# How do you express sin(pi/ 4 ) * sin( ( 1 9 pi) / 12 )  without using products of trigonometric functions?

Sep 20, 2016

$\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{19 \pi}{12}\right) = \frac{1}{2} \cos \left(\frac{4 \pi}{3}\right) - \frac{1}{2} \cos \left(\frac{11 \pi}{6}\right)$

#### Explanation:

We know that

$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$ and

$\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$. Subtracting latter from former, we get

$2 \sin A \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)$

or $\sin A \sin B = \frac{1}{2} \cos \left(A - B\right) - \frac{1}{2} \cos \left(A + B\right)$

Hence $\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{19 \pi}{12}\right) = \sin \left(\frac{19 \pi}{12}\right) \sin \left(\frac{\pi}{4}\right)$

= $\frac{1}{2} \cos \left(\frac{19 \pi}{12} - \frac{\pi}{4}\right) - \frac{1}{2} \cos \left(\frac{19 \pi}{12} + \frac{\pi}{4}\right)$

= $\frac{1}{2} \cos \left(\frac{19 \pi}{12} - \frac{3 \pi}{12}\right) - \frac{1}{2} \cos \left(\frac{19 \pi}{12} + \frac{3 \pi}{12}\right)$

= $\frac{1}{2} \cos \left(\frac{16 \pi}{12}\right) - \frac{1}{2} \cos \left(\frac{22 \pi}{12}\right)$

= $\frac{1}{2} \cos \left(\frac{4 \pi}{3}\right) - \frac{1}{2} \cos \left(\frac{11 \pi}{6}\right)$

Nov 6, 2016

$= - \frac{1}{4} \left(\sqrt{3} + 1\right)$

#### Explanation:

$\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{19 \pi}{12}\right)$

$= \sin \left(\frac{\pi}{4}\right) \sin \left(\pi + \frac{7 \pi}{12}\right)$

$= - \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{7 \pi}{12}\right)$

$= - \sin \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3} + \frac{\pi}{4}\right)$

=-sin(pi/4)[sin(pi/3)cos(pi/4)+cos(pi/3)sin(pi/4))]

$= - \frac{1}{\sqrt{2}} \left[\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}}\right]$

$= - \frac{1}{4} \left(\sqrt{3} + 1\right)$