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I am trying to figure out for what values of $n$, the numbers $\sin\left(\frac{2\pi k}{n}\right)$, for $k = 1,\dots,n-1$, are linearly independent over the rationals.

Any thoughts on how I may want to approach this problem? It would be greatly appreciated.

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Do you know if they are linearly independent over the reals? E.g. Fourier series. –  Calvin Lin Jun 20 '13 at 14:54
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$1\cdot \sin(2\pi\cdot 0/n)=0$ shows not linearly independent over the rationals. I think you want to start your $k$ at $1$. –  coffeemath Jun 20 '13 at 15:58
    
oh yes, I think the k must be started at 1. Sorry. –  metallicmural Jun 20 '13 at 17:27
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Don't these numbers sum to $0$ by any chance? (I mean, $\sum_k \sin 2k \pi/n = \Im \sum_k e^{2 k i \pi / n } = \Im 0 = 0$) –  Feanor Jun 20 '13 at 19:30
    
yes they do!!! didn't notice that. So they are not linearly independent ! thank you ! –  metallicmural Jun 20 '13 at 20:43

1 Answer 1

It so happens that the sum of the mentioned numbers sum up to $0$. It follows that, in particular, they cannot be linearly independent.

In the general case, to compute the sum of the type: $\sum_{k=0}^{m-1} \sin \frac{2 \pi k}{n}$ we can proceed as follows: $$\sum_{k=0}^{m-1} \sin \frac{2 \pi k}{n} = \sum_{k=0}^{m-1} \Im(e^{2 \pi ik/n}) = \Im \frac{e^{2 \pi im/n} -1}{e^{2 \pi i/n}-1} $$ When you plug in $m = n$, it turns out that: $$\sum_{k=0}^{m-1} \sin \frac{2 \pi k}{n} = \Im \frac{e^{2 \pi i} -1}{e^{2 \pi i/n}-1} = \Im \frac{1 -1}{e^{2 \pi i/n}-1} = 0 $$


As Gerry Myerson rightly points out, this is not quite the simplest solution ;).

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Don't these sines just cancel out in pairs? –  Gerry Myerson Jun 21 '13 at 9:38
    
@GerryMyerson: yes, they do! my bad –  Feanor Jun 21 '13 at 9:41
    
I am now struggling with the case when n = prime and k = 0 ... (p-1)/2 ... –  metallicmural Jun 21 '13 at 15:17

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