Answer
$2\displaystyle \cos\frac{13x}{2}\cdot\cos\frac{3x}{2}$
Work Step by Step
Sum-to-Product:
$\displaystyle \cos A+\cos B=2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$
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$\displaystyle \frac{A+B}{2}=\frac{5x+8x}{2}=\frac{13x}{2}$
$\displaystyle \frac{A-B}{2}=\frac{5x-8x}{2}=-\frac{3x}{2}$
$\displaystyle \cos 5x+\cos 8x=2\cos\frac{13x}{2}\cdot\cos(-\frac{3x}{2})$
...Even/odd identity: $\cos(-A)=\cos A$
$=2\displaystyle \cos\frac{13x}{2}\cdot\cos\frac{3x}{2}$