How to prove that $\cos(\pi÷11)+\cos(3\pi÷11)+\cos(5\pi÷11)+\cos(7\pi÷11)+\cos(9\pi÷11)=0.5$?

Question

I need to prove that $$\cos\dfrac{\pi}{11}+\cos\dfrac{3\pi}{11}+\cos\dfrac{5\pi}{11}+\cos\dfrac{7\pi}{11}+\cos\dfrac{9\pi}{11}=\dfrac{1}{2}$$

How to do it?

Answer 1

Consider an $11$-sided regular polygon inscribed in the unit circle, with one vertex at $(-1,0)$. The centre of gravity of the eleven vertices is the origin. Looking at the $x$-coordinate, this shows that $$0 = -1 + \sum_{k = 0}^4 \cos\frac{(2k+1)\pi}{11} + \sum_{k = 0}^4 \cos\frac{-(2k+1)\pi}{11}.$$ But in view of the fact that the cosine function is even, the two sums appearing above are in fact equal. Therefore, each (and hence the first, which is what the question is about) is equal to half.