Solve this

Question:

(i) $\sin ^{-1} \frac{1}{2}-2 \sin ^{-1} \frac{1}{\sqrt{2}}$

(ii) $\sin ^{-1}\left\{\cos \left(\sin ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$

Solution:

(i)

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

$=\sin ^{-1} \frac{1}{2}-\sin ^{-1} \sqrt{2} \times \frac{1}{\sqrt{2}}$

$=\sin ^{-1} \frac{1}{2}-\sin ^{-1} 1$

$=\sin ^{-1}\left(\sin \frac{\pi}{6}\right)-\sin ^{-1}\left(\sin \frac{\pi}{2}\right)$

$=\frac{\pi}{6}-\frac{\pi}{2}$

$=-\frac{\pi}{3}$

(ii)

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

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

$=\sin ^{-1}\left\{\frac{1}{2}\right\}$

$=\sin ^{-1}\left\{\sin \frac{\pi}{6}\right\}$

$=\frac{\pi}{6}$

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