20. INVERSE TRIGONOMETRIC FUNCTIONS
1. INTRODUCTION TO INVERSE TRIGONOMETRY
The inverse trigonometric functions are the inverse functions of the trigonometric functions. They are sometimes referred to as cyclometric functions.
2. IMPORTANT DEFINITIONS
Given two non-empty sets X and Y, let f:X → Y be a function, such that y = f(x). The set X is called as the domain of f while the set Y is called as the co-domain of f. The set {f(x): x ∈X} is called as range of f.A map f: A → B is said to be one-one or injective, if and only if, distinct elements of A have distinct images in B, i.e. if, and only if,
x12≠⇒ x f(x 1 ) ≠ f(x 2 ) , for all x12 ,x∈ A Onto map or Surjective map: A map f: A → B is said to be an onto map or Surjective map if, and only if, each element of B is the image of some element of A, i.e. if, and only if, Range of f = co-domain of f. Objective map: A map f : A → B is an objective map if, and only if, it is both one – one and onto.
3. INVERSE FUNCTIONS
If f : X → Y is one-to-one and onto (i.e. f is objective), then, we can define a unique function g : Y → X, such that g(y) = x, where x ∈X is such that y = f (x). Thus, the domain of g = range of f and range of g = domain of f. The function is called the inverse of f and is denoted by f-1. (a) Trigonometric functions are many-one functions but these become one-one, onto, if we restrict the domain of trigonometric functions. Similarly, co-domain is equated to range to make it an onto function. We can say that the inverse of trigonometric functions are defined within restricted domains of corresponding trigonometric functions. (b) Inverse of sin (sine functions) is denoted by sin-1 (arc sine function). We also write it as sin-1 x. Similarly, other inverse trigonometric functions are given by cos-1 x, tan-1 x, sec-1 x, cot-1x and cosec-1 x. 1 (c) Note that sin−1 x ≠ and ( sin−1 x)² ≠ sin−2 x, Also sin−−11 x≠ (sinx) sinx (d) Domain and Range of Inverse Trigonometric Functions:
Function Domain Range (Principal value branch)
(i) −1 ππ y= sin x −≤≤1x1 − ≤≤y 22 20.2 | Inverse Trigonometric Functions
Function Domain Range (Principal value branch)
(ii) −1 y= cos x −≤≤1x1 0y≤ ≤π
(iii) −∞ (iv) ππ y= cosec−1 x x≥ 1 or x ≤− 1 − ≤≤y ,y ≠ 0 22 (v) π y= sec−1 x x≥ 1 or x ≤− 1 0≤ y ≤π ,y ≠ 2 (vi) −1 y= cot x −∞ (e) The principal value of an inverse trigonometric function is the value of that inverse trigonometric function which lies in the range of principal branch. MASTERJEE CONCEPTS If no branch of an inverse trigonometric function is mentioned, then it can be implied that the principal value branch of that function. You can remember range as set of angles that have the smallest absolute values satisfying for all the values of domain. Vaibhav Gupta (JEE 2009 AIR 54) 4. TRANSFORMATION OF TRIGONOMETRIC FUNCTIONS TO INVERSE TRIGONOMETRIC FUNCTIONS 4.1 sin x to sin-1 x The graph of an inverse trigonometric function can be obtained from the graph of the original by interchanging x and y axes. Note: It can be shown that the graph of an inverse function can be obtained from the corresponding graph of original function as the mirror image in the line y = x. Y 1 3 -5p -p p 2 2 2 XX’ O 5 -2 p p p 2p p p -1 2 2 Y’ Figure 20.1 (a) y = sin x, x ∈ R and y1≤ ; y= sin−1 x, x ≤ 1, y ∈ [ −π / 2, π / 2] Mathematics | 20.3 4.2 cos x to cos-1 x (b) y=∈≤ cosx,x Rand y 1 y= cos−1 x,x ∈− [ 1,1]andy ∈ [0, π ] Y Y 5p 5p 2 2 2p 3p 3p 2 2 y=x p p p 2 -p -2 2 -1 1 X’ p p X X’ X - O 2 O p p p - p -x 2 p 2 2 2 - p p -3 -3p