# How do you use the half angle formula to determine the exact values of the sine, cosine, and tangent (19pi)/12?

Feb 5, 2018

$\sin \left(\frac{19 \pi}{12}\right) = - \frac{\sqrt{2 + \sqrt{3}}}{2}$
$\cos \left(\frac{19 \pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$

#### Explanation:

First, call $\left(\frac{t}{2}\right) = \frac{19 \pi}{12}$ -->
$t = \frac{38 \pi}{12} = \frac{2 \pi}{12} + 3 \pi = \frac{\pi}{6} + 3 \pi$
Next, use the half angle formulas:
$\sin \left(\frac{t}{2}\right) = \pm \sqrt{\frac{1 - \cos t}{2}}$
$\cos \left(\frac{t}{2}\right) = \pm \sqrt{\frac{1 + \cos t}{2}}$
Note that
$\cos t = \cos \left(\frac{\pi}{6} + 3 \pi\right) = - \cos \left(\frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$.
Therefor, since $\left(\frac{t}{2}\right)$ is in Quadrant 4 -->
$\sin \left(\frac{t}{2}\right) = - \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = - \frac{\sqrt{2 + \sqrt{3}}}{2}$
By the same process, with $\cos \left(\frac{t}{2}\right)$ positive -->
$\cos \left(\frac{t}{2}\right) = + \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$
From there -->
$\tan \left(\frac{t}{2}\right) = \frac{\sin}{\cos} = - \frac{2 + \sqrt{3}}{2 - \sqrt{3}}$