# How do you calculate the cos[2 sin^ -1 (1/2)]?

Feb 26, 2016

$\frac{1}{2}$

#### Explanation:

First, focus on just ${\sin}^{-} 1 \left(\frac{1}{2}\right)$. This is another way of asking, for what angle is sine equal to $\text{1/2}$? Recall that

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

Thus, since $\sin$ and $\arcsin$ are inverse functions,

${\sin}^{-} 1 \left(\frac{1}{2}\right) = \frac{\pi}{6}$

Substitute this into the original expression:

$\cos \left[2 {\sin}^{-} 1 \left(\frac{1}{2}\right)\right] = \cos \left(2 \cdot \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$