Inverse Trigonometric Functions — Class 12 Maths Solution

We have,
${\sin ^{ - 1}}\left( {\cos \frac{{33\pi }}{5}} \right) = {\sin ^{ - 1}}\left[ {\cos \left( {6\pi + \frac{{3\pi }}{5}} \right)} \right] = {\sin ^{ - 1}}\left[ {\cos \left( {\frac{{3\pi }}{5}} \right)} \right]$

$= {\sin ^{ - 1}}\left[ {\cos \left( {\frac{\pi }{2} + \frac{\pi }{{10}}} \right)} \right] = {\sin ^{ - 1}}\left( { - \sin \frac{\pi }{{10}}} \right)$
$= - {\sin ^{ - 1}}\left( {\sin \frac{\pi }{{10}}} \right)$

$= - \frac{\pi }{{10}}$