# How do you use half angle formula to find tan (5pi)/12?

Apr 28, 2015

$\tan \left(5 \pi\right) \equiv \tan \left(\pi\right) = 0$
and there is no point in using the half angle formula to evaluate
$\frac{\tan \left(5 \pi\right)}{12}$

Therefore, let's assume that you really want to evaluate
$\tan \left(\frac{5 \pi}{12}\right)$

Half angle formula (for tan):
${\tan}^{2} \left(\frac{\theta}{2}\right) = \frac{1 - \cos \left(\theta\right)}{1 + \cos \left(\theta\right)}$

${\tan}^{2} \left(\frac{5 \pi}{12}\right) = \tan \left(\frac{\frac{5 \pi}{6}}{2}\right)$

$= \frac{1 - \cos \left(\frac{\pi}{6}\right)}{1 + \cos \left(\frac{\pi}{6}\right)}$

$= \frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}$

$= {\left(1 + \frac{\sqrt{3}}{2}\right)}^{2} / \left(1 - \frac{3}{4}\right)$

$= 4 {\left(1 + \frac{\sqrt{3}}{2}\right)}^{2}$

That is
${\tan}^{2} \left(\frac{5 \pi}{12}\right) = {\left(2 \left(1 + \frac{\sqrt{3}}{2}\right)\right)}^{2}$

Since $\frac{5 \pi}{12}$ is in Quadrant 1, the tan is positive
and
$\tan \left(\frac{5 \pi}{12}\right) = 2 + \sqrt{3}$