Question
Write the value of $\sin^{-1}\Big(\cos\frac{\pi}{6}\Big).$
✓
Answer
$\sin^{-1}\Big(\cos\frac{\pi}{6}\Big)=\sin^{-1}\Big\{\sin\Big(\frac{\pi}{2}-\frac{\pi}{9}\Big)\Big\}$ $\Big[\because\ \cos\text{x}=\sin\Big(\frac{\pi}{2}-\text{x}\Big)\Big]$
$=\sin^{-1}\Big\{\sin\Big(\frac{7\pi}{18}\Big)\Big\}$
$=\frac{7\pi}{18}$ $[\because\ \sin^{-1}(\sin\text{x})=\text{x}]$
$\therefore\ \sin^{-1}\Big(\cos\frac{\pi}{9}\Big)=\frac{7\pi}{18}$
$=\sin^{-1}\Big\{\sin\Big(\frac{7\pi}{18}\Big)\Big\}$
$=\frac{7\pi}{18}$ $[\because\ \sin^{-1}(\sin\text{x})=\text{x}]$
$\therefore\ \sin^{-1}\Big(\cos\frac{\pi}{9}\Big)=\frac{7\pi}{18}$
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