# How do you calculate cos^-1(sqrt3/2)?

Since the range of the function $y = \arccos x$ is $\left[0 , \pi\right]$ and the value positive $\frac{\sqrt{3}}{2}$, the angle is in the first quadrant.
${\cos}^{-} 1 \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.