p 2 2 -2p -5p -5p 2 2 Y’ Y’ y-sin x and y-sin-1(x) y-sin-1(x) Figure 20.2 Y 5p 2 2p 3p 2 p p Y 2 1 X’ -1 1 X -5p p 5p O - 2 X’ 2 p p 2 X -3 - O 3 2 -p -2p p p p p 2 2 2 -1 2 -p Y’ -3p y=cos x 2 -2p -5p 2 Y’ y=cos-1 (x) Figure 20.3 20.4 | Inverse Trigonometric Functions 4.3 tan x to tan-1 x π −1 ππ (c) y= tanx,x ∈ R − x : x = (2 π+ 1) ,n ∈ Z and y ∈ R y= tan x,x ∈ R and y ∈− , 2 22 Y Y 2 p 3p 2 p 2 p p 3p 2 - 1 2 2 2 X’ p p X O O -2 -1 -p p X’ X 2 -1 -2 12 -p 2 -p Y’ Y’ y=tan x Figure 20.4 4.4 cot x to cot-1 x (d) y= cot x,x ∈− R{ x:x =π n , n ∈ Z} and y ∈ R y= cot−1 x,x ∈ R and y ∈π (0, ) Y Y 2 p 3p 2 p 2 p p 3p 2 - 1 2 2 2 X’ p p X O O -2 -1 -p p X’ X 2 -1 -2 12 -p 2 -p Y’ Y’ y=tan x y=tan-1 x Figure 20.5 4.5 sec x to sec-1 x π −1 π (e)(v) y= secx,x ∈− R x : x = (2n + 1) ,n ∈ z andy ∈−− R ( 1,1) y= sec x,x ∈ R −− ( 1,1)andy ∈ [0, π ] 2 2 Mathematics | 20.5 2 Y p 3p 2 2 p p - 1 2 2 X’ p p p X - O 3 p -1 p p X’ X 2 -2 2 2 -2 -1 1 2 -p 2 - Y’ p y = sec x Y’ y = sec-1 x Figure 20.6 4.6 cosec x to cosec-1 x −1 ππ (f) y= cosecx,x ∈ {x : x = n π ,n ∈ Z} andy ∈−− R ( 1,1) y= cosec x,x ∈ R −− ( 1,1)andy ∈− , − {0} 22 Y 2p Y 3p 2 2 p - 3 p p 1 p 2 - 2 2 p p 2p X’ X X O X’ -2 -1 p -1 2 O 12 -2 -p 2 - Y’ p y = cosec x Y’ y = cosec-1 x Figure 20.7 20.6 | Inverse Trigonometric Functions Illustration 1: Find the domain of definition of the function f(x) = 3cos−1 (4 x) −π. (JEE MAIN) Sol: Use the condition that the expression inside the square root is ≥ zero. For domain of f(x) = 3cos−1 (4 x) −π, we must have π 11 4x≥ cos ⇒ 4x ≥ ⇒≥ x …..(i) 3 28 −11 Also −≤1 4x ≤ 1 ⇒ ≤ x ≤ .….(ii) 44 −11 ∴ From (i) and (ii), we get x'∈ 48 MASTERJEE CONCEPTS In case of confusion, try solving problems by replacing inverse functions with angles and applying trigonometric identities. Shrikant Nagori (JEE 2009 AIR 30) Illustration 2: If 0< cos−1 x < 1and 1 + sin(cos − 1 x) + sin 21 (cos −− x) + sin 31 (cos x) + ...... ∞= 2, then find the value of x. (JEE MAIN) Sol: Use summation of infinite GP series. We have 1+sin(cos-1 x) + sin²(cos-1 x) + ……… ∞=2 11 1π 3 ⇒ =⇒=−2 1 sin(cos−−11 x) ⇒ sin(cos x) =⇒cos−1 x = ⇒= x 1− sin(cos−1 x) 2262 2 Illustration 3: Let f(x)= (sin−−−111 [x] ++ tan [x] cot [x]) , where [x] denotes the greatest integer less than or equal to π x. If A and B denote the domain and range of f(x) respectively, find the number of integers in ∪A B. (JEE ADVANCED) π Sol: Use tan−−11 [x]+= cot [x] and proceed. 2 For domain of f(x), we must have −1 ≤ [x] ≤ 1 ⇒− 1 ≤ x < 2, so set A = [-1,2) 2 −1 π −−11π f(x)= sin [x] + As tan [x]+ cot [x] = , ∀∈ x A π 2 2 So, set B = {0, 1, 2} = Range of f(x). Now, A∪=− B [ 1,2) ∪ {0,1,2} =− [ 1,2] Hence, number of integers in ( AB∪ )=4 5.PROPERTIES/IDENTITIES OF INVERSE TRIGONOMETRIC FUNCTIONS 5.1 Complementary Angles π (a) sin−−11 x+ cos x = , ∀ x ∈− [ 1,1] 2 Mathematics | 20.7 π (b) tan−−11 x+ cot x = , ∀∈ x R 2 π (c) sec−−11 x+ cosec x = , ∀ x ∈ ( −∞ , − 1] ∪ [1, ∞ ) 2 5.2 Negative Arguments (a) sin−−11 (− x) =− sin x, ∀ x ∈− [ 1,1] (b) cos−−11 (− x) =π− cos x, ∀ x ∈− [ 1,1] (c) tan−−11 (− x) =− tan x, ∀∈ x R (d) cot−−11 (− x) =π− cot x, ∀ x ∈ R (e) sec−−11 (− x) = π − sec x, ∀ x ∈ ( −∞ , − 1] ∪ [1, ∞ ) (f) cosec−−11 (− x) = − cosec x ∀ x ∈ ( −∞ , − 1) ∪ [1, ∞ ) 5.3 Reciprocal Arguments 1 (a) cosec−−11 x= sin ; x ≥ 1 (Both the functions are identical) x 1 and sin−−11 x= cosec ; x ≤≠ 1,x 0 (Both the functions are not identical) x 1 (b) sec−−11 x= cos ; x ≥ 1 (Both the functions are identical) x 1 and cos−−11 x= sec ; x ≤ 1 (Both the