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

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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