Answer
$\sqrt {1-4x^2}$
Work Step by Step
Suppose $\theta =\sin^{-1} (2x)$
This gives: $\sin \theta=2x $
Since, $ r=\sqrt {x^2+y^2}$
This implies that $ a=\sqrt {(\dfrac{1}{2})^2-x^2}=\dfrac{\sqrt {1-4x^2}}{2}$
Therefore, we have $\cos [\sin^{-1} (2x)]=\sqrt {1-4x^2}$