functions are not identical) x −−111 −1 1 (c) tan x= cot , x∈∞ (0, ) = − π+cot , x∈ ( −∞ ,0) , x x −−111 −1 1 and cot x= tan , x ∈∞ (0, ) = π+ tan , x∈ ( −∞ ,0) x x 5.4 Forward Inverse Identities − − (a) y= sin(sin1 x) = x,x ∈− [ 1,1],y ∈− [ 1,1], y is (b) y= cos(cos1 x) = x,x ∈− [ 1,1],y ∈− [ 1,1], y is aperiodic aperiodic y y 1 1 y=x y=x o o -1 45 x -1 45 x O 1 O 1 -1 -1 Figure 20.8 20.8 | Inverse Trigonometric Functions (c) y = tan ( tan−1 x )=x, x ∈ R, y ∈ R, y is aperiodic (d) y = cot (cot−1 x) = x, x ∈ R, y ∈ R, y is aperiodic y y y=x y=x 45o 45o x x O O Figure 20.9 (e) y = cosec ( cosec−1 x )=x, x≥≥ 1, y 1 ,y is aperiodic (f) y = sec (sec−1 x) = x, x≥≥ 1, y 1 , y is aperiodic y y y=x y=x o 1 1 45 -1 -1 x x O 1 O 1 -1 -1 y=x y=x Figure 20.10 Also, cos(sin−12 x)= 1 − x sin(cos−12 x)= 1 − x −1 1 2 cos tan x = −1 1x− ( ) 2 tan(cos x) = 1x+ x x x sin( tan−1 x) = tan( sin−1 x) = 1x+ 2 1x− 2 5.5 Inverse Forward Identities −1 ππ (a) y= sin (sinx) = x, x ∈ R ,y ∈− , , Periodic with period 2 π 22 y p/2 y=( y= +x p p+x) y=x p-x y=2 -/2 45o 3/2 y=x-2 p p p x - /2 2 -2p -3p/2 p O p p p -/p 2 Figure 20.11 Mathematics | 20.9 (b) y= cos−1 (cosx) = x,x ∈ R, y ∈π [0, ], Periodic with period 2 π y y = -x p y=2 -x /2 y=x p p y=x+2p x -2 - 2 p p -/p 2 O p/2 p p Figure 20.12 −1 π ππ (c) y= tan (tanx) = x,x ∈− R (2n − 1) n ∈ I , y ∈− , , Periodic with period π 2 22 y p/2 - y=x 3 p y=x- p -2 y=x+p 2 p 2 p x -3 O 2 p -p p p p 2 2 p/2 Figure 20.13 (d) y= cot−1 (cotx) = x,x ∈− R {n π }, y ∈ (0, π ), periodic with π y p y=x y=x+2p y=x+ y=x- 2 pO p p p p 2 x -2 O 2 p -p p p Figure 20.14 −1 ππ (e) y= cosec (cosec x)=∈− x R {n π ,n ∈ I}, y ∈− ,0 ∪ 0, y is periodic with period 2π 22 y p/2 y=( y= p+x) y=x p-x o -/2 45 3/2 y=x-2p p p x -3 /2 - O /2 p p p p 2p -/p 2 Figure 20.15 20.10 | Inverse Trigonometric Functions (f) y= sec−1 (sec x)= x, y is periodic, y p y = y= - x 2 - y=x p x y=x+2p - x -2 -3 - p O p p 3 2 p p p 2 p p 2 2 2 Figure 20.16 π ππ x∈− R (2n − 1) n ∈ I ,y ∈ 0, ∪ , π with period 2π 2 22 1 (i) tan−1 (cotx)= π− x for x ∈ [0, π ] 2 1 (ii) sin−1 (cosecx)= π− x for x ∈ [0, π ] 2 −1 11 (iii) sec (cosx)= π− x for x ∈ 0, π . 22 5.6 Sum of Angles −1 2 2 22 sin x 1−+ y y 1 − x if x ≥ 0; y ≥ 0and x +≤ y 1 −−11 (a) sin x+= sin y −1 2 2 22 π−sin x 1 − y + y 1 − x if x ≥ 0; y ≥ 0and x + y > 1 −−11 − 1 2 2 (b) sin x− sin y = sin x 1 −− y y 1 − x x > 0; y > 0 −−1 1 − 1 22 (c) cos x± cos y = cos [xy 1 −− x 1 y ] if x, y >0 and x² + y² ≤ 1 −−1 1 − 1 22 (d) cos x± cos y =π− cos [xy 1− x 1 − y ] if x, y > 0 and x²+y² ≤ 1 −1xy+ −−11π tan x> 0 y > 0 and xy <⇒ 1 0 < tan x + tan y < −−11 1− xy 2 (e) tan x+= tan y xy+ π π−tan−1 x> 0 y > 0andxy > 1 ⇒ < tan−−11 x + tan y <π 1− xy 2 xy− (f) x > 0 & y > 0 then tan−−11 x−= tan y tan − 1 (with no other restriction) 1+ xy −−−111 − 1x++− y z xyz −−11 − 1xy− (g) tan x++= tan y tan z tan ; tan−= tan y tan 1−−− xy yz zx 1+ xy Mathematics | 20.11 MASTERJEE CONCEPTS The above results can be generalized as follows: −−11 − 1 − 1S135−+− S S ...... tan x12+ tan x ++ ... tan x n = tan 1−+−+ S246 S S ...... where Sk denotes the sum of products of x12 ,x , ,x n taken k at a time Rohit Kumar (JEE 2012 AIR 78) −1 15 Illustration 4: Evaluate: sin tan (JEE MAIN) 8 15 Sol: Convert tan−1 to sin−1 . 