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# The value of $\sin^{-1}\bigg(\sin\frac{3\pi}{5}\bigg)$ is ________________.

$\begin{array}{1 1} \frac{3 \pi}{5} \\ \frac{-3 \pi }{5} \\ \frac{2 \pi}{5} \\ \frac{-2\pi}{5} \end{array}$

Toolbox:
• Principal interval of sin is $[\large\frac{\pi}{2},-\large\frac{\pi}{2}$]
• $sin(\pi-\theta)=sin\theta$
Ans : $\large \frac{2\pi}{5}$

$\large\frac{3\pi}{5}$ is not in the principal interval of sin.
Therefore, reduce it to its principal interval

$\large\frac{3\pi}{5}= \pi - \large\frac{2\pi}{5}$
$sin\large\frac{3\pi}{5} =sin(\pi- \large\frac{2\pi}{5})=sin\large\frac{2\pi}{5}$
$\large\frac{2\pi}{5}$ is within the principal interval

$\Rightarrow\:sin^{-1}sin\large\frac{3\pi}{5}=sin^{-1}sin\large\frac{2\pi}{5}=\large\frac{2\pi}{5}$

edited Mar 16, 2013