Finding product of trigonometric values (cos/sin) using complex number/roots of unity Using the ninth root of unity of 1, show that $cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$.

Here is my solution.    
Let $\omega=cos\theta+isin\theta$,  
As $\omega^9=1$,  
$$cos9\theta+isin9\theta=1$$
so $$9\theta=2nπ {(n\in Z)}$$
$$\theta=\frac{2nπ}{9} {(n\in Z)}$$
$$\theta=\frac{\pm2π}{9},  \frac{\pm4π}{9}, \frac{\pm6π}{9}, \frac{\pm8π}{9},$$
$$\omega=cis\frac{\pm2π}{9},  cis\frac{\pm4π}{9}, cis\frac{\pm6π}{9}, cis\frac{\pm8π}{9},$$
therefore $$\omega^9-1=0$$
$$(\omega-1)(1+\omega +\omega^2 +\omega^3+\omega^4+\omega^5+\omega^6+\omega^7+\omega^8)=0$$ 
and because $\omega$ is not real,
therefore $$1+\omega +\omega^2   +\omega^3+\omega^4+\omega^5+\omega^6+\omega^7+\omega^8=0$$
through factorisation,
$$(\omega - cis\frac{2π}{9})(\omega - cis\frac{-2π}{9})...(\omega - cis\frac{8π}{9})(\omega - cis\frac{-8π}{9})=1+\omega +\omega^2   +...+\omega^7+\omega^8$$
$$(\omega^2 - 2cos\frac{2π}{9}\omega +1)...(\omega^2 - 2cos\frac{8π}{9}\omega +1)=1+\omega +\omega^2 +...+\omega^7+\omega^8$$
divide both sides by $\omega^4$,
$$(\omega - 2cos\frac{2π}{9} +\frac{1}{\omega})...(\omega - 2cos\frac{8π}{9} +\frac{1}{\omega})=\frac{1}{\omega^4}+\frac{1}{\omega^3}+...  +\omega^3+\omega^4$$
Let $x=\omega+\frac{1}{\omega}$,
$$\omega^2+\frac{1}{\omega^2}=(\omega+\frac{1}{\omega})^2-2=x^2-2$$
similarly,
$$\omega^3+\frac{1}{\omega^3}=x^3-3x$$
$$\omega^4+\frac{1}{\omega^4}=x^4-4(x^2-2)-6=x^4-4x^2+2$$
therefore, substituting in
$$x^4+x^3-3x^2-2x+1=(x - 2cos\frac{2π}{9})...(x - 2cos\frac{8π}{9})$$
$$2^4cos\frac{2π}{9}cos\frac{4π}{9}cos\frac{6π}{9}cos\frac{8π}{9}=1$$
as $cos\frac{8π}{9}=-cos\frac{π}{9}$ and $cos\frac{6π}{9}=-\frac{1}{2}$
$$cos\frac{π}{9}cos\frac{2π}{9}cos\frac{4π}{9}=\frac{1}{8}$$

I believe that I might have over-complicated this question. Any other thoughts?
 A: Root of Unity Approach
Using
$$
\omega=\cos\left(\frac\pi9\right)+i\sin\left(\frac\pi9\right)
$$
we get $\omega^9=-1$
$$
\begin{align}
\overbrace{\left(\omega-\omega^8\right)}^{2\cos\left(\frac\pi9\right)}\overbrace{\left(\omega^2-\omega^7\right)}^{2\cos\left(\frac{2\pi}9\right)}\overbrace{\left(\omega^4-\omega^5\right)}^{2\cos\left(\frac{4\pi}9\right)}
&=\omega^7-\omega^8-\omega^{12}+\omega^{13}-\omega^{14}+\omega^{15}+\omega^{19}-\omega^{20}\\
&=\omega^7-\omega^8+\omega^3-\omega^4+\omega^5-\omega^6+\omega^1-\omega^2\\[3pt]
&=1
\end{align}
$$

Trigonometric Identity Approach
We can also use $\sin(2x)=2\sin(x)\cos(x)$,
$$
\begin{align}
\sin\left(\frac{\pi}9\right)\cos\left(\frac{\pi}9\right)\cos\left(\frac{2\pi}9\right)\cos\left(\frac{4\pi}9\right)
&=\frac12\sin\left(\frac{2\pi}9\right)\cos\left(\frac{2\pi}9\right)\cos\left(\frac{4\pi}9\right)\\
&=\frac14\sin\left(\frac{4\pi}9\right)\cos\left(\frac{4\pi}9\right)\\
&=\frac18\sin\left(\frac{8\pi}9\right)
\end{align}
$$
Now divide by $\sin\left(\frac{\pi}9\right)=\sin\left(\frac{8\pi}9\right)$.
