Question:
Write the value of $\sin ^{-1}\left(\cos \frac{\pi}{9}\right)$.
Solution:
Consider,
$\sin ^{-1}\left(\cos \frac{\pi}{9}\right)=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\frac{\pi}{9}\right)\right\} \quad\left[\because \cos x=\sin \left(\frac{\pi}{2}-x\right)\right]$
$=\sin ^{-1}\left\{\sin \left(\frac{7 \pi}{18}\right)\right\}$
$=\frac{7 \pi}{18} \quad\left[\because \sin ^{-1}(\sin x)=x\right]$
$\therefore \sin ^{-1}\left(\cos \frac{\pi}{9}\right)=\frac{7 \pi}{18}$