# How do you solve the inverse trig function arcsin (sin 5pi/6)?

May 20, 2015

When $- 1 \le x \le 1$, $\arcsin \left(x\right)$ is defined as the value $\theta$

in the range $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ such that $\sin \theta = x$.

The value $\frac{5 \pi}{6}$ does not fall into this range.

$\frac{\pi}{2} < \frac{5 \pi}{6} < \pi$
Use the equality $\sin \left(\pi - \theta\right) = \sin \theta$ and note that
$\frac{5 \pi}{6} = \pi - \frac{\pi}{6}$
So $\sin \left(\frac{\pi}{6}\right) = \sin \left(\pi - \frac{\pi}{6}\right) = \sin \left(\frac{5 \pi}{6}\right)$
So $\arcsin \left(\sin \left(\frac{5 \pi}{6}\right)\right) = \frac{\pi}{6}$