Answer
$\dfrac{\sqrt 3}{2}$
Work Step by Step
Suppose $\theta =\sin^{-1} (\dfrac{1}{2})$
This gives: $\sin \theta=\dfrac{1}{2}$
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ x=\sqrt {r^2-y^2}=\sqrt {2^2-(1)^2}=\sqrt 3$
Therefore, we have $\cos \theta =\cos [\sin^{-1} (\dfrac{1}{2})]=\dfrac{\sqrt 3}{2}$
