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Show the following result:$$\sum_{m=1}^{99}{\frac{\sin{\left(\frac{17 m \pi}{100}\right)} \sin{\left(\frac{39 m \pi}{100}\right)}}{1+\cos{\left( \frac{m\pi}{100} \right) }}}=1037$$

The source of this problem is unknown to me.

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  • $\begingroup$ A masochist would love this problem. Looks like it will involve repeated use of the trigonometric sum/product formulas. Rewrite the sine product in terms of the difference of cosines. Write the $m=50$ term separately and pair the $m=50-k$ and $m=50+k$ terms for $k=1,49$ to get terms involving product of cosines and sums of cosines. Rewrite those using the sum/product formulas and pray for a simplification. Gruesome! $\endgroup$ Commented Mar 12, 2016 at 5:36

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Not that tedious if viewed algebraically. Let $\Omega_n=\{\omega : \omega^n=1,\omega\neq 1\}$ and $$S(n,m,k)=\sum_{\omega\in\Omega_n}\omega^{-k}(1-\omega)^{-m}$$ (the "$\color{red}{-}k$" is taken for further convenience); then the given sum is equal to $$\frac{1}{4}\big(f(55)+f(-57)-f(21)-f(-23)\big),\qquad f(k)=S(200,2,k).$$ Since $S(n,m,k)-S(n,m,k-1)=S(n,m-1,k)$, we have $$S(n,m,k)=S(n,m,0)+\sum_{d=1}^{k}S(n,m-1,d),\qquad k\geqslant 0,$$ and $S(n,m,0)$ can then be obtained from $$\sum_{k=0}^{n-1}S(n,m,k)\left[=\sum_{k=1}^{n}S(n,m,k)=S(n,m+1,n)-S(n,m+1,0)\right]=0.$$ This way we find, for $0\leqslant k<n$, \begin{align}S(n,0,k)&=n\delta_{0,k}-1=\begin{cases}n-1,&k=0\\\hfill-1,&k\neq 0\end{cases},\\S(n,1,k)&=-k+\frac{n-1}{2},\\S(n,2,k)&=-\frac{k(k+1)}{2}+\frac{n-1}{2}k-\frac{(n-1)(n-5)}{12},\end{align} which is sufficient.

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