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# linear independence of $\sin(k \pi / m)$

I have tried searching the literature for a result like the following, but have not found anything.

For a positive integer $m$, is it known that $$\{ \sin (k \pi / m): 1 \leq k \leq m/2, (k,m)=1 \}$$ is linearly independent over the rationals?

References or a proof would be greatly appreciated.

$$\sin \frac{5\pi}{36}+\sin \frac{7\pi}{36}-\sin \frac{17\pi}{36}=0.$$ This may be otained by multiplying $2\sin\frac{\pi}6-1=0$ by $\cos \frac{\pi}{36}$.

We have $$\sin\frac{\pi}{9}+\sin\frac{2\pi}9-\sin\frac{4\pi}9=\sin\frac{2\pi}{18}+\sin\frac{4\pi}{18}-\frac{8\pi}{18}=\sin\frac{2\pi}{18}-\frac{8\pi}{18}+\sin\frac{14\pi}{18},$$ and the latter, denoting $\xi_{18}=\exp\frac{2\pi i}{18}$, is the imaginary part of $$\xi_{18}-\xi_{18}^4+\xi_{18}^7=\xi_{18}(1-\xi_{18}^3+\xi_{18}^6)=0.$$ Thus, your conjecture is wrong.

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# linear independence of $\sin(k \pi / m)$

I have tried searching the literature for a result like the following, but have not found anything.

For a positive integer $m$, is it known that $$\{ \sin (k \pi / m): 1 \leq k \leq m/2, (k,m)=1 \}$$ is linearly independent over the rationals?

References or a proof would be greatly appreciated.

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$$\sin \frac{5\pi}{36}+\sin \frac{7\pi}{36}-\sin \frac{17\pi}{36}=0.$$ This may be otained by multiplying $2\sin\frac{\pi}6-1=0$ by $\cos \frac{\pi}{36}$.

How do you come up with this? - Nik Weaver Jul 19 at 16:46

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