# How do you express sin(pi/ 4 ) * sin( ( 15 pi) / 8 )  without using products of trigonometric functions?

Mar 10, 2016

$\sin \left(\frac{5 \pi}{4}\right) \sin \left(\frac{15 \pi}{8}\right) = \frac{1}{2} \left(\cos \left(\frac{5 \pi}{8}\right) + \cos \left(\frac{\pi}{8}\right)\right)$

#### Explanation:

For this one uses the identity

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

Subtracting upper identity from lower, one gets

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

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

Hence $\sin \left(\frac{5 \pi}{4}\right) \sin \left(\frac{15 \pi}{8}\right)$

= $\frac{1}{2} \left(\cos \left(\frac{15 \pi}{8} - \frac{5 \pi}{4}\right) - \cos \left(\frac{15 \pi}{8} + \frac{5 \pi}{4}\right)\right)$

= $\frac{1}{2} \left(\cos \left(\frac{5 \pi}{8}\right) - \cos \left(\frac{25 \pi}{8}\right)\right)$

= $\frac{1}{2} \left(\cos \left(\frac{5 \pi}{8}\right) - \cos \left(3 \pi + \frac{\pi}{8}\right)\right)$

= $\frac{1}{2} \left(\cos \left(\frac{5 \pi}{8}\right) + \cos \left(\frac{\pi}{8}\right)\right)$