Question
Using principal value, evaluate the following:$\sin^{-1} \bigg( \sin \frac{3{\pi}}{5}\bigg)$

Answer

$\frac{3{\pi}}{5} = \pi - \frac{2{\pi}}{5}$$\therefore \sin^{-1} \bigg( \sin \frac{3{\pi}}{5}\bigg)$
$= \sin^{-1} \Bigg[ \sin\bigg( \pi - \frac{2{\pi}}{5}\bigg)\Bigg]$
$= \sin^{-1} \bigg[ \sin\frac{2\pi}{5}\bigg] = \frac{2\pi}{5} \in \bigg[ - \frac{\pi}{2}, \frac{\pi}{2}\bigg]$
 

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