Question

Prove: $\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}$

Answer 1

$L.H.S.=\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}$

$=\frac{9}{4}(\frac{\pi }{2}-{\sin }^{-1}\frac{1}{3})$

$=\frac{9}{4}\left({\cos }^{-1}\frac{1}{3}\right).....\left(1\right)$

Now, let ${\cos }^{-1}\frac{1}{3}=x$.

Then, $\cos x=\frac{1}{3}\Rightarrow \sin x=\sqrt{1-{\left(\frac{1}{3}\right)}^2}=\frac{2\sqrt{2}}{3}$

$\therefore x={\sin }^{-1}\frac{2\sqrt{2}}{3}\Rightarrow {\cos }^{-1}\frac{1}{3}={\sin }^{-1}\frac{2\sqrt{2}}{3}$

$\therefore \text{L.H.S.}=\frac{9}{4}{\sin }^{-1}\frac{2\sqrt{2}}{3}=\text{R.H.S.}$