8 We know that sin (sin−1 x)= x , for all x ∈ [-1, 1], So, will convert each expression in the −1 form sin (sin x) by using h=17 p=15 b p pppb cos−1= sin − 1 , tan −−−− 1111 = sin ,cot = sin etc. h h bhbh b=8 Where b, p and h denote the base, perpendicular and hypotenuse of a right triangle. Figure 20.17 −−1115 15 15 sin tan = sin sin = 8 17 17 −1 13 Illustration 5: Evaluate: cos cosec (JEE MAIN) 12 h=13 p=12 Sol: Write cosec-1 in terms of cos-1. b=5 −113 − 1 5 5 −− 11 13 5 cos cosec = cos cos = cosec = cos 12 13 13 12 13 Figure 20.18 Illustration 6: Find the principal value of cot-1 (- 3 ) (JEE MAIN) Sol: The principal value of cot-1 x lies in between 0 to π . Let cot−1 (−=θ 3) π Then cotθ= − 3 = − cot 6 Since principal value branch of cot-1 x is 0 <θ<π. Therefore, we want to find the value of θ such that 0 <θ<π. π ππ5 Now, cotθ=− cot = cot π− = cot 6 66 5π Therefore, principal value of cot−1 (−= 3) . 6 20.12 | Inverse Trigonometric Functions −1 10ππ 10 Illustration 7: sin sin = (JEE MAIN) 77 10π 3π Sol: Write as π+ and expand. 7 7 −−1110π 3 π − 1 3 ππ 3 =sin sin = sin − sin = sin sin −=− 7 7 77 −1 π Illustration 8: cos sin− (JEE MAIN) 9 −−11ππ 11π 11 π Sol: =cos cos += cos cos = 2 9 18 18 −1 13π Illustration 9: sin cos (JEE MAIN) 10 Sol: Similar to previous example. −−1113π 3π − 1 5ππ 2 −−11π ππ =sin cos =−=−− sin cos sin sin =sin − sin = sin sin −=− 10 10 10 10 5 55 −1 1 Illustration 10: Find the principal value of sin . (JEE MAIN) 2 −1 1 1 −1 1 Sol: Let sin= y. Then sin y = ⇒=y sin 2 2 2 −1 ππ π 1 We know that, the range of the principal value branch of sin is−= , and sin . 22 4 2 −1 1 π Therefore, principal value of sin is . 2 4 211−− Illustration 11: Find the integral solution of the inequality 3x+< 8x 2sin (sin4) − cos (cos 4) . (JEE ADVANCED) Sol: Use inverse forward identities to simplify the equation. 3x2 + 8x <− 4 ⇒ 3x2 + 8x +< 4 0 ⇒ 3x2 + 6x + 2x +< 4 0 ⇒ 3x(x++ 2) 2(x +< 2) 0 -2 -1 -2/3 (x+ 2)(3x +< 2) 0 x=-1 Figure 20.19 Illustration 12: Find the largest integral value of k, for which (k-2)x² + 8x + k + 4 > sin-1(sin12) +cos-1(cos 12), for all x ∈ R. (JEE ADVANCED) Sol: Use inverse forward identities. sin−−11 (sin 12)= sin (sin(12 −π= 4 )) 12 −π 4 Mathematics | 20.13 cos−−11 (cos 12)= cos (cos(4π− 12)) = 4 π− 12 2 0 ∴(k − 2)x + 8x + k + 4 > 0, ∀∈ x R -6 4 If k = 2, then 8x+4>0, (not possible) and if k2≠ , then k-2>0 ⇒ k>2 2 and 64− 4(k − 2)(k +<⇒ 4) 0 16 <+− k2 2k 8 Figure 20.20 ⇒ k2 + 2k − 24 > 0 ⇒ (k + 6)(k −> 4) 0 K = 5 1 Illustration 13: Find domain of f(x) = . (JEE MAIN) ln( cot−1 x) − Sol: Find the range of x for which ln(cot1 x)> 0 ⇒cot−1 x > 1 ⇒ x < cot 1 ⇒ x ∈ ( −∞ ,cot1) Illustration 14: Evaluate the following: (JEE MAIN) π π 7π 3 π (i) sin−1 sin (ii) tan−1 tan (iii) cos−1 cos (iv) cos cos−1 + 3 4 6 26 ππ Sol: Recall that, sin−−11 (sinθ ) =θ , if − ≤θ≤ ,cos (cosθ ) =θ , if 0 ≤θ≤π and 22 ππ tan−1 (tanθ ) =θ , if − <θ< . Therefore, 22 −1 ππ −1 ππ (i) sin sin = (ii) tan tan = 33 44 −1 77ππ 7 π (iii) cos cos≠π , because does not lie between 0 and . 66 6 −−1175ππ 75ππ−1 5π 5π Now, cos cos= cos cos 2π− =2 π− = cos cos [ cos(2π−θ ) = cos θ ] = 6666 6 6 35π ππ 35π (iv) cos cos−1 +=cos + cos−1 = 2 6 6626 Illustration 15: Evaluate the following: −1 3 π −1 1 −1 (i) sin cos (ii) sin−− sin (iii) sin(cot x) (JEE MAIN) 5 22 3 34 Sol: (i) Let cos−1 =θ . Then, cosθ= ⇒ sin θ= 5 55 −1 34 ∴ sin cos= sin θ= 55 20.14 | Inverse Trigonometric Functions π−1 1 ππ 23 π (ii) sin− sin − = sin −− =sin = 2 2 2 6 32 (iii) Let cot−1 x=θ=θ , Then, x cot 1 1 Now, cotθ= x ⇒ sin θ= ∴sin(cot−1 x) = sin θ= 1x+ 2 1x+ 2 Illustration 16: Evaluate the following: (i) sin−1 (sin5) (ii) cos−1 (cos10) (JEE MAIN) Sol: Notice that the angle is in radians. π π (i) Here, θ=5 radians. Clearly, it does not lie between - and . But 2 2 π π 25π− and 52−π both lie between - and such that 2 2 sin (5− 2 π ) = sin( − (2 π− 5)) =− sin(2 π− 5) =− ( − sin5) = sin5 ⇒sin−−11 (sin 5) = sin (sin(5 −π=−π 2 )) 5 2 . (ii) We know that cos−1 (cosθ=θ ) , if 0.