# Prove that 9π/8 - 9/4 sin^{- 1} 1/3 = 9/4 sin^{- 1} 2√2/3

**Solution:**

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios

Here the basic trigonometric function of Sin θ = y, can be changed to θ = sin^{-1} y

LHS =

9π / 8 - 9/4 sin^{- 1} 1/3

= 9/4 (π/2 - sin^{- 1} 1/3)

= 9/4 (cos^{- 1} 1/3) ....(1)

Now, let cos^{- 1} 1/3 = x

⇒ cos x = 1/3

Therefore,

sin x = √ [1 - (1/3)^{2}]

= 2√2 / 3

x = sin^{- 1} 2√2 / 3

cos^{- 1} 1/3 = sin^{- 1} 2√2/3 ....(2)

Thus, by using (1) and (2)

9π/8 - 9/4 sin^{- 1} 1/3

= 9/4 sin^{- 1} 2√2/3

Hence Proved

NCERT Solutions for Class 12 Maths - Chapter 2 Exercise ME Question 12

## Prove that 9π/8 - 9/4 sin^{- 1} 1/3 = 9/4 sin^{- 1} 2√2/3

**Summary:**

Hence we have proved by using inverse trigonometric functions that 9π/8 - 9/4 sin^{- 1} 1/3 = 9/4 sin^{- 1} 2√2/3

Math worksheets and

visual curriculum

visual curriculum