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

Toolbox:
• $$sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$$
• $$sin^{-1}1=\large\frac{\pi}{2}$$
Given $\frac{9 \pi}{8} - \large\frac{9}{4} sin ^{-1}\large\frac{1}{3}= \large\frac{9}{4} sin^{-1}\large\frac{2\sqrt 2}{3}$

Rearrranging the terms, we need to prove that $$\large\frac{\pi}{8}=\large\frac{1}{4} \bigg[ sin^{-1}\large\frac{2\sqrt 2}{3}+sin^{-1}\large\frac{1}{3} \bigg]$$

R.H.S. = $$\large\frac{1}{4} \bigg[ sin^{-1}\large\frac{2\sqrt 2}{3}+sin^{-1}\large\frac{1}{3} \bigg]$$
We know that $$sin^{-1}x+sin^{-1}y=sin^{-1} \bigg[ x\sqrt{1-y^2}+y\sqrt{1-x^2} \bigg]$$

By taking $$x=\large\frac{2\sqrt 2}{3}\:and\:y=\large\frac{1}{3}$$
$$\sqrt{1-y^2}=\sqrt{1-\large\frac{1}{9}}=\sqrt{\large\frac{8}{9}}=\large\frac{2\sqrt{2}}{3}$$
$$\sqrt{1-x^2}=\sqrt{1-\large\frac{8}{9}}=\large\frac{1}{3}$$

$$\Rightarrow\:$$ R.H.S. = $$\large\frac{1}{4} \bigg[ sin^{-1} \bigg(\large \frac{2\sqrt 2}{3} \sqrt{1-\large\frac{1}{9}} +\large\frac{1}{3} \sqrt{1-\frac{8}{9}} \bigg) \bigg]$$
$$=\large\frac{1}{4}sin^{-1}\bigg(\large\frac{2\sqrt{2}}{3}.\large\frac{2\sqrt{2}}{3}+\large\frac{1}{3}.\large\frac{1}{3}\bigg)$$
$$= \large\frac{1}{4}sin^{-1} \bigg(\large\frac{8}{9}+\large\frac{1}{9}\bigg)$$

$$\large\frac{1}{4}sin^{-1}1=\large\frac{1}{4}.\large\frac{\pi}{2}=\large\frac{\pi}{8}$$ =L.H.S

edited Mar 15, 2013