# How do you express cos( (4 pi)/3 ) * cos (( 7 pi) /4 )  without using products of trigonometric functions?

Oct 4, 2016

$\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{7 \pi}{4}\right) = \frac{1}{2} \cos \left(\frac{13 \pi}{12}\right) + \frac{1}{2} \cos \left(\frac{5 \pi}{12}\right)$

#### Explanation:

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

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

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

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

Hence $\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{7 \pi}{4}\right)$

= $\frac{1}{2} \left[\cos \left(\left(\frac{4 \pi}{3}\right) + \left(\frac{7 \pi}{4}\right)\right) + \cos \left(\left(\frac{4 \pi}{3}\right) - \left(\frac{7 \pi}{4}\right)\right)\right]$

= $\frac{1}{2} \left[\cos \left(\frac{4 \times 4 \pi + 3 \times 7 \pi}{12}\right) + \cos \left(\frac{4 \times 4 \pi - 3 \times 7 \pi}{12}\right)\right]$

= $\frac{1}{2} \left[\cos \left(\frac{16 \pi + 21 \pi}{12}\right) + \cos \left(\frac{16 \pi - 21 \pi}{12}\right)\right]$

= $\frac{1}{2} \left[\cos \left(\frac{37 \pi}{12}\right) + \cos \left(\frac{- 5 \pi}{12}\right)\right]$

= $\frac{1}{2} \left[\cos \left(2 \pi + \frac{13 \pi}{12}\right) + \cos \left(\frac{5 \pi}{12}\right)\right]$

= $\frac{1}{2} \cos \left(\frac{13 \pi}{12}\right) + \frac{1}{2} \cos \left(\frac{5 \pi}{12}\right)$