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# Prove the following : $\large\frac{9\pi}{8}-\frac{9}{4}sin^{-1}\bigg(\large\frac{1}{3} \bigg) = \frac{9}{4}sin^{-1}\bigg(\large \frac{2\sqrt 2}{3} \bigg)$

This is Q.No. 12 Misc. of chapter 2

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 18, 2013