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.
Best Answer
Note that\begin{multline}\cos\left(\frac{2\pi}9\right)+\cos\left(\frac{4\pi}9\right)+\cos\left(\frac{6\pi}9\right)+\cos\left(\frac{8\pi}9\right)=\\=\frac12\left(e^{2\pi i/9}+e^{-2\pi i/9}+e^{4\pi i/9}+e^{-4\pi i/9}+e^{6\pi i/9}+e^{-6\pi i/9}+e^{8\pi i/9}+e^{-8\pi i/9}\right)\end{multline}But this is half the sum of all ninth roots of unity other than $1$. So, it's half the sum of the roots of$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x$$and that sum is $-1$.