# How do you evaluate cos^-1(-sqrt3/2) without a calculator?

Sep 26, 2016

${\cos}^{-} 1 \left(- \frac{\sqrt{3}}{2}\right) = \frac{5 \pi}{6.}$

#### Explanation:

Let ${\cos}^{1} \left(- \frac{\sqrt{3}}{2}\right) = \theta$.

Then, by Defn. of ${\cos}^{-} 1$ fun., we have, $\cos \theta = - \frac{\sqrt{3}}{2}$, and,

$\theta \in \left[0 , \pi\right] = \left[0 , \frac{\pi}{2}\right] \cup \left[\frac{\pi}{2} , \pi\right] .$ But, as $\cos \theta < 0 , \theta \in \left[\frac{\pi}{2} , \pi\right] .$

We know that, $\cos \left(\frac{5 \pi}{6}\right) = \cos \left(\pi - \frac{\pi}{6}\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$.

Thus, $\cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2} , \mathmr{and} , \left(5 \frac{\pi}{6}\right) \in \left[\frac{\pi}{2} , \pi\right] \subset \left[0 , \pi\right]$

Hence, by Defn., ${\cos}^{-} 1 \left(- \frac{\sqrt{3}}{2}\right) = \frac{5 \pi}{6.}$