Question

Find the values of the given expression.

$si{n}^{-1}\left(sin\frac{2\pi }{3}\right)$

Answer 1

$si{n}^{-1}\left(sin\frac{2\pi }{3}\right)=si{n}^{-1}\left[sin(\pi -\frac{\pi }{3})\right]=si{n}^{-1}\left(sin\frac{\pi }{3}\right)[\because sin(\pi -\theta )=sin\theta ]$
$=\frac{\pi }{3}$ which lies in the range $[-\frac{\pi }{2},\frac{\pi }{2}]$.

Note $si{n}^{-1}\left(sin\theta \right)=\theta$ is only when $\theta \epsilon [-\frac{\pi }{2},\frac{\pi }{2}]$