# Explain why tan pi=0 does not imply that arctan0=pi?

Sep 10, 2015

$\pi$ is not in the interval used for $\arctan$

#### Explanation:

We want $\arctan$ to be a function. We want it to give one, never two, values for a single input.

By definition:

$y = \arctan x$ if and only if ($\tan y = x$ and $- \frac{\pi}{2} < y < \frac{\pi}{2}$)

Since $\pi$ is not in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ there is no $x$ for which $\arctan x = \pi$

(The situation is similar to: Explain why ${\left(- 3\right)}^{2} = 9$ does not imply that $\sqrt{9} = - 3$. It ($- 3$) is the wrong kind of number to be a principal square root.)