# How do you evaluate arccos(cos(5pi/4))?

Jul 29, 2015

Evaluate $\arccos \left(\cos \left(\frac{5 \pi}{4}\right)\right)$
Ans:$\pm \frac{3 \pi}{4}$

#### Explanation:

$\cos \left(\frac{5 \pi}{4}\right) = \cos \left(\frac{\pi}{4} + \pi\right) = - \cos \frac{\pi}{4} = - \frac{\sqrt{2}}{2}$

$a r c x = \arccos \left(\frac{- \sqrt{2}}{2}\right)$ ---> $x = \pm \frac{3 \pi}{4}$

Jul 29, 2015

$\frac{3 \pi}{4}$

#### Explanation:

$\arccos x$ can be thought of as an angle that measures between $0$ and $\pi$ radians whose cosine is x.

(It can also be thought of as simply a number between $0$ and $\pi$ whose cosine is $x$.)

The restriction to angles between $0$ and $\pi$ makes $\arccos$ a function.

$\arccos \left(\cos \left(\frac{5 \pi}{4}\right)\right)$ is an angle between $0$ and $\pi$ whose cosine is the same as the cosine of $\frac{5 \pi}{4}$.

The angle we want is $\frac{3 \pi}{4}$

We know that $\cos \left(\frac{5 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$
and the Quadrant II angle with cosine equal to $- \frac{\sqrt{2}}{2}$
is $\frac{3 \pi}{4}$