≤θ ≤π Here, θ=10 radians. Clearly, it does not lie between 0 and π such that, (4π− 10) = cos10 ⇒ cos−−11 (cos 10)= cos (cos(4π− 10)) = 4 π− 10 Illustration 17: Evaluate the following: −−11 −1 1 π (i) sin (2 sin 0.8) (ii) tan 2tan − (JEE MAIN) 54 Sol: Write the term inside the brackets in (i) and (ii) as sin-1 and tan-1 respectively. (i) We know that: 2sin−−11 x= sin (2x 1− x 2 ) ∴=2 sin−−11 0.8 sin (2x0.8x 1 − 0.64) ⇒==sin−−11 (2sin 0.8) sin{sin−1 (0.96)} 0.96 −1 1 π (ii) tan 2tan − 54 −1 5 π −−1115 = tan tan − From (ii) we have, 2tan= tan 12 4 5 12 −−115 −−11 − 1xy− −1 −77 = tan tan− tan 1 tan x− tan y = tan if xy>− 1 = tan tan = − 12 1+ xy 17 17 Illustration 18: Write the following in their simplest forms: 1− cosx (i) tan−1 (ii) sin [cot−−11 {cos(tan x)}] (JEE ADVANCED) 1+ cosx Sol: (i) Use the formula 1− cosx = 2sin2 x / 2 and 1+ cosx = 2cos2 x / 2 Mathematics | 20.15 (ii) Write the term inside the square bracket in terms of sin-1. 1− cosx 2sin2 x / 2 x x (i) tan−−11=tan = tan − 1 tan = 1+ cosx 2cos2 x / 2 2 2 (ii) sin[cot−−11 {cos(tan x)}] 2 −−111 −1 1 −1 1x+ −−111 = sin cot cos cos = sin cot = sin sin cot x= sin 2 2 2 2 1x+ 1x+ 2x+ 1x+ 1x+ 2 = 2x+ 2 −1 cosx ππ Illustration 19: Express tan ,− << x in the simplest form. (JEE ADVANCED) 1− sinx 2 2 x Sol: Convert the term inside the bracket in terms of tan and proceed. 2 22xx cos− sin −−11cosx 22 We write, ⇒=tan tan 1− sinx x x xx cos22+− sin 2sin cos 2 2 22 xx xx cos+− sin cos sin xx x cos++ sin 1 tan 22 22 −−1122 2 −1 ππxx = = tan =tan =tan tan +=+ 2 xx x 4 2 42 xx cos−− sin 1 tan cos− sin 22 22 2 Alternatively, π π−2x sin− x sin −−11cosx 22 − 1 tan =tan = tan 1− sinx π π−2x 1−− cos x 1 − cos 22 π−2x π− 2x 2sin cos −144−−11π−2x π π−2x −1 ππxx = tan =tan cot = tan tan − =tan tan +=+ 2 π−2x 4 24 42 42 2sin 4 −−111 Illustration 20: If sin sin+ cos x =1, find the value of x. (JEE MAIN) 5 −−111 π Sol: From the question, we have sin+= cos x and proceed. 52 −−111 We have sin sin+ cos x =1 5 20.16 | Inverse Trigonometric Functions 1 1 π ⇒sin−1 += cos −− 11 x sin 1 ⇒ sin−−11+= cos x 5 52 −−π 1 −−1 1 ⇒=−cos11 x sin ⇒ cos11 x= cos ⇒=x 25 5 5 Illustration 21: Find the value of cos (sec−−11 x+≥ cosec x), x 1. (JEE MAIN) π Sol: Use sec−−11 x+= cosec x 2 −−11π We have cos( sec x+== cosec x) cos 0 2 22 Illustration 22: Find maximum & minimum values of (sec−−11 x) + ( cosec x) . (JEE ADVANCED) π Sol: Apply the identity sec−−11 x+= cosec x and then use suitable substitution to form a quadratic. 2 y= (sec−−12 x) + (cosec 12 x) =+−(sec−1 x cosec − 12 x ) 2sec −− 1 x cosec 1 x π put t= sec−1 x ; sec−−11 x+= cosec x 2 22 ππ2 π y= − 2t − t = 2t −π t + 42 4 2 2222 22 2 ππ π π π π ππ5 y= 2t −+ t = 2 t − + =+2t − ∴=y ; y = at t =π 2 8 4 16 8 4 min 84max Illustration 23: Find the range of f(x) = sin−−−111 x++ tan x sec x . (JEE MAIN) Sol: Find the domain of the given function and then find the range. f(x)=++ sin−−−111 x tan x sec x Here domain is only x = 1 or −1; So range will contain only 2 elements {3ππ /4, /4} π Illustration 24: Find the number of solutions of the equation tan−−13 x+= cot 1 (e x ) . (JEE ADVANCED) 2 π Sol: Use tan−−11 A+= cot A to simplify the given equation and then take 2 the help of graph to find the number of solution. y=1 π cot−1x (e )=−= tan −−13 (x ) cot 13 (x ) ⇒=⇒ex x 3 xe 3x− = 1 O 3 2 y= xe3x− Plotting the graph of y = l and we can see that the line intersects Figure 20.21 the curve at two points. Hence there are 2 solutions for the above equation. Mathematics | 20.17 Illustration 25: Find the number of values of x satisfying the equation xx35 xx23 π tan−−11 x − + − +cot x + + − =for 0 << x 2. (JEE ADVANCED) 4 16 2 4 2 π Sol: Use tan−−11 A+= cot A . 