# How do you express sin(pi/ 6 ) * cos( ( 9 pi) / 8 )  without using products of trigonometric functions?

Mar 25, 2016

$= - \frac{1}{2} \sqrt{\frac{1}{2} \cdot \left(1 + \frac{1}{\sqrt{2}}\right)}$

#### Explanation:

$\sin \left(\frac{\pi}{6}\right) \cdot \cos \left(\pi + \frac{\pi}{8}\right) = - \frac{1}{2} \cdot \cos \left(\frac{\pi}{8}\right)$
$= - \frac{1}{2} \sqrt{\frac{1}{2} \cdot \left(1 + \cos \left(\frac{\pi}{4}\right)\right)} = - \frac{1}{2} \sqrt{\frac{1}{2} \cdot \left(1 + \frac{1}{\sqrt{2}}\right)}$

It may also be expressed as sum of two trigonometric functions
using formula
$\sin A \cos B = \frac{1}{2} \cdot \left(\sin \left(A + B\right) + \sin \left(A - B\right)\right)$

$\sin \left(\frac{\pi}{6}\right) \cdot \cos \left(\pi + \frac{\pi}{8}\right) = - \sin \left(\frac{\pi}{6}\right) \cdot \cos \left(\frac{\pi}{8}\right)$

$= - \frac{1}{2} \cdot \left(\sin \left(\frac{\pi}{6} + \frac{\pi}{8}\right) + \sin \left(\frac{\pi}{6} - \frac{\pi}{8}\right)\right)$

=-1/2*(sin(7pi/24)+sin(pi/24)

Mar 25, 2016

$- \left(\frac{1}{2}\right) \left(\cos \frac{\pi}{8}\right)$

#### Explanation:

$P = \sin \left(\frac{\pi}{6}\right) . \cos \left(\frac{9 \pi}{8}\right)$
Trig table --> $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$
$\cos \left(\frac{9 \pi}{8}\right) = \cos \left(\frac{\pi}{8} + \pi\right) = - \cos \left(\frac{\pi}{8}\right)$
The product P can be expressed as:
$P = - \left(\frac{1}{2}\right) \left(\cos \frac{\pi}{8}\right)$

If being asked, we can find P by evaluating $\cos \left(\frac{\pi}{8}\right)$, using the trig identity: $\cos 2 a = 2 {\cos}^{2} a - 1.$