# How do you find the exact value of sin^-1(-0.5)?

Apr 15, 2015

${\sin}^{-} 1 \left(- 0.5\right) = {\sin}^{-} 1 \left(- \frac{1}{2}\right)$

Is there a special angle with $\sin t = \frac{1}{2}$?

Yes, $t = \frac{\pi}{6}$.

But we want $\sin t = - \frac{1}{2}$, so we want an answer in quadrant 3 or 4.

${\sin}^{-} 1 \left(x\right)$ always gives us a $t$ between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$, so we want $t = - \frac{\pi}{6}$

${\sin}^{-} 1 \left(- 0.5\right) = \frac{\pi}{6}$