# Question #c5acc

May 26, 2017

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

#### Explanation:

$\cos \left(\frac{3 \pi}{4} - \frac{\pi}{3}\right) = \cos \left(\frac{9 \pi - 4 \pi}{12}\right) = \cos \left(\frac{5 \pi}{12}\right)$
Find $\cos \left(\frac{5 \pi}{12}\right)$ by using trig identity:
$2 {\cos}^{2} a = 1 + \cos 2 a$
In this case, trig table gives:
$\cos 2 a = \cos \left(\frac{10 \pi}{12}\right) = \cos \left(5 \frac{\pi}{6}\right) = - \frac{\sqrt{3}}{2}$
Call $\cos \left(\frac{5 \pi}{12}\right) = \cos t$, we get:
$2 {\cos}^{2} t = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$
${\cos}^{2} t = \frac{2 - \sqrt{3}}{4}$
$\cos t = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$
Since $\frac{5 \pi}{12}$ is in Quadrant 1, its cos is positive -->
$\cos \left(\frac{5 \pi}{12}\right) = \cos t = \frac{\sqrt{2 - \sqrt{3}}}{2}$