How to prove $\cos \frac{2\pi}{9} + \cos \frac{4\pi}{9} + \cos \frac{8\pi}{9}$=0 I tried letting $z=cis \frac{2\pi}{9}$
and using trigonometric identities i got $\frac{1}{2z^4} [\frac{z^9-1}{z-1}]-\frac{1}{2z^3} - \frac{1}{2}$
and since i assumed that z then $z^9=cis (2\pi) =1$
but i ended up with $\cos \frac{2\pi}{n} + \cos \frac{4\pi}{n} + \cos \frac{8\pi}{n} = \frac{(-1+i\sqrt(3))}{4}$

$\cos \frac{2\pi}{9} + \cos \frac{4\pi}{9} + \cos \frac{8\pi}{9}=0$
Let $\frac{2\pi}{9}=t$, then
$\cos t + \cos 2t + \cos 4t=\cos t+\cos ^2t-\sin ^2t+\cos^2(2t)-\sin ^2(2t)=$
$=\cos t+\cos ^2t-\sin ^2t+(\cos ^2t-\sin ^2t)^2-(2 \sin t \cos t)^2=0$
$=\cos t+\cos ^2t-\sin ^2t+\cos ^4t+\sin ^4t-2 \cos ^2t \sin ^2t -4 \cos ^2t \sin ^2t =$
Try this. It should help you.
