<s>[INST] Prove $\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$ with complex numbers (roots of unity) I want to prove that 
Using $z^9=1$ and the fact that $1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8=0$ where $w=cis(\frac{2π}{9})$, that $$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$$
I am able to do this by using $w^n+w^{-n}=2\cos \frac{2nπ}{9}$ and hence expanding the LHS but it is laborious and, so, I was wondering is there is a more efficient or elegant method of proving this this identity using the ninth root of unity. 
Thanks very much 
 [/INST] Note
$$\frac{1-z^9}{1-z}= \prod_{k=1}^8(w^k-z)
$$
Set $z=-1$ and use $\prod_{k=1}^8w^{k}=1$
\begin{align}
1&=\prod_{k=1}^8(w^k+1)= (\prod_{k=1}^8w^{k})^{1/2}\prod_{k=1}^8(w^{k/2}+ w^{-k/2})
= 2^8 \prod_{k=1}^8\cos\frac{k\pi}9
=\left( 2^4 \prod_{k=1}^4\cos\frac{k\pi}9 \right)^2
\end{align}
Then, substitute  $\cos\frac{3\pi}9=\frac12$ to obtain
$$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac18$$
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