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.

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

Find the exact value without a calculator :

Example 6 :

cos (sin⁻¹(1/2))

Solution :

= cos (sin⁻¹(1/2))

For sin 30 degree we get 1/2 and the angle lies in the interval [-π/2, π/2]

= cos (30)

Value of cos 30 = √3/2

Example 7 :

sin (cos⁻¹(√2/2))

Solution :

= sin (cos⁻¹(√2/2))

For sin 45 degree we get √2/2 and the angle lies in the interval  [0, π].

= sin (45)

Value of sin (cos⁻¹(√2/2)) = √2/2

Example 8 :

sin^-1 (cos(π/3))

Solution :

= sin^-1 (cos(π/3))

cos (π/3) = √3/2

Applying the value of cos (π/3), we get

= sin^-1 (√3/2)

= π/3

So, the value of sin^-1 (cos(π/3)) is π/3.

Example 9 :

cos^-1 (sin (π/6))

Solution :

= cos^-1 (sin (π/6))

sin (π/6) = 1/2

Applying the value of sin (π/6), we get

= cos^-1 (1/2)

= π/3

So, the value of cos^-1 (sin (π/6)) is π/3.

Example 10 :

sin^-1 (sin (7π/4))

Solution :

= sin^-1 (sin (7π/4))

sin (7π/4) = sin (π + 3π/4)

= -sin (3π/4)

= -sin (π - π/4)

= - sin (π/4)

sin (7π/4) = -√2/2

Applying the value of sin (7π/4), we get

= sin^-1 (-√2/2)

= -π/4

So, the value of sin^-1 (sin (7π/4)) is π/4.

Example 11 :

sin (tan^-1 (√3))

Solution :

= sin (tan^-1 (√3))

tan^-1 (√3) = π/3

= sin (π/3)

= √3/2

So, the value of sin (tan^-1 (√3)) is √3/2.

Example 12 :

cos (tan^-1 (-1))

Solution :

= cos (tan^-1 (-1))

tan^-1 (-1) = -π/4

= cos (π/4)

= √2/2

So, the value of cos (tan^-1 (-1)) is √2/2.

Example 13 :

tan^-1 (cos π)

Solution :

= tan^-1 (cos π)

cos π = -1

Applying the value of cos π, we get

= tan^-1 (-1)

= -π/4

So, the value of tan^-1 (cos π) is -π/4.

Find the inverse of each function and list the domain and range of 𝒇^−𝟏(𝒙)

Example 14 :

𝑓(𝑥) = 2 sin 𝑥 + 1 for −𝜋/2 ≤ 𝑥 ≤ 𝜋/2

Solution :

𝑓(𝑥) = 2 sin 𝑥 + 1

Let y = 𝑓(𝑥) = 2 sin 𝑥 + 1

y = 2 sin x + 1

2 sin x = y - 1

sin x = (1/2) y - 1

x = sin^-1 [(y - 1)/2]

f^-1(x) = sin^-1 [(x - 1)/2]

Domain for sin^-1 function -1 ≤ 𝑥 ≤ 1

Domain of the function sin^-1 [(x - 1)/2] is -1 ≤ 𝑥 ≤ 3.

Range is −𝜋/2 ≤ 𝑥 ≤ 𝜋/2 (domain of the original function).

Example 15 :

𝑓(𝑥) = (1/3) cos 2𝑥 for 0 ≤ 𝑥 ≤ 𝜋/2

Solution :

𝑓(𝑥) = (1/3) cos 2𝑥

Let y = (1/3) cos 2𝑥

cos 2x = 3y

2x = cos^-1(3y)

x = (1/2) cos^-1(3y)

f^-1(x) = (1/2) cos^-1(3x)

Domain for cos^-1 function -1 ≤ 𝑥 ≤ 1

Domain of the function (1/2) cos^-1(3x) is -1/3 ≤ 𝑥 ≤ 1/3.

Range is 0 ≤ 𝑥 ≤ 𝜋/2 (domain of the original function).

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