Answer
$cos~x+cos~4x=2~sin~\frac{5x}{2}~sin~\frac{3x}{2}$
Work Step by Step
$cos~x+cos~4x=-2~sin(\frac{x+4x}{2})~sin(\frac{x-4x}{2})=-2~sin~\frac{5x}{2}~sin~(-\frac{3x}{2})=-2~sin~\frac{5x}{2}~(-sin~\frac{3x}{2})=2~sin~\frac{5x}{2}~sin~\frac{3x}{2}$
Remember that: $sin(-x)=-sin~x$