2 xx35 xx23 We must have xx−+− =+++ 4 16 2 4 x x 4x 2x ⇒ = ⇒ = ⇒2x2 (x += 2) 0 x22x 4x+ 2x− 1 + 1 − 4 2 ∴x = 0, − 2 (As 0 << x 2) Clearly no value of x satisfies given equation. 271 Illustration 26: Prove that tan−−11+= tan tan − 1 (JEE MAIN) 11 24 2 −−11 − 1xy+ Sol: Use the formula tan x+= tan y tan 1− xy 27 −−11 + tan x+ tan y 27 =+=−−−11111 24 We have, tan tan tan −1 xy+ 11 24 27 = tan If xy< 1 1x− 1− xy 11 24 −148+ 77 −− 11 125 1 = tan = tan = tan 264− 14 250 2 Illustration 27: If tan−−−111 4+=λ tan 5 cot ( ) then find λ . (JEE MAIN) Sol: Write the L.H.S. in terms of cot-1 and compare. 4+ 5 9 19 We have tan−−−111 4+ tan 5 = tan =π−tan−1 =π−cot − 1 1− 20 19 9 −1 19 19 =cot − ⇒λ=− 99 1x− 1y−−yx Illustration 28: Prove that: tan−−11−= tan sin − 1 (JEE MAIN) 1x++ 1y 22 1x++ 1y −1xy+ −− 11 Sol: Use the formula tan =tan x + tan y . 1− xy 1x− 1y− We have, LHS= tan−−11 − tan =(tan−−−−1111 1 −− tan x) (tan 1 − tan y) = tan −− 11 y − tan x 1x++ 1y 20.18 | Inverse Trigonometric Functions −−11yx−−yx −1 yx− = tan = sin = sin = RHS 1+ yx 22 22 (1+ yx) +− (y x) (1++ x )(1 y ) 1 12 33 8π Illustration 29: Prove that: (i) tan−−11+= tan tan − 1 (ii) tan−1+− tan −− 11 tan = 7 13 9 4 5 19 4 1111π (iii) tan−−−−1111+ tan + tan += tan (JEE ADVANCED) 57384 Sol: Same as above. 11 (i) LHS= tan−−11 + tan 7 13 11 + −17 13 −−11 − 1xy+ −−1120 2 =tan tan x+= tan y tan if xy< 1 = tan= tan = R.H.S. 11 1− xy 90 9 1x− 7 13 33 8 (ii) L.H.S.= tan−1 +− tan −− 11 tan 4 5 19 33 + −−1133 − 1 8 − 145 −18 =+−tan tan tan = tan −tan 4 5 19 33 19 1x− 45 27 8 − −127 −− 11 8 11 19 −−11425 π =tan −= tan tan =tan = tan 1 = = R.H.S. 11 19 27 8 425 4 1x+ 11 19 11 1 1 (iii) L.H.S.=++ tan−−11 tan tan − 1 + tan − 1 57 3 8 11 11 ++ −−1111 −− 11 11 − 157−1 38 =+++tan tan tan tan =tan +tan 57 38 11 11 1x−− 1x 57 38 6 11 + −−116 11 − 117 23 −1325 −1 π =+==tan tan tan tan = tan= = R.H.S. 17 23 6 11 325 4 1x− 17 23 1 23 Illustration 30: Show that tan−−11+= tan tan − 1. (JEE MAIN) 2 11 4 12 + 1 2 15 3 Sol: We have, L.H.S.=+= tan−−11 tan tan − 12 11 =tan−1 == tan − 1 R.H.S. 2 11 12 20 4 1x− 2 11 −1 acosx− bsinx a Illustration 31: Simplify tan if tanx>− 1. (JEE MAIN) bcosx+ asinx b Mathematics | 20.19 Sol: Divide the numerator and denominator inside the bracket by bcos x and expand. acosx− bsinx a −tanx −−11acosx− bsinx bcosx − 1b −−11aa − 1 We have, tan =tan = tan =−=−tan tan (tanx) tan x bcosx+ asinx bcosx− asinx a bb 1+ tanx bcosx b Illustration 32: Solve the following equations: x1− x1 +π (i) tan−−11+= tan (ii) 2tan−−11 (cosx)= tan (2cosec x) (JEE ADVANCED) x2−+ x2 4 π Sol: Write as tan-1 1 and simplify. 