Question

Prove: $\frac {9\pi}{4} - \frac 94 sin^{-1}\frac 13 = \frac 94sin^{-1}\frac {2\sqrt 2}{3}$

Answer 1

L.H.S = $\frac {9\pi}8{}$ - $\frac 94$sin-1$\frac 13$
=$\frac 94$($\frac {\pi}{2}$ - sin-1$\frac 13$)
=$\frac 94$(cos-1$\frac 13$) ……....(1) [sin-1x + cos-1x = $\frac {\pi}{2}$]
Now, let cos-1$\frac 13$ = x. Then, cos x = $\frac 13$ $\implies$sin x = $\sqrt {1-(\frac 13)^2}$ = $\frac {2\sqrt 2}{3}$.
Therefore x = sin-1$\frac {2\sqrt 2}{3}$
Therefore L.H.S = $\frac 94$sin-1$\frac {2\sqrt 2}{3}$
=R.H.S