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

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?

• I think this was done yesterday. @MartinR found the question... I think we should consider this a duplicate. Jan 16, 2015 at 9:44
• More generally here: math.stackexchange.com/questions/17966/… Jan 16, 2015 at 9:45
• The linked question asks for a proof using Euler's formula. This one doesn't. Jan 16, 2015 at 9:51
• Please no \dfrac in titles (except when absolutely needed for readability).
– Did
Jan 16, 2015 at 10:23

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.