Answer
The exact value of expression is $\frac{\pi }{3}$.
Work Step by Step
By using the inverse property,
${{\sin }^{-1}}\left( \sin x \right)=x$
Therefore, the above property is applicable for every $x$ in $\left[ \frac{-\pi }{2},\frac{\pi }{2} \right]$.
Then, simplify ${{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)$. We see that $\frac{2\pi }{3}$ is not in the interval.
So, first simplify $\left( \sin \frac{2\pi }{3} \right)$. Use the property:
$\left( \sin \frac{\pi }{2}+\theta \right)=\cos \theta $
Put the $\left( \sin \frac{2\pi }{3} \right)$ in the above property.
So,
$\begin{align}
& \left( \sin \frac{\pi }{2}+\frac{\pi }{6} \right)=\cos \frac{\pi }{6} \\
& =\frac{\sqrt{3}}{2}
\end{align}$
Finally, simplify
${{\sin }^{-1}}\frac{\sqrt{3}}{2}=\frac{\pi }{3}$
Thus, ${{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)=\frac{\pi }{3}$.
