Write the value

Question:

Write the value of $\sin ^{-1}\left(\cos \frac{\pi}{9}\right)$.

Solution:

Consider,

$\sin ^{-1}\left(\cos \frac{\pi}{9}\right)=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\frac{\pi}{9}\right)\right\} \quad\left[\because \cos x=\sin \left(\frac{\pi}{2}-x\right)\right]$

$=\sin ^{-1}\left\{\sin \left(\frac{7 \pi}{18}\right)\right\}$

$=\frac{7 \pi}{18} \quad\left[\because \sin ^{-1}(\sin x)=x\right]$

$\therefore \sin ^{-1}\left(\cos \frac{\pi}{9}\right)=\frac{7 \pi}{18}$

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