How do you evaluate arcsin(sin 2)?

May 19, 2015

The range of arcsin is $- \frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.

$2 > \frac{\pi}{2}$ so it lies outside the range.

Use $\sin \left(\pi - \theta\right) = \sin \theta$

Then $\sin 2 = \sin \left(\pi - 2\right)$.

$\pi - 2 \cong 1.14$ so $0 < \pi - 2 < \frac{\pi}{2}$

So $\arcsin \left(\sin \left(2\right)\right) = \pi - 2$