2 + y2 + z2 + 2xyz = 1 Solution: We have cos-1 x + cos-1 y = - cos z. On taking cos of both sides we get cos(cos-1 x) cos(cos-1 y) – sin(cos-1 x) sin(cos-1 y) = cos( - cos-1 z) xy 1 x 2 1 y 2 z xy z 1 x 2 1 y 2 On squaring this yields x2 + y2 + z2 = 1 + 2xyz 4. Show that if a Solution: 1 1 the equation (sin-1 x)3 + (cos-1 x)3 = a3 has no solution if a 32 32 Since sin-1 x + cos-1 x = /2. The equation reduces to t 3 t a3 where t = sin-1 x 2 3 which is same as 12 t2 - 62t + (1 – 8a) 3 = 0 which will not have real roots. If 36 4 – 48 4(1 – 8a) < 0 which reduces to a 5. 1 . 32 Prove that sin cot 1 cos 2 tan 1 cos 2 tan 2 0 2 Thus sin cot 1 cos 2 tan 1 cos 2 sin 2 tan 1 cos 2 2 2 1 tan 1 cos 2 sin 2 cos 2 tan 2 2 2 1 cos 2 1 tan x 3 1 x2 1 6. Evaluate cos 1 x cos 1 where 1 x 2 2 2 1 Solution: Let cos-1 x = . Then x = cos as 1 x 2 3 Solution: Let tan 1 cos 2 . Then cos 2 = tan2 Text Book of Trigonometry 53 does not Inverse Trigonometric Functions x 3 1 x 2 cos 3 1 cos 2 2 2 2 2 2 cos cos sin sin 3 sin 1 cos 3 3 cos 3 x 3 1 x2 cos 1 cos 1 cos 2 2 3 3 2 3 0 3 3 x 3 1 x2 Hence cos 1 x cos 1 2 2 3 3 pq 1 qr 1 rp 1 7. Prove that cot 1 cot 1 cot 1 , if p > q > r. pq qr rp pq 1 Solution: Since p > q, cot 1 (*) cot 1 q cot 1 p p 1 qr 1 Since q r , cot 1 (**) cot 1 r cot 1 q qr rp 1 Again since p r , therefore , cot 1 (***) cot 1 p cot 1 r rp pq 1 qr 1 rp 1 Thus form (*), (**) and (***) , cot 1 cot 1 cot 1 pq qr rp 1 1 1 1 1 1 (cot q cot p) (cot r cot q) ( cot p cot r ) . Thus 8. Prove that sin 1 x sin 1 y sin 1 x 1 y 2 y 1 y 2 if x, y [-1, 0], x2 + y2 > 1 ,0, ,0 2 2 , 2 2 Solution: sin-1 x = , sin-1 y = , then Also , 0 , x, y 1, 0 The condition x2 + y2 > 1 yields sin2 + sin2 > 1 cos 2 cos 2 0 cos 0 cos cos 0 (*) cos 0 Now sin ( + ) = sin cos + cos sin either belongs to , or , 0 2 2 But from (*) cos(+)<0. thus , and then ( ) , 0 . 2 2 Text Book of Trigonometry 54 Inverse Trigonometric Functions = x 1 y 2 y 1 x2 Now (**) becomes sin( ( )) + = - - sin -1 [ x 1 y 2 y 1 x 2 ] etc . x y if x, y < 0, xy > 1. 1 xy Solution: tan 1 x , tan 1 y , then , , 0 2 Also xy> 1 tan tan 1 sin sin cos cos ( cos cos 0) cos(+ )<0 + , ( initially ( , 0)) 2 tan tan x y x y Now tan ( ) tan ( ) 0 etc . 1 tan tan 1 xy 1 xy (iv) Put x = tan , then x > -1 , 4 2 and LHS tan 1 (tan ) tan 1 = , liein , 4 4 4 4 2 2 Again if x < - 1 , then , 2 4 9. Prove that tan-1 x + tan-1 y = tan 1 (*) LHS tan 1 (tan ) tan 1 tan 4 3 Now the first term is still but as , , tan 1 tan 4 2 4 4 4 tan 1 tan if , whence RHS of 2 3 = . 