Answer
$\dfrac{\pi}{4}$
Work Step by Step
Suppose $\theta = \sin^{-1} (\dfrac{\sqrt 2}{2})$
This gives: $\sin \theta= \dfrac{\sqrt 2}{2}$
Now, $\theta = \sin^{-1} (\dfrac{\sqrt 2}{2})=\dfrac{\pi}{4}$
Therefore, $\sin^{-1} (\dfrac{\sqrt 2}{2})=\dfrac{\pi}{4}$
