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 

Start out as by noticing that 
$$ 
\omega = cis(\frac{2\pi}{9}),\quad \omega^{8}=cis(\frac{-2\pi}{9})\quad \\
\omega^2 = cis(\frac{4\pi}{9}),\quad\omega^7=cis(\frac{-4\pi}{9})
\\ etc.
$$
Now see, 
$$ \omega\ + \omega^8 = 2\ cos(\frac{2\pi}{9}) \\
\omega^2\ + \omega^7= 2\ cos(\frac{4\pi}{9})\\
\omega^3 + \omega^6 = 2\ cos(\frac{6\pi}{9})\\
\omega^4\ + \omega^5= 2\ cos(\frac{8\pi}{9})\\
$$
If we multiply these we get: 
$$
[2\ cos(\frac{2\pi}{9})][2\ cos(\frac{4\pi}{9})][2\ cos(\frac{6\pi}{9})][2\ cos(\frac{8\pi}{9})] = (\omega\ + \omega^8)(\omega^2\ + \omega^7)(\omega^3 + \omega^6)(\omega^4\ + \omega^5)
$$
If we carefully pull out the $\omega$s, we can get the R.H.S to equal $ \ \omega^{10}(1\ + \omega)(1\ + \omega^3)(1\ + \omega^5)(1\ + \omega^7)$
We also know that $ \ \omega^{10} = \omega$, so we can simplify it to be $\ \omega(1\ + \omega)(1\ + \omega^3)(1\ + \omega^5)(1\ + \omega^7) $
Using the fact: $\ \omega\ + \omega^2\ + \omega^3\ + \omega^4\ + \omega^5\ + \omega^6\ + \omega^7\ + \omega^8\ = -1 $, the expression simplifies to -1. 
Now simplifying the L.H.S. we get: $16[cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{6\pi}{9})\cos(\frac{8\pi}{9})]$. Knowing that $cos(\frac{2\pi}{3}) = -\frac{1}{2}$, we can finally get that:
$$
-8\ [cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{8\pi}{9})] = -1 \\
[cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})\cos(\frac{8\pi}{9})] = \frac{1}{8} \\ 
[cos(\frac{\pi}{9})\cos(\frac{2\pi}{9})\cos(\frac{4\pi}{9})] = \frac{1}{8}
$$