4 4 ak Prove that tan bc 1 1 1 ck 2 10. tan where k = a+b+c, a, b, c > 0 ab ak bk k bc ac Solution: The sum of first two terms = + tan 1 ( x, y 0, xy 1 as xy 1) k c 1 c ck whence the result follows. tan 1 ab x 3 3x 2 1 11. Prove that cos 1 x cos 1 is x ,1 2 2 2 3 1 Text Book of Trigonometry 2 bk tan ac (*) 1 2 1 55 Inverse Trigonometric Functions Solution: Put x = cos, then 0, = ( x(1/ 2, 1)) 3 1 3 LHS cos 1 (cos ) cos 1 cos sin 2 2 = cos 1 cos cos sin sin cos 1 cos 3 3 3 0, 3 3 3 3 3x 4x Solve sin 1 sin 1 sin 1 x 5 5 12. Solution: On taking sine of both sides , we get 3x 16 x 2 9 x2 1 1 x 5 25 25 Or x(3 25 16 x 2 4 25 9 x 2 ) 25 x One root is obviously x = 0 the other roots can be obtained by solving the equation 3 25 16 x 2 4 25 9 x 2 25 by means of substitutions x2 = t. We will finally obtain x = 1. Roots are to be checked. 13. Find the latest and greatest value of (sin-1 x)3 + (cos-1 x)3 Solution: Put cos 1 x t , then 0 < t < and the function ( Say f(t)) is given by f (t ) t t 3 f (t ) 3t 2 3 t 3 t 2 2 4 f is MD in 0, and MI in , 4 4 There is an absolute minimum at and for absolute maximum we can compare the 4 3 2 values of f(t) at boundary points t = 0 , t = , we easily obtain 3 f (min) f , 4 32 73 . 8 1 x 1 x2 14. If 0 < x < 1 and A 2 tan 1 . Show that A + B = , B sin 1 1 x 1 x2 ( 0 x 1) Solution: On putting x tan , then 0 4 A 2 tan 1 tan , B sin 1 (cos 2) 4 A = 2 , B 2 Note that both and 2 0, 2 2 4 2 2 f (max) f () whence A + B = . Text Book of Trigonometry 56 Inverse Trigonometric Functions 1 1 3 x x if 0 x or x 3 tan 1 3x 2 3 2x 1 1 15. Prove that tan x tan 3 1 x2 1 1 3 x x if x 1 tan 2 1 3x 3 1 Solution: If 0 x , then x = tan 0 . 6 3 1 1 LHS tan (tan ) tan (2) 2 3 RHS tan 1 (tan 3) 3 LHS RHS . 2 3 If x 3, then x = tan 2 , 3 3 2 3 2 1 1 LHS tan (tan ) tan (2) = + tan-1 tan(2 - ) =3 - ( 2 0 ) 3 RHS tan 1 (tan 3) tan 1 (tan 3 ) 3 0 3 2 LHS RHS . Exercise 1. Prove that (i) (ii) (iii) (iv) (v) (vi) 2. sin-1 x + sin-1 y = cos 1 1 x 2 1 y 2 xy if x, y 0, x2 + y2 1. 2 2 sin-1 x + sin-1 y = sin x 1 y y 1 x if xy > 0, x2 + y2 > 1. 