Answer
$\dfrac{\sqrt 2}{2}$
Work Step by Step
Suppose $\theta =\cos^{-1} (\dfrac{\sqrt 2}{2})$
This gives: $\cos \theta=\dfrac{\sqrt 2}{2}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ y=\sqrt {r^2-x^2}=\sqrt {2^2-(\sqrt 2)^2}=\sqrt 2$
Therefore, we have $\sin \theta =\sin [\cos^{-1} (\dfrac{\sqrt 2}{2})]=\dfrac{\sqrt 2}{2}$
