# How do you evaluate arcsin(sin3)?

Aug 21, 2015

$\arcsin \left(\sin \left(3\right)\right) = \pi - 3$

#### Explanation:

Note that $\arcsin$ maps $\left[- 1 , 1\right]$ to $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$.

We need to find an angle $x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ where $\sin \left(x\right) = \sin \left(3\right)$.

Since $\sin \left(x\right)$ is symmetric about $x = \frac{\pi}{2}$ we have $\sin \left(3\right) = \sin \left(\pi - 3\right)$. $\pi - 3$ is in $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, therefore we have:

$\arcsin \left(\sin \left(3\right)\right) = \pi - 3$