# How do you find the value for tan^-1[tan(5pi/7)]?

May 28, 2015

The formula
${\tan}^{- 1} \left[\tan \left(a\right)\right] = a$
works for a in (-pi/2;pi/2)

In our case $\frac{5 \pi}{7} > \frac{\pi}{2}$ but we can use the periodicity of $\tan$:
$\tan \left(\frac{5 \pi}{7}\right) = \tan \left(\frac{5 \pi}{7} - \pi\right) = \tan \left(- \frac{2 \pi}{7}\right)$
${\tan}^{- 1} \left[\tan \left(\frac{5 \pi}{7}\right)\right] = {\tan}^{- 1} \left[\tan \left(- \frac{2 \pi}{7}\right)\right] = - \frac{2 \pi}{7}$

Btw, a similar formula
$\tan \left[{\tan}^{- 1} \left(a\right)\right] = a$
works for all $a \in \mathbb{R}$.