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.
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 ;).