2 if x 0 1 tan 1 x tan 1 x if x 0 2 if x 1 4 1 1 1 x tan x tan 1 x 3 if x 1 4 1 2 2x 2x 1 1 x cos tan 1 if x 1 2 2 1 x 1 x 1 x2 1 2 2 x2 1 sin sin 1 x 2 2 4 2 tan 1 x sin 1 Prove the following numerical equalities Text Book of Trigonometry 57 Inverse Trigonometric Functions 1 63 1 sin sin 1 8 2 4 3 4 sin sin 1 sin 1 1 5 5 (i) (iii) sin-1 (sin 10) = 3 - 10 (v) (iv) (vi) 1 2 2 1 sin sin 1 2 3 3 3 8 36 sin 1 sin 1 sin 1 5 17 85 2 cos-1 (cos 20) = 20 - 6 46 6 33 1 sin 1 sin cos cos 7 7 7 1 1 1 1 13 (ix) 4 tan 1 tan 1 cos 1 cos 1 cos 1 5 239 4 2 7 14 1 1 1 1 tan 1 tan 1 tan 1 tan 1 3 5 7 8 4 1 1 cos 2 tan 1 sin 4 tan 1 7 3 (vii) (viii) (x) (xi) 3. (ii) Solve the following equation 52 8 (i) sin 1 1 x 2sin 1 x 2 (ii) tan 1 x cot 1 x (iii) sin-1 x – cos-1 x = sin-1 (3x – 2) (iv) 2tan-1 (2x + 1) = cos-1 x (v) cos 1 x sin 1 x (vi) tan-1(x + 1) + tan-1(x – 1) = tan-1 6 2 2 8 31 x2 1 2x 2 tan 1 2 2 x 1 x 1 3 [Hint: Put x = tan, consider cases x (, 1) , x (1,1), x (1, ) ] Solve cos 1 (vii) 4. Find the greatest and least value of tan 1 x provided |x| 1.] 2 1 x 1 1 tan 1 x. tan 1 , 0 x 1 [Hint: Function tan 1 tan x 4 1 x sin-1 x + cos-1 x + tan-1 x [Hint: Function (i) (ii) 5. Use the fact that tan 1 x is monotonically increa sin g. on R.] If sin-1 x + sin-1 y + sin-1 z = prove that x4 + y4 + z4 + 4x2y2z2 = 2(x2y2 + y2z2 + z2x2) [Hint: Solution: We have sin 1 x sin 1 y sin 1 z take cos of both sides , we get 1 x2 6. 7. 8. 1 y 2 xy 1 z 2 1 x2 1 y2 xy 1 z 2 . Now square twice to obtain the required result.] x cos 1 cos Reduce to its simplest form tan 1 [Ans. ] cot 1 x sin x sin If tan-1 y = 4 tan-1 x, find y as an algebraical function of x. Hence prove that tan 22 30 is a root of the equation x4 – 6x2 + 1 = 0 m tan a n tan 1 nm If , then Prove that a tan 1 tan a 2 2 2 cos cos a nm Text Book of Trigonometry 58 Inverse Trigonometric Functions 9. 10. pq 1 qr 1 rq 1 cot 1 cot 1 2, if p q r pq qr rp [Hint: LHS cot 1 q cot 1 p cot 1 r cot 1 q cot 1 p cot 1 r 2. ] pq qr rp If p > 0 > q > r and pq > -1 > pr, show that tan 1 tan 1 tan 1 1 pq 1 qr 1 rp Prove that cot 1 11. If sin-1 x + sin-1 y + sin-1 z = prove that x 1 x 2 y 1 y 2 z 1 z 2 2 xyz 12. [Hint: Put sin 1 x ,sin 1 y ,sin 1 y and use sin 2 A sin 2 B sin 2C 4sin A sin B sin C ] Using the principal values, express the following as a single angle: 142 1 1 3tan 1 2 tan 1 sin 1 2 5 65 5 142 142 1 11 1 5 [Hint: Change sin 1 to tan 1 . Also 3 tan 1 tan 1 , 2 tan 1 tan 1 . ] 2 2 5 12 31 65 5 13. 1 1 2 3 4 1 1 1 3 4 1 1 cot tan , and 2 3 6 3 3 5 3 2 1 1 1 2 2 1 1 tan . 3 36 3 3 Prove that (i) (ii) 1 1 tan 2 Answers 3. (i) (v) 4. (i) 0 ½ 3 , 4 4 Text Book of Trigonometry (ii) -1 (ii) ,0 4 (iii) 59 1, ½ (iv) 0