# How do you express sin(pi/ 6 ) * cos( ( pi) / 12 )  without using products of trigonometric functions?

Jan 29, 2016

Either evaluate or use the product to sum identity.

#### Explanation:

Evaluate
We have:

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

and

$\cos \left(\frac{\pi}{12}\right) = \cos \left(\frac{\pi}{4} - \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{6}\right) + \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$.

So $\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{12}\right) = \frac{1}{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{8}$

Use the product to sum identity.

$\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{12}\right) = \frac{1}{2} \left[\sin \left(\left(\frac{\pi}{6}\right) + \left(\frac{\pi}{12}\right)\right) + \sin \left(\left(\frac{\pi}{6}\right) - \left(\frac{\pi}{12}\right)\right)\right]$

$= \frac{1}{2} \left[\sin \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{12}\right)\right]$