If $x= \frac{\pi}{10}$, what is the value of $\cos(8x) +\cos(4x)$? 
If $x= \dfrac{\pi}{10}$, what is the value of $\cos(8x) +\cos(4x)$?

My try:
$5x=\dfrac{\pi}{2}$ and $10x=\pi \to \cos(8x)=-\cos(2x)\;$  & 
$\;\cos(4x)=\sin(x) $ .
How can I complete? 

We can apply sum-to-product identities, i.e. $\quad \cos \alpha + \cos \beta = 2\cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}$.
So, we have: $$\cos (8x) + \cos (4x) = 2\cos(6x)\cos(2x) = 2\cos \frac{6\pi}{10} \cos \frac{2\pi}{10} = 2 \cos \frac{3\pi}{5} \cos\frac{\pi}{5}.$$
Next, we know that holds (see https://www.math-only-math.com/exact-value-of-cos-36-degree.html): $$\cos \frac{\pi}{5} = \frac{\sqrt{5} + 1}{4} \quad \quad (\text{hint:} \quad \frac{\pi}{5}=36°).$$
Similarly, we know that holds:
$$\cos \frac{3\pi}{5} = \frac{1 - \sqrt{5}}{4}.$$
Finally, we have:
$$\cos (8x) + \cos (4x) = 2 \frac{1 - \sqrt{5}}{4} \frac{\sqrt{5} + 1}{4} = ... =-\frac{1}{2}.$$
