# What is the Inverse cos(cos(-pi/3)?

Aug 27, 2015

$\arccos \left(\cos \left(- \frac{\pi}{3}\right)\right) = \frac{\pi}{3}$

#### Explanation:

In order that $\arccos$ be a well-defined function, its range is defined to be $\left[0 , \pi\right]$.

So if $\theta = \arccos \left(\cos \left(- \frac{\pi}{3}\right)\right)$, then $\cos \left(\theta\right) = \cos \left(- \frac{\pi}{3}\right)$ and $\theta \in \left[0 , \pi\right]$

For any $\theta$ we have $\cos \left(\theta\right) = \cos \left(- \theta\right)$, so we can see that in our case $\theta = \frac{\pi}{3}$ satisfies the required conditions to be $\arccos \left(\cos \left(- \frac{\pi}{3}\right)\right)$

Here's the graph of $\arccos \left(\theta\right)$

graph{arccos(x) [-5.168, 4.83, -0.86, 4.14]}