<s>[INST] Evaluate $\cos \frac{\pi}{7} \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}$ Evaluate $$\cos \frac{\pi}{7} \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}.$$
The first thing i noticed was that $$\cos \frac{\pi}{7}=\frac{\zeta_{14}+\zeta_{14}^{-1}}{2},$$ where $\zeta_{14}=e^{2\pi i/14}$ is the 14-th root of unity.
Substituting this into the expression and simplifying, the expression is then transformed to $$\frac{1}{4}Re (\zeta_{14}+\zeta_{14}^3+\zeta_{14}^5+\zeta_{14}^7).$$
But then i have no idea how to get the real part of that thing. I only observed that $\zeta_{14}^7=-1$ and all the terms are primitive 14-th roots of unity. Please helps. Thanks!
 [/INST] Multiply by $8\sin\tfrac{\pi}{7}$ and simplify using $2\sin x \cos x = \sin 2x$ three times:
$$8\sin\tfrac{\pi}{7} \cos \tfrac{\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7}
= 4 \sin\tfrac{2\pi}{7} \cos \tfrac{2\pi}{7}\cos \tfrac{4\pi}{7} 
= 2 \sin\tfrac{4\pi}{7} \cos \tfrac{4\pi}{7} 
= \sin\tfrac{8\pi}{7}$$
But you also have:
$$\sin\tfrac{8\pi}{7} = \sin\left( \pi - \tfrac{8\pi}{7} \right) = -\sin\tfrac{\pi}{7}$$
Now divide again by $8\sin\tfrac{\pi}{7}$.
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