# How do you evaluate arc cos(sin((pi/3))?

$\arccos \left(\sin \left(\frac{\pi}{3}\right)\right) = \arccos \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$
The restriction for the range of arccos x is $\left[0 , \pi\right]$ and since $\frac{\pi}{3}$ is in the restriction we find the ratio for $\sin \left(\frac{\pi}{3}\right)$ which is $\frac{\sqrt{3}}{2}$. Then $\arccos \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$