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.

The roots of unity for $z^9=1$ are $e^{i \frac{2\pi k}9},\>k=0, \pm1,\pm2,\pm3,\pm4$ and their sum is equal to zero, i.e.
$$0=1+ e^{i\frac{2\pi k}9} + e^{-i\frac{2\pi k}9}
 + e^{i\frac{4\pi k}9} + e^{-i\frac{4\pi k}9}
 + e^{i\frac{6\pi k}9} + e^{-i\frac{6\pi k}9}
 + e^{i\frac{8\pi k}9} + e^{-i\frac{8\pi k}9}
$$
which leads to
$$\cos\frac{2\pi}{9}+
 \cos\frac{4\pi}{9}+ \cos\frac{6\pi}{9} + \cos\frac{8\pi}{9} =- \frac{1}{2}$$
