Question

Prove that $\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\left(\frac{1}{3}\right)=\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$

Answer 1

$L.H.S.$

$\frac{9\pi }{8}-\frac{9}{4}si{n}^{-1}\left(\frac{1}{3}\right)$

$=\frac{9}{4}(\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{3}\right))\therefore {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

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

$=\frac{9}{4}\left({\sin }^{-1}\sqrt{1-{\left(\frac{1}{3}\right)}^2}\right)\therefore {\cos }^{-1}x={\sin }^{-1}\sqrt{1-x^2}$

$=\frac{9}{4}\left({\sin }^{-1}\sqrt{\frac{8}{9}}\right)$

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

$R.H.S.$