A question of Roots of unity 
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
 \cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.

I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.

Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
The roots of $$\dfrac{z^9-1}{z-1}=0$$ are $e^{2\pi ir/9}; r=1,2,3,4,5,6,7,8$
Dividing both sides $z^4,$
$$z^4+\dfrac1{z^4}+z^3+\dfrac1{z^3}+z^2+\dfrac1{z^2}+z+\dfrac1z+1=0$$
$z+\dfrac1z=2\cos\dfrac{2\pi r}9=2y$(say)
Use
$\left(z+\dfrac1z\right)^2=z^2+\dfrac1{z^2}+2$
$\left(z+\dfrac1z\right)^3=z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$
$z^4+\dfrac1{z^4}=\left(z^2+\dfrac1{z^2}\right)^2-2=\left(\left(z+\dfrac1z\right)^2-2\right)^2-2$
to form a bi-quadratic equation in $y$
Now use Vieta's formula
See also: factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
