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.



linear independence of $\sin(k \pi / m)$
Note: Fedor and Vladimir have already answered the question, but this is a partial answer in the other direction, under a stronger hypothesis. (This answer, which I had earlier deleted, has been edited in response to some helpful comments.) 


If $m$ is odd and square-free, then the claim of the OP holds.


Let $S$ be the set $\{k: 1 \leq k \leq m/2, (k, m) = 1\}$. If $\sum\_{k \in S} a\_k \sin(k\pi/m) = 0$ for some rationals $a\_k$, then 


$$\sum\_{k \in S} a\_k(e^{k\pi i/m} - e^{-k\pi i/m}) = 0.$$ 


Here $N = 2m$ is square-free, and in that case the primitive $N$-th roots of unity are linearly independent over $\mathbb{Q}$ (see this mathstackexchange discussion: <https://math.stackexchange.com/questions/87290/basis-of-primitive-nth-roots-in-a-cyclotomic-extension>). 


For $N = 2m$, any odd $k \in S$ is prime to $N$ and hence $e^{k \pi i/m}, e^{-k\pi i/m}$ are primitive $N$-th roots of unity. If $k \in S$ is even, then $m + k$ is odd and prime to $m$ and thus to $N$, so $e^{(m + k)\pi i/m} = -e^{k\pi i/m}$ is also primitive $N$-th root of unity, as is its conjugate $e^{(m - k)\pi i/m}$; notice the $m-k$ lie in $\{j: m/2 \leq j \leq m, \gcd(j,m) = 1\}$ which is disjoint from $S$. Then 


$$\begin{array}{lll}
\sum\_{k \in S} a\_k(e^{k\pi i/m} - e^{-k\pi i/m}) & = & \sum\_{\text{odd}\; k \in S} a\_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum\_{\text{even}\; k \in S} a\_k(e^{k\pi i/m} - e^{-k\pi i/m}) \\ 
& = & \sum\_{\text{odd}\; k \in S} a\_k(e^{k\pi i/m} - e^{-k\pi i/m}) + \sum\_{\text{even}\; k \in S} a\_k(e^{(m-k)\pi i/m} - e^{-(m-k)\pi i/m})
\end{array}$$ 


where all the primitive roots of unity appearing in the last expression are manifestly distinct. By linear independence of the primitive roots, if that linear combination is zero, then $a\_k = 0$ for all $k$, as required. 


