# How do you express cos( (3 pi)/ 2 ) * cos (( 13 pi) / 8 )  without using products of trigonometric functions?

Mar 27, 2016

Answer is $0$. Please follow the following process.

#### Explanation:

We know $\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
and $\cos \left(\alpha - \beta\right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$

Adding two we get $\cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right) = 2 \cos \alpha \cos \beta$ or

$\cos \alpha \cos \beta = \frac{1}{2} \left\{\cos \left(\alpha + \beta\right) + \cos \left(\alpha - \beta\right)\right\}$

Hence $\cos \left(\frac{3 \pi}{2}\right) \cos \left(\frac{13 \pi}{8}\right) = \frac{1}{2} \left\{\cos \left(\frac{3 \pi}{2} + \frac{13 \pi}{8}\right) + \cos \left(\frac{3 \pi}{2} - \frac{13 \pi}{8}\right)\right\}$

= $\frac{1}{2} \left\{\cos \left(\frac{12 \pi}{8} + \frac{13 \pi}{8}\right) + \cos \left(\frac{12 \pi}{8} - \frac{13 \pi}{8}\right)\right\}$

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

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

= $\frac{1}{2} \left\{- \cos \left(\frac{\pi}{8}\right) + \cos \left(\frac{\pi}{8}\right)\right\} = 0$

Mar 27, 2016

Zero

#### Explanation:

Trig table -->$\cos \left(\frac{3 \pi}{2}\right)$ = 0#, therefor, the product is equal to zero.