Inverse Trigonometric Functions — Class 12 Maths Solution

${\sin ^1}\left( {\sin \frac{{2\pi }}{3}} \right) \ne \frac{{2\pi }}{3}$ as the principal value branch of
${\sin ^{ - 1}}\;is\;\left[ { - \frac{\pi }{2},\;\,\frac{\pi }{2}} \right]$

So, ${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left( {\sin \left( {\pi - \frac{\pi }{3}} \right)} \right)$

$= {\sin ^{ - 1}}\left( {\sin \frac{\pi }{3}} \right) = \frac{\pi }{3} \in \left[ { - \frac{\pi }{2},\;\frac{\pi }{2}} \right]$

Hence, ${\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right) = \frac{\pi }{3}$