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

Toolbox:
• Principal interval of sin is $[-\frac{\pi}{2},\frac{\pi}{2}]$
• $sin(\pi-\theta)=sin\theta$
$\frac{3\pi}{5}\notin\:[-\frac{\pi}{2},\frac{\pi}{2}]$ which is principal interval of sin
Therefore, reduce it within the principal interval.
We know that $sin(\pi-\theta)=sin\theta$ By taking $\theta=\frac{3\pi}{5}$ we get
$sin\frac{3\pi}{5}=sin(\pi-\frac{3\pi}{5})$
$=sin\frac{2\pi}{5}$
$\frac{2\pi}{5}\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$\Rightarrow\:sin^{-1}sin\frac{3\pi}{5}=sin^{-1}sin\frac{2\pi}{5}$
=$\frac{2\pi}{5}$