1 INVERSE TRIGONEMETRIC FUNCTION 09-04-2020 We know that all trigonometric functions are periodic in nature and hence are not one- one and onto i.e they are not bijections and therefore they are not invertible. But if we restrict their domain and co-domain in such a way that they become one –one and onto, then they become invertible and they have an inverse. Consider the function f:R → R given by f(x) = sinx. The domain of sinx is R and the codomain is π 1 5π 1 R. It is many one(because many values of x give the same image i.e f = and f = 6 2 6 2 and into function because range [-1,1] is the subset of codomain R. If we restrict the domain π π to − , and make the co-domain equal to range[-1,1], sine function becomes one -one and 2 2 onto i.e bijective and hence it becomes invertible. Actually sine function restricted to any of 3π π π π π3 π the intervals − , , − , , , is one –one and its range is [-1,1].We can define the 2 2 2 2 2 2 inverse of sine function in each of these intervals. Corresponding to each such intervals, we get π π a branch of sine inverse function. The branch with range − , is called principal value 2 2 π π branch .Thus f: − , → [-1,1] given by f(x) =sinx is invertible. Similarly all trigonometric 2 2 functions are invertible when their domain and co domain are restricted. Here are the list of all inverse trigonometric functions ( principal value branch ) with their domains and ranges. Function Domain Range sin−1 x [-1,1] π π − , 2 2 cos−1 x [-1,1] [0, π] tan−1 x R π π − , 2 2 cot−1 x R (0, π) sec−1 x R-(-1,1) π [0, π] -{ } 2 cos ec−1 x R-(-1,1) π π − , -{0} 2 2 NOTE: 2 1. The value of inverse trigonometric functions which lies in the range of principal value branch is called the principal value of that inverse trigonometric function. 2. First quadrant is common to all inverse trigonometric functions. 3. Third quadrant is not at all used in inverse trigonometric functions i.e the principal value of an inverse trigonometric functions never lies in the third quadrant. 4. All inverse trigonometric functions represents an angle or angles and not ratios. For 1 1 example, sin −1 means an angle or angles whose sin is . 2 2 5. sin−1 x ≠ (sinx) -1.In the symbol sin−1 x , -1 is not an algebraic exponent with -1 as an index, as it 1 would mean i.e cosecx which is ratio and not an angle. sin x 6. sin−1 x is also denoted as Arc sinx . 7. Principal branch of sin−1 x is denoted as arc sinx . Questions on principal value of the ITF: − 1 Example -Find the principle value of (i) sin 1 − (CBSE 2010) 2 −1 1 1 π π π π −1 1 π Soln: Let sin − = θ ⇒ sin θ = − ⇒ θ = − because − ∈ − , .Hence sin − = − 2 2 6 6 2 2 2 6 QUESTIONS: Find the principle value of the following functions: − 1 (i) cos 1 − (NCERT) 2 − 4π (ii) cos1 sin 3 −1 3 (iii) cos (NCERT) 2 −1 − 1 3 (iv) sin cos sin (NCERT EXEMPLAR) 2 Question on domains of ITF : Example (i) Find the domain of (i) f(x) = cos−1 (2x− 1) . 3 The domain of cos−1 x is [-1,1].Therefore f(x) = cos−1 (2x− 1) is defined for all values of x satisfying -1 ≤ 2x-1 ≤ 1 ⇒ 0 ≤ 2x ≤ 2 ⇒0 ≤ x ≤ 1. Hence the domain of f(x) = cos−1 (2x− 1) is [0,1] Example(ii) Find the domain of f(x) = sin−1 x− 1 The domain of sin−1 x is [-1,1].Therefore f(x) = sin−1 x− 1 is defined for all value of x satisfying - -1 ≤ x− 1 ≤ 1 ⇒0 ≤ x− 1 ≤ 1 ⇒0 ≤ x-1 ≤ 1 ⇒1 ≤ x ≤ 2. Hence the domain of f(x) = sin−1 x− 1 is [1,2] QUESTIONS: Find the domain of the following functions: 1.f(x) = cos−1 (2x− 1) 2. f(x) = cos−1 (x 2 − 4) 3.f(x) = 2 cos−1 2x +cosx 4. f(x) = sin−1 (2x− 3) 5. f(x) = sin−1 (− x 2 ) 6. f(x) = sin−1 x− 1 7. f(x) = sin−1 x 2 − 1 8. f(x) = sin−1 (x 2 ) ........................................................................................................................................................... .