HOW TO FIND PRINCIPAL VALUE OF INVERSE TRIGONOMETRIC FUNCTIONS

Principal Angle :

The inverse functions sin−1 x, cos−1 x, tan−1 x, cosec−1(x), sec−1(x), cot−1(x) are called inverse circular functions. For the function y = sinx, there are infinitely many angles x which satisfy sin x = t, −1 ≤ t ≤ 1. Of these infinite set of values, there is one which lies in the interval [-π/2, π/2].

This angle is called the principal angle and denoted by sin−1 t. The principal value of an inverse function is that value of the general value which is numerically least. It may be positive or negative.

When there are two values, one is positive and the other is negative such that they are numerically equal, then the principal value is the positive one.

Inverse function

Domain

Range

sin-1 x

cos-1 x

tan-1 x

cot-1 x

cosec-1 x

sec-1 x

[-1, 1]

[-1, 1]

(-∞, ∞)

(-∞, ∞)

R-(-1, 1)

R-(-1, 1)

[-π/2, π/2]

[0, π]

(-π/2, π/2)

(0, π)

[-π/2, π/2]

[0, π] - {π/2}

Find the principal value of each of the following : 

Example 1 :

sin−1(1/√2)

Solution :

Let x = sin−1(1/√2).

sin x =  1/√2

sin x =  sin 45

x = π/4, where x ∊ [-π/2, π/2]

Example 2 :

cos−1(√3/2)

Solution :

Let x = cos−1(√3/2).

cos x =  √3/2

cos x =  cos π/6

x = π/6, where x ∊ [0, π]

Example 3 :

csc−1(-1)

Solution :

Let x = csc−1(-1).

Range of csc-1(x) is[-π/2, π/2], so the required angle lies in the above interval. 

cosec x = cosec(-π/2) 

x = -π/2, where x ∊ [-π/2, π/2]

Example 4 :

sec−1(-√2)

Solution :

Let x = sec−1(-√2).

Range of sec-1(x) is [0, π] - {π/2},  so the required angle lies in the above interval. 

The required angle lies in the 2nd quadrant.

sec x = sec(π - π/4) 

sec x = sec(3π/4) 

x = 3π/4, where x ∊ [0, π] - {π/2}

Example 5 :

tan−1(√3)

Solution :

Let x = tan−1(√3).

Range of tan-1(x) is (-π/2, π/2), so the required angle lies in the above interval. 

tan x = tan (π/3)

x = π/3

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