4 x1− x1 +π (i) tan−−11+= tan x2−+ x2 4 x1−+ x1 x1−x1 + ⇒+=⇒=−tan−1 tan − 1 tan − 1 1 tan − 1 tan −− 11 1 tan x2−+ x2 x2−x2 + x1+ 1 − x1− x1− x2x1 +−− ⇒=tan−−11 tan x2+ ⇒tan−1 = tan − 1 x2− x1+ x2− x2x1 +++ 1 + x2 + x1−− 1 x1 1 ⇒tan−−11 = tan ⇒ = ⇒(2x + 3)(x −=− 1) x 2 x2− 2x3 + x2 −+ 2x3 1 ⇒2x22 +−=−⇒ x 3 x 2 2x −=⇒=± 1 0 x 2 (ii) 2tan−−11 (cosx)= tan (2cosecx) 2cosx ⇒=tan−−11tan (2cosecx) 2 1− cos x 2cosx π ⇒ =2cosec x ⇒ cosx = sinx ⇒ tanx =⇒= 1 x sin2 x 4 3 8 84 Illustration 33: Prove that: sin−−11−= sin cos − 1 (JEE MAIN) 5 17 85 Sol: Covert the L.H.S. in terms of cos-1. −−1138−−114 15 −−113 4 − 1 8 − 1 15 We have sin− sin =cos − cos sin= cos & sin= cos 5 17 5 17 5 5 17 17 22 −1 4 15 4 15 −14 15 3 8 −− 1160 24 84 =cos x +− 1 x 1 − =cos x + x = cos +=cos 5 17 5 17 5 17 5 17 85 85 85 4 5 16 π Illustration 34: Prove that: sin−−11++ sin sin − 1 = (JEE MAIN) 5 13 65 2 20.20 | Inverse Trigonometric Functions −14 − 1 5 − 1 16 −− 11 4 5 − 1 16 Sol: We havesin sin−1 −+=+ sin sin sin sin + sin 5 13 65 5 13 65 22 −−114 5 5 4 16 =sin 1 − +−1 + sin 5 13 13 5 25 −−114 12 5 3 16 −−1163 16 =sin x ++ x sin =sin + sin 5 13 13 5 25 65 25 2 −−1116 16 − 1 63 − 163−1 16 =cos + sin sin= cos 1 −= cos 65 25 65 65 65 ππ−−11 =sin x += cos x 22 6. SIMPLIFICATION OF INVERSE FUNCTIONS BY ELEMENTARY SUBSTITUTION (a) 2sin−−11 x= sin (2x 1 − x 2 ) if −≤≤ 1 x 1 (b) 2cos−−1 x= cos 12 (2x − 1) if −≤≤ 1 x 1 −1 2x tan −≤≤1 x 1 1x− 2 2x (c) 2tan−−11 x= sin 0≤≤ x 1 2 1x+ 1x− 2 cos−1 0≤ x <∞ 2 1x+ 2tan−1 x−≤≤ 1 x 1 2x (d) sin−−11= π−2tan x x≥ 1 2 1x+ −π − 2tan−1 x x≤ − 1 1x− 2 2tan−1 x x≥ 0 (e) cos−1 = 2 −1 1x+ −<2tan x x 0 π+2tan−1 x x<− 1 2x (f) tan−−11=2tan x −<< 1 x 1 2 1x− 2tan−1 x−π x > 1 2 −−1 12− 1x −11x− −11 −11 (g) sin x= cos 1 −= x tan =cot =sec =cosec 22xx 1x−−1x Mathematics | 20.21 2 −−1 12 − 11x− − 1 x − 1 1 − 1 1 (h) cos x= sin 1 −= x tan = cot = sec = cosec xx22 1x−− 1x 2 −−11x − 1 1 − 1 1 − 12 − 1 1x+ (i) tan x= sin = cos = cot = sec 1 += x cosec 22xx 1x++ 1x 2x (j) f(x)= sin−−11 + 2tan x =π≥ , if x 1 1x+ 2 2x (k) f(x)= sin−−11 + 2tan x =−π if x ≤− 1 1x+ 2 −π+( 3sin−1 x) − 1 ≤ x ≤ 1 / 2 −−13 1 (l) sin (3x− 4 x ) = 3 sin x − 1 / 2 ≤≤ x 1 / 2 π−3sin−1 x 1 / 2 ≤ x ≤ 1 3cos−1 x− 2 π − 1 ≤ x ≤− 1 / 2 −−13 1 (m) cos (4 x− 3x) = 2 π− 3cos x − 1 / 2 ≤ x ≤ 1 / 2 3cos−1 x 1 / 2≤≤ x 1 −1 11 3tan x−< x 33 3x− x3 1 (n) tan−−11= −π + 3tan x x > 2 1− 3x 3 1 π+3tan−1 x x <− 3 MASTERJEE CONCEPTS While writing inverse trigonometric functions in their simplest forms, we use the following substitutions. • For ax22− , we substitute x =a sinθ= or x a cos θ • For ax22+ , we substitute x =a tanθ= or x a cot θ • For xa22− , we substitute x = a secθ= or x a cosec θ • For ax+ and ax− occurring together or separately, we substitute x = a cos θ Rohit Kumar (JEE 2012 AIR 78) Illustration 35: Solve for x: sin( 2cos−−11() cot(2tan x) = 0 (JEE ADVANCED) nπ Sol: The R.H.S. is equal to zero implies cos−−11 cot(2tan x) = and proceed accordingly to find the value of x. () 2 20.22 | Inverse Trigonometric Functions 0 if n= 0 1 −−11nππ−1 cos( cot(2tan x)) ===if n 1⇒= cot( 2tan x) 0 22 −1 π=if n 2 π n3ππ π π nπ+ +−, 4 28 8 8 −−−111π nππ π π ⇒2tan x = n π+ ⇒ tan x = + ⇒tan x = , − 2 24 44 π n3ππ ππ n,π− − − 4 2 8 88 ⇒=±±−±+x 1, ( 2 1), ( 2 1) 2 Illustration 36: Solve the system of inequalities involving inverse circular functions arc tan x− 3 arc tanx+> 2 0 −−11 and [sin x]> [cos x] where [ ] denotes the greatest integer function. (JEE ADVANCED) Sol: Substitute tan-1 x equal to t. ⇒−(t2) ( t1 −>) 0 ⇒>t 2 or t > 1 ⇒>tan−−11 x 2 or tan x > 1 x∈( −∞ , tan 1) x> tan1 Again [sin−−11 x]> [cos x] [sin−1 x] can take the values {−− 2, 1,0,1} And [cos−1 x] can take the values {0,1,2,3} Hence [sin−1 x] can be greater than [cos−1 x] only If [sin−1 x]= 1 and [cos−1 x]= 1 Now, [sin−−11 x]= 1 ⇒ 1 ≤ sin x ≤π / 2 (1 ≤ sin − 1 x < 2) sin1≤≤ x 1 And [cos−−11 x]=⇒≤ 0 0 cos x < 1 cos1<≤ x 1 Now, x must satisfy From this x∈ [sin1,1] Mathematics | 20.23 PROBLEM-SOLVING TACTICS •• Making a habit of writing angle values