I'm trying to understand this proof of the theorem:
$$\cos({\pi\over 11}) + \cos({3\pi\over 11}) + \cos({5\pi\over 11}) + \cos({7\pi\over 11})+ \cos({9\pi\over 11}) = {1\over 2}$$
I'll recreate it briefly here.
Consider the sum:
$$S_1 = \cos({\pi\over 11}) + \cos({3\pi\over 11}) + \cos({5\pi\over 11}) + \cos({7\pi\over 11}) + \cos({9\pi\over 11})$$
As $\cos$ is an even function, we may re-express this as:
$$S_2 =\cos({-\pi\over 11}) + \cos({-3\pi\over 11}) + \cos({-5\pi\over 11}) + \cos({-7\pi\over 11}) + \cos({-9\pi\over 11})$$
Together with:
$$\cos({11\pi\over 11}) = \cos\pi = -1$$
These give the real part of all $11$ eleventh roots of unity, which sum to $0$.
$$S_1 + S_2 - 1 = 0 = 2S_1 -1$$
Hence $S_1 = 1/2$
There are a couple key steps I don't follow. Firstly, how are $\cos({\pi\over 11}), \cos({3\pi\over 11}), \ldots \cos({-9\pi\over 11})$ found to correspond to the eleventh roots of unity? In general, the real part $n^{th}$ roots of unity are given by:
$$\cos({2\pi k \over n})\;\;\;\;\;\;\;\; k = 0,1,\ldots n-1$$
In the case $n=11$, this yields:
$$\cos(0), \;\cos({2\pi \over 11}), \;\cos({4\pi k \over n}), \ldots\;\cos({20\pi k \over n})$$
Further, the expression within the cosine is never negative. How, then, do $\cos({-\pi\over 11}), \ldots$ also correspond to roots of unity?
First, there is a mistake in the language. The roots are not roots of unity, but of $-1$. Also the roots of $-1$ are symmetric with respect to the real line in the complex plane. Note that your angles go from $0$ to $\pi$, so you get only half of the roots. To get all the roots you need to add another angle interval of length $\pi$. You can go from $\pi$ to $2\pi$, or, as in this case, from $-\pi$ to $\pi$.