Can we show that
$$\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$\cos\frac{\pi}{5}+\cos\frac{2\pi}{5}+\cos\frac{3\pi}{5}+\cos\frac{4\pi}{5}=0$$
as well, so just out of curiosity, is it true that $$\sum_{k=1}^{n-1} \cos\frac{k\pi}{n} = 0$$
for all $n$ odd?
Best Answer
Note that $$\cos(\pi - \alpha)= - \cos(\alpha)$$ Therefore $$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{5\pi}{7})+\cos(\frac{6\pi}{7})=$$
$$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})-\cos(\frac{3\pi}{7})-\cos(\frac{2\pi}{7})-\cos(\frac{\pi}{7})=0$$
The same goes for other natural numbers $n$ instead of $7$.