# How do you evaluate arcsin(sin((7*pi)/3))?

Jun 8, 2018

$\arcsin \sin \left(\frac{7 \pi}{3}\right) = \frac{7 \pi}{3} + 2 \pi k \mathmr{and} - \frac{\pi}{3} + 2 \pi k \quad$ integer $k .$

$\textrm{A r c} \textrm{\sin} \setminus \sin \left(\frac{7 \pi}{3}\right) = \frac{\pi}{3}$

#### Explanation:

If we treat $\arcsin$ as multivalued then

$x = \arcsin \sin a$

is the same as

$\sin x = \sin a$

which has supplementary angle solutions

$x = a + 2 \pi k \quad \mathmr{and} \quad x = \left(\pi - a\right) + 2 \pi k \quad$ for integer $k$

$x = \arcsin \sin \left(\frac{7 \pi}{3}\right)$

$\sin x = \sin \left(\frac{7 \pi}{3}\right)$

$x = \frac{7 \pi}{3} + 2 \pi k \mathmr{and} x = \left(\pi - \frac{7 \pi}{3}\right) + 2 \pi k$

$x = \frac{7 \pi}{3} + 2 \pi k \mathmr{and} x = - \frac{\pi}{3} + 2 \pi k$

$\arcsin \sin \left(\frac{7 \pi}{3}\right) = \frac{7 \pi}{3} + 2 \pi k \mathmr{and} - \frac{\pi}{3} + 2 \pi k \quad$ integer $k .$

The principal value for the inverse sine is in the range $- \frac{\pi}{2}$ to $\frac{\pi}{2}$ so we take $k = - 1$ in the first clause and get

$\textrm{A r c} \textrm{\sin} \setminus \sin \left(\frac{7 \pi}{3}\right) = \frac{\pi}{3}$

[Had this one sitting around in my browser; I forget to hit post.]