# How do you express sin(pi/ 8 ) * cos(( 5 pi)/6 )  without products of trigonometric functions?

Feb 9, 2016

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

#### Explanation:

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

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