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

Dec 13, 2016

$\frac{\pi}{6} + 2 k \pi$
$\frac{5 \pi}{6} + 2 k \pi$

#### Explanation:

${\sin}^{-} 1 \left(\frac{1}{2}\right) - \to \arcsin \left(\frac{1}{2}\right)$
Trig table of special arcs gives -->
$\sin x = \frac{1}{2}$ --> arc $x = \frac{\pi}{6}$
Unit circle gives another arc that has the same sin value (1/2) -->
arc $x = \pi - \frac{\pi}{6} = \frac{5 \pi}{6}$
$x = \frac{\pi}{6} + 2 k \pi$
$x = \frac{5 \pi}{6} + 2 k \pi$