# How do you evaluate arctan [tan ((-2pi)/3)]?

Jul 19, 2015

$\arctan \left[\tan \left(\frac{- 2 \pi}{3}\right)\right] = \frac{\pi}{3}$

#### Explanation:

For a ratio $r$, $\arctan \left(r\right) = \theta$
$\textcolor{w h i t e}{\text{XXXX}}$where $\theta \epsilon \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$
and
$\textcolor{w h i t e}{\text{XXXX}}$$\tan \left(\theta\right) = r$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(definition)

The question therefore becomes:
$\textcolor{w h i t e}{\text{XXXX}}$For what angle $\theta \epsilon \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$ is
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$\tan \left(\theta\right) = \tan \left(\frac{- 2 \pi}{3}\right)$ ?

Note that the angle $\left(\frac{- 2 \pi}{3}\right)$ occurs in the third quadrant and is a positive value (based on CAST or recognition that both the "rise" and the "run" are negative, so the ratio $\tan = r i s \frac{e}{r} u n$ is positive). The required angle $\theta$ must occur in the first quadrant (since $\theta \epsilon \left[- \frac{\pi}{2} , \pi , 2\right]$ and $\tan \left(\theta\right) \ge 0$)

Note that the reference angle (see diagram above) for $\left(- \frac{2 \pi}{3}\right)$ is $\frac{\pi}{3}$.