# How do you evaluate sin^-1(sin((19pi)/10))?

Jul 26, 2017

$- \frac{\pi}{10}$

#### Explanation:

${\sin}^{-} 1 \left(x\right)$ is a number (or angle), $t$ in $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
with $\sin \left(t\right) = x$

So,

${\sin}^{-} 1 \left(\sin \left(\frac{19 \pi}{10}\right)\right)$ is a number (or angle), $t$ in $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
with $\sin \left(t\right) = \sin \left(\frac{19 \pi}{10}\right)$

$\frac{19 \pi}{10}$ is almost $2 \pi$, so it is a fourth quadrant angele.

The reference angle for $\frac{19 \pi}{10}$ is $\frac{\pi}{10}$.

So $t = - \frac{\pi}{10}$