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

Question

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?

https://math.stackexchange.com/questions/827540/proving-trigonometric-equation-cos36-circ-cos72-circ-1-2 — lab bhattacharjee, Mar 03 '20 at 12:24
Does this answer your question? How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. — Lê Thành Đạt, Mar 04 '20 at 05:59

score 2 · Accepted Answer · answered Mar 03 '20 at 13:15

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: $$\cos (8x) + \cos (4x) = 2 \frac{1 - \sqrt{5}}{4} \frac{\sqrt{5} + 1}{4} = ... =-\frac{1}{2}.$$

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