Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$

Question

Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$

This question came after an exercise involving finding the $7$th roots of $-1$. The roots were $\operatorname{cis}\frac{\pi}{7},\operatorname{cis}\frac{3\pi}{7},\dots$

This made me wonder if I could somehow use those roots, along with the geometry of complex numbers, to compute the expression. Any insight would be helpful. Thanks!

This is the question about the same sum: How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Although there the OP does not ask specifically for solution based on complex numbers. (So perhaps the two questions should not be considered duplicates based on this difference.) I will add that I found the other question using Approach0. — Martin Sleziak, Apr 28 '18 at 15:17

Delta-u · Answer 1 · 2018-04-28T14:47:50.053

2

As $\cos(\pi-x)=-\cos(x)$ this is the same as to compute: $$\cos \left( \frac{\pi}{7} \right)+\cos \left( \frac{3 \pi}{7} \right)+\cos \left(5 \frac{\pi}{7} \right)$$

But denoting by $\omega_i$ $i\in{0,\ldots 6}$ the $7$th roots of $-1$.

You have: $$\sum_{i} \omega_i=0$$ and by taking the real part: $$-1+2\left( \cos \left( \frac{\pi}{7} \right)+\cos \left( \frac{3 \pi}{7}\right)+ \cos \left(5 \frac{\pi}{7} \right) \right)=0$$

The geometric interpretation is that the center of mass of a regular heptagon (the $7$th roots of $-1$) is $0$ so it must be the same for the center of mass of the projections on the $x$-axis of it vertices.

edited Apr 28 '18 at 14:47

answered Apr 28 '18 at 14:25

Delta-u

6,038

There is a typographical error in the first line: You meant to say $\cos(\pi - x) = \color{red}{-}\cos x$. – N. F. Taussig Apr 28 '18 at 14:44
Indeed, thank you :-). – Delta-u Apr 28 '18 at 14:47

Compute $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}$

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