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I have the following sum:

$$\sum_{\sigma=1,\text{odd}}^{\frac{p}{2}-2}\frac{\sin \frac{r \sigma \pi}{p}}{\sin \frac{q \sigma \pi}{p}}$$

where $p\equiv2\pmod4$, $p$ and $q$ are coprime numbers such that $p-q\ge1$, and $r\in \{1,\dots,p-1\}$ is odd. I know the result is of the form $$\pm \frac{\frac{p}{2}\pm 1}{2},$$ but the signs depend on the relation of $r, p, q$, and it is not easy to guess.

Is there some formula for this type of sums?

I tried to reduce it using exponentials as the following:

$$\sum_{\sigma=1,odd}^{\frac{p}{2}-2}e^{i\pi\frac{(q-r)\sigma}{p}}\frac{e^{2i\pi\frac{r\sigma}{p}}-1}{e^{2i\pi\frac{q\sigma}{p}}-1}$$

and define $\omega=e^{2 i\pi\frac{q}{p}}$ which is a primitive root of unity but I don't know how this can help.

Please help! Thanks!!!

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    $\begingroup$ Hey so r is a multiple of q modulo p/(p,\sigma) so replace r with aq and now it’s easy to divide the denominator into the numerator and sum the resulting geometric series. $\endgroup$ – alpoge May 17 '19 at 16:02
  • $\begingroup$ @alpoge, it should be taking $r \equiv a q \pmod{2p}$, not just $r \equiv a q \pmod p$, right? Otherwise it seems there is a sign issue. $\endgroup$ – LSpice May 17 '19 at 17:22
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    $\begingroup$ Hi Prof. @LSpice! I’m was referring to the last display (in order to get rid of the denominator) but I absolutely agree! It’d have been easier to just start that way (note that q is still prime to 2p). $\endgroup$ – alpoge May 17 '19 at 18:27
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    $\begingroup$ @alpoge, now I want to see if I can get the sign on my office changed to Prof. @‍LSpice. :-) $\endgroup$ – LSpice May 17 '19 at 18:36

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