in radians rather in degrees makes the calculation of inverse trigonometric functions easier. •• Try to remember graphs of inverse trigonometric functions. Sometimes it is easier to approximate answers using graphical methods. •• Always verify whether the results are in the range or domain of the respective function. •• In some cases, constructing a right angled triangle for the given inverse function and then solving using properties of triangle is much helpful. •• In case of identities in inverse circular functions, principal values should be taken. As such signs of x, y, etc., will determine the quadrant in which the angles will fall. In order to bring the angles of both sides in the same quadrant, one should make an adjustment by π . FORMULAE SHEET 1. If y = sin x, then x = sin‒1 y, similarly for other inverse T-functions. 2. Domain and Range of Inverse T-functions: Function Domain(D) Range (R) −1 ππ sin x −≤1x1 ≤ − ≤θ ≤ 22 cos−1 x −≤1x1 ≤ 0 ≤θ ≤ π −1 −∞ cot−1 x −∞ −1 π sec x x≤− 1, x ≥ 1 0,≤θ ≤ π θ≠ 2 −1 ππ cosec x x≤− 1, x ≥ 1 − ≤ θ ≤,0 θ≠ 22 3. Properties of Inverse T-functions: ππ (i) sin−1 (sinθ ) =θ provided − ≤θ≤ 22 cos−1 (cosθ ) =θ provided 0 ≤θ≤π ππ tan−1 (tanθ ) =θ provided − <θ< 22 cot−1 (cotθ ) =θ provided 0 <θ<π 20.24 | Inverse Trigonometric Functions ππ sec−1 (secθ ) =θ provided 0≤θ< or <θ≤π 22 ππ cosec−1 (cosecθ ) =θ provided − ≤θ< 0 or 0 < θ ≤ 22 (ii) sin (sin−1 x)= x provided −≤ 1 x ≤ 1 cos (cos−1 x)= x provided −≤ 1 x ≤ 1 tan (tan−1 x)= x provided−∞ < x < ∞ cot (cot−1 x)= x provided−∞ < x < ∞ sec (sec−1 x)= x provided−∞ < x ≤− 1 or 1 ≤ x < ∞ cosec (cosec−1 x)= x provided−∞ < x ≤− 1 or 1 ≤ x < ∞ −−11 (iii) sin (−=− x) sin x, π (iv) sin−−11 x+ cos x = , ∀ x ∈− [ 1, 1] 2 cos−−11 (− x) =π− cos x −−11π −−11−=− tan x+ cot x = , ∀∈ x R tan ( x) tan x 2 −−11 cot (− x) = π− cot x π sec−−11 x+ cosec x = , ∀ x ∈ ( −∞ , − 1] ∪ [1, ∞ ) cosec−−11 (−=− x) cosec x 2 sec−−11 (− x) = π− sec x 4. Value of one inverse function in terms of another inverse function: 2 −−1 12 − 1x −1 1x− (i) sin x= cos 1 −= x tan =cot 1x− 2 x 11 =sec−−11 =cosec , 0≤≤ x 1 1x− 2 x 1x− 2 x (ii) cos−−1 x= sin 12 1 −= x tan − 1 =cot −1 x 1x− 2 11 =sec−−11 = cosec , 0≤≤ x 1 x 1x− 2 2 x 11−−12 11x+ (iii) tan−−11 x= sin = cos − 1= cot − 1=sec 1 += x cosec , x≥ 0 22x x 1x++ 1x −−111 (iv) sin= cosec x,∀ x ∈ ( −∞ ,1] ∪ [1, ∞ ) x −−111 (v) cos= sec x, ∀ x ∈ ( −∞ ,1] ∪ [1, ∞ ) x 1 cot−1 x for x> 0 (vi) tan−1 = −1 x −π+ cot x for x< 0 Mathematics | 20.25 5. Formulae for sum and difference of inverse trigonometric function: −−11 − 1xy+ (i) tan x+ tan y = tan ; if x>> 0, y 0, xy < 1 1− xy −−11 − 1xy+ (ii) tan x+ tan y =π+ tan ; if x> 0, y > 0, xy > 1 1− xy −−11 − 1xy+ (iii) tan x− tan y = tan ; if xy>− 1 1+ xy −−11 − 1xy− (iv) tan x− tan y =π+ tan ; if x> 0, y < 0, xy <− 1 1+ xy −−−−1111x++− y z xyz (v) tan x+ tan y += tan z tan 1−−− xy yz zx −−11 − 1 2 2 (vi) sin x± sin y = sin x 1 −± y y 1 − x ; If x,y,≥ 0 & x22 +≤ y 1 −−11 − 1 2 2 (vii) sin x± sin y =π− sin x 1 − y ± y 1 − x ; If x,y,≥ 0 & x22 +> y 1 −−1 1 − 1 22 (viii) cos x± cos y = cos xy 1 −− x 1 y ; If x,y,> 0 & x22 +≤ y 1 −−1 1 − 1 22 (ix) cos x± cos y =π− cos xy 1− x 1 − y ; If x,y,> 0 & x22 +> y 1 6. Inverse trigonometric ratios of multiple angles (i) 2sin−−11 x= sin (2x 1 − x 2 ), if −≤ 1 x ≤ 1 (ii) 2cos−−1 x= cos 12 (2x − 1), if −≤ 1 x ≤ 1 2x 2x 1− x2 (iii) 2tan−−11 x= tan =sin − 1=cos − 1 22 2 1x−+ 1x 1x + (iv) 3sin−−11 x= sin (3x − 4x 3 ) (v) 3cos−−1 x= cos 13 (4 x − 3x) 3x− x3 (vi) 3tan−−11 x= tan 2 1− 3x