# How do you simplify using the half angle formula cos(-(7pi)/12)?

Apr 5, 2018

$- \frac{\sqrt{2 - \sqrt{3}}}{2}$

#### Explanation:

$\cos \left(\frac{- 7 \pi}{12}\right) = \cos \left(\frac{- \pi}{12} - \frac{6 \pi}{12}\right) = \sin \left(\frac{- \pi}{12}\right) =$
$= - \sin \left(\frac{\pi}{12}\right)$
To find $\sin \left(\frac{\pi}{12}\right)$ use trig half angle identity:
$\sin \left(\frac{t}{2}\right) = \pm \sqrt{\frac{1 - \cos t}{2}}$
In this case, $\cos t = \cos \left(\frac{2 \pi}{12}\right) = \cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
We get, with $\left(\frac{\pi}{12}\right)$ being in Quadrant 1,
$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{1 - \frac{\sqrt{3}}{2}}}{2} = \frac{\sqrt{2 - \sqrt{3}}}{2}$
Finally,
$\cos \left(\frac{- 7 \pi}{12}\right) = - \sin \left(\frac{\pi}{12}\right) = - \frac{\sqrt{2 - \sqrt{3}}}{2}$
Check by calculator.
$- \sin \left(\frac{\pi}{12}\right) = - \sin {15}^{\circ} = - 0.258$
$- \frac{\sqrt{2 - \sqrt{3}}}{2} = - \frac{0.517}{2} = - 0.258$. Proved.