# How do you express sin(pi/ 6 ) * cos( ( 15 pi) / 12 )  without using products of trigonometric functions?

Mar 26, 2016

$- \frac{\sqrt{2}}{4}$

#### Explanation:

$P = \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{15 \pi}{12}\right) .$
Trig table --> $\sin \frac{\pi}{6} = \frac{1}{2}$
$\cos \left(\frac{15 \pi}{12}\right) = \cos \left(\frac{- 9 \pi}{12} + 2 \pi\right) = \cos \left(\frac{- 3 \pi}{4}\right) =$
$= \cos \left(\frac{3 \pi}{4}\right) = - \frac{\sqrt{2}}{2.}$
$P = \left(\frac{1}{2}\right) \left(- \frac{\sqrt{2}}{2}\right) = - \frac{\sqrt{2}}{4}$

Apr 4, 2016

$\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{15 \pi}{12}\right) = \frac{1}{2} \sin \left(\frac{17 \pi}{12}\right) - \frac{1}{2} \sin \left(\frac{13 \pi}{12}\right)$

#### Explanation:

$2 \sin A \cos B = \sin \left(A + B\right) + \sin \left(A - B\right)$
$\sin A \cos B = \frac{1}{2} \left(\sin \left(A + B\right) + \sin \left(A - B\right)\right)$
$A = \frac{\pi}{6} , B = \frac{15 \pi}{12}$
$\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{15 \pi}{12}\right) = \frac{1}{2} \left(\sin \left(\frac{\pi}{6} + \frac{15 \pi}{12}\right) + \sin \left(\frac{\pi}{6} - \frac{15 \pi}{12}\right)\right)$
$= \frac{1}{2} \left(\sin \left(\frac{17 \pi}{12}\right) + \sin \left(- \frac{13 \pi}{12}\right)\right)$
$= \frac{1}{2} \sin \left(\frac{17 \pi}{12}\right) - \frac{1}{2} \sin \left(\frac{13 \pi}{12}\right)$