Question

Prove that: `(9π)/8 - 9/4 sin^-1 (1/3) = 9/4 sin^-1 ((2sqrt(2))/3)`

Answer 1

To prove: `(9π)/8 - 9/4 sin^-1 (1/3) = 9/4 sin^-1 ((2sqrt(2))/3)`

L.H.S. = `(9π)/8 - 9/4 sin^-1  1/3`

= `9/4 [π/2 - sin^-1  1/3]`

`["Using", sin^-1x + cos^-1x = π/2 ⇒ cos^-1x = π/2 - sin^-1x]`

= `9/4 cos^-1 (1/3)`   ...(i)

Let `cos^-1  1/3 = θ`

`cos θ = 1/3`

sin²θ + cos²θ = 1

`sin^2θ + (1/3)^2 = 1`

`sin^2θ = 1 - 1/9`

`sin^2θ = 8/9`

`sin θ = (2sqrt(2))/3`

`θ = sin^-1  (2sqrt(2))/3`

`cos^-1 (1/3) = sin^-1 ((2sqrt(2))/3)`

Now, from equation (i) = `9/4 sin^-1 ((2sqrt(2))/3)`

= R.H.S.

Hence proved.