# How do you solve and find the value of sin(sin^-1(1/2))?

Dec 16, 2016

$\sin \left({\sin}^{- 1} \left(\frac{1}{2}\right)\right) = \frac{1}{2}$
${\sin}^{- 1} \left(\frac{1}{2}\right)$ means an angle $\theta$ so that $\sin \theta = \frac{1}{2}$
As we have $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$, ${\sin}^{- 1} \left(\frac{1}{2}\right) = \frac{\pi}{6}$
and $\sin \left({\sin}^{- 1} \left(\frac{1}{2}\right)\right) = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$