# How do you express cos( (15 pi)/ 8 ) * cos (( 5 pi) /3 )  without using products of trigonometric functions?

Mar 13, 2016

$\frac{1}{2} \cos \left(\frac{11 \pi}{24}\right) + \frac{1}{2} \cos \left(\frac{5 \pi}{24}\right)$

#### Explanation:

As $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ and $\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$, adding them gives us

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

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

$\cos \left(15 \frac{\pi}{8}\right) \cos \left(5 \frac{\pi}{3}\right) = \frac{1}{2} \left\{\cos \left(\frac{15 \pi}{8} + \frac{5 \pi}{3}\right) + \cos \left(\frac{15 \pi}{8} - \frac{5 \pi}{3}\right)\right\}$

= $\frac{1}{2} \left\{\cos \left(\frac{45 \pi + 40 \pi}{24}\right) + \cos \left(\frac{45 \pi - 40 \pi}{24}\right)\right\}$

= $\frac{1}{2} \left\{\cos \left(\frac{85 \pi}{24}\right) + \cos \left(\frac{5 \pi}{24}\right)\right\}$

= $\frac{1}{2} \left\{\cos \left(\frac{96 \pi - 11 \pi}{24}\right) + \cos \left(\frac{5 \pi}{24}\right)\right\}$

= $\frac{1}{2} \left\{\cos \left(4 \pi - \frac{11 \pi}{24}\right) + \cos \left(\frac{5 \pi}{24}\right)\right\}$

= 1/2cos((11pi)/24)+1/2cos((5pi)/24)}