Question

Show 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$=\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}$

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

using ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

${\sin }^{-1}x=\frac{\pi }{2}-{\cos }^{-1}x$

Replace $x$ by $\frac{1}{3}$ we get

${\sin }^{-1}\frac{1}{3}=\frac{\pi }{2}-{\cos }^{-1}\frac{1}{3}$ .....$\left(2\right)$

Substituting $\left(2\right)$ in $\left(1\right)$ we get

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

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

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

Let $a={\cos }^{-1}\frac{1}{3}$

$\Rightarrow \cos a=\frac{1}{3}$

Now,${\sin }^{2}a=1-{\cos }^{2}a$

$\Rightarrow \sin a=\sqrt{1-{\cos }^{2}a}$

$=\sqrt{1-{\left(\frac{1}{3}\right)}^2}=\sqrt{1-\frac{1}{9}}=\frac{9-1}{9}=\frac{8}{9}=\frac{2\sqrt{2}}{3}$

$\therefore \sin a=\frac{2\sqrt{2}}{3}$

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

Now, from $\left(3\right)$

$\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\frac{1}{3}=\frac{9}{4}{\cos }^{-1}\frac{1}{3}$

Putting the value

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

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

Hence proved.