# How do you solve the inverse trig function sin(sin^-1 (1/3))?

${\sin}^{-} 1 \left(\frac{1}{3}\right)$ or $\arcsin \left(\frac{1}{3}\right)$ is the angle $\theta$ for which $\sin \left(\theta\right) = \frac{1}{3}$
Therefore $\sin \left({\sin}^{-} 1 \left(\frac{1}{3}\right)\right)$
$= \sin \left(\theta\right)$ for the value of $\theta$ for which $\sin \left(\theta\right) = \frac{1}{3}$
$\sin \left({\sin}^{-} 1 \left(\frac{1}{3}\right)\right) = \frac{1}{3}$