# How do you express sin(pi/ 8 ) * cos(( ( 5 pi) / 12 )  without using products of trigonometric functions?

##### 1 Answer
Feb 9, 2016

$\frac{\left(\sqrt{2 - \sqrt{2}}\right) . \left(\sqrt{2 + \sqrt{3}}\right)}{4}$

#### Explanation:

Product $P = \sin \left(\frac{\pi}{8}\right) . \cos \left(\frac{5 \pi}{12}\right)$.
a. Find $\sin \left(\frac{\pi}{8}\right)$ by trig identity: $\cos 2 x = 1 - 2 {\sin}^{2} x$
$\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = 1 - 2 {\sin}^{2} \left(\frac{\pi}{8}\right)$
${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{4}$
$\sin \left(\frac{\pi}{8}\right) = \frac{\sqrt{2 - \sqrt{2}}}{2}$ --> (sin $\frac{\pi}{8}$ is positive)
b. Find $\cos \left(\frac{5 \pi}{12}\right) = \cos t$ by identity: $\cos 2 t = 2 {\cos}^{2} t - 1$
$\cos 2 t = \cos \left(\frac{10 \pi}{12}\right) = \cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2} = 2 {\cos}^{2} t - 1$
$2 {\cos}^{2} t = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$
${\cos}^{2} t = \frac{2 + \sqrt{3}}{4}$
$\cos t = \cos \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{2 + \sqrt{3}}}{2}$--> (cos t is positive)

Finally: $P = \frac{\left(\sqrt{2 - \sqrt{2}}\right) . \left(\sqrt{2 + \sqrt{3}}\right)}{4}$