Compute $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ using roots of unity . 
Compute $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$.

The hint says to take $\zeta=\text {cos}(20^\circ)+ \text {i sin} (20^\circ)$ and then asks us to write $\cos(20^\circ)$  is terms of $\zeta$ .
Now, I took $\zeta=\text {cos}(20^\circ)+ \text {i sin} (20^\circ)$ be the 18 root of unit. So $\zeta^2=\text {cos}(40^\circ)+ \text {i sin} (40^\circ)$ and $\zeta^4=\text {cos}(80^\circ)+ \text {i sin} (80^\circ)$.
But I didn't understand , how to write $\cos(20^\circ)$  is terms of $\zeta$ .
Any hints ? Thanks in advance !
 A: Try to substitute the following:
$$
\begin{align}
2\cos{(20)}&=\zeta-\zeta^{8}\\
2\cos{(40)}&=\zeta^{2}-\zeta^{7}\\
2\cos{(80)}&=\zeta^{4}-\zeta^{5}\\
\\
\zeta^{9}&=-1
\end{align}
$$
Then use the equation
$$
\begin{align}
\zeta^{8}-\zeta^{7}+\zeta^{6}-\zeta^{5}+\zeta^{4}-\zeta^{3}+\zeta^{2}-\zeta+1&=0
\end{align}
$$
A: Notice that $\text{cis}(-20^{\circ}) = \cos(20^{\circ})-i\sin(20^{\circ})$, which is $e^{-20^{\circ}} = \frac{1}{\zeta}$. Now we can use elimination to get that $2\cos(20^{\circ}) = \zeta+\frac{1}{\zeta}$, and the rest follows from here.
In particular, you can try multiplying both sides by $\zeta-\frac{1}{\zeta}$, which is identical to the standard trigonometric solution, but simple expansion works just as well using the fact that the 9th roots of unity sum to 0.
A: Here's an alternative method, which is rather neat as it only uses the double angle formulae which are quite basic. Here we go :)
$$\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$$
Applying this twice more, we have
$$\sin x=2^3\sin\frac{x}{8}\cos\frac{x}{8}\cos\frac{x}{4}\cos\frac{x}{2}$$
You may see where this is going. Let $\frac{x}{2}=80$ degrees:
$$\frac{x}{2}=80\implies x=160$$
$$\implies \sin160=2^3\sin20\cos20\cos40\cos80$$
But we know that $\sin20=\sin(180-20)=\sin160$ so we have
$$\sin160=2^3\sin160\cos20\cos40\cos80$$
$$\implies2^3\cos20\cos40\cos80=1\implies \cos20\cos40\cos80=\frac{1}{8}$$
I hope you find that helpful or interesting, or maybe even both :))
