Answer
$$\cos4x+\cos8x=2\cos6x\cos 2x$$
Work Step by Step
$$A=\cos4x+\cos8x$$
The sum-to-product identity that will be applied here is $$\cos X+\cos Y=2\cos(\frac{X+Y}{2})\cos(\frac{X-Y}{2})$$
Therefore, A would be $$A=2\cos[\frac{4x+8x}{2}]\cos[\frac{4x-8x}{2}]$$ $$A=2\cos(\frac{12x}{2})\cos(\frac{-4x}{2})$$ $$A=2\cos 6x\cos(-2x)$$
We know that $\cos(-X)=\cos X$, which means $\cos(-2x)=\cos 2x$. Therefore, $$A=2\cos6x\cos 2x$$