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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?

10GeV
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    Is there a + sign between the last two cosines? – Andrei Apr 05 '22 at 15:58
  • @Andrei Yes, thanks for pointing that out. – 10GeV Apr 05 '22 at 16:01
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    They're actually $-1$ times the (real parts of the) eleventh roots of unity. Since those will add up to $0$ anyway, it doesn't really matter. Note that $\cos \frac{9\pi}{11} = -\cos \frac{2\pi}{11}, \cos \frac{7\pi}{11} = -\cos \frac{4\pi}{11}$, etc, and $\cos (-\theta) = \cos \theta$, so $S_1 = S_2$. – Brian Tung Apr 05 '22 at 16:06

1 Answers1

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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 $\pi$.

Andrei
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