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Numerically, I have found that the following formula seems to be independent of $N$ for any choice of $a$ and $b$ at large $N$: $$\sum_{k=1}^N \frac{\sin^2\phi_k}{N(a^2+b^2-2ab\cos\phi_k)^2}$$ with $\phi_k = \frac{2\pi k}{N}$. This seems somehow related to the cosine and sine formulas (at least for positive $a$ and $b$); however, I have not managed to find an analytical proof so far for this claim. I know this is true for extreme cases (such as $a\ll b$), but I would like a proof for the general case.

Could anyone help me solve this issue, or at least point me in the good direction?

eje eje
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    What you are seeing is due to this being a Riemann sum, i.e. it's an approximation for the integral $\int_0^{2\pi}\frac{\sin^2 \theta}{2\pi(a^2+b^2-2ab\cos \theta)^2}d\theta$. The larger $N$ gets, the closer the sum gets to this integral (it converges as $N\to\infty$). What you need to look at is what the conditions are for a Riemann sum to converge and then check that this function has this property. – Winther May 07 '23 at 09:51

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The sum can be evaluated in closed form (like I did it here or here).

Assume $0<|a|<|b|$. Then the sum is $\boxed{S_N(a/b)/(ab)^2}$, where $$S_N(z)=\frac1N\sum_{k=0}^{N-1}\left(\frac{z\sin\phi_k}{1-2z\cos\phi_k+z^2}\right)^2.$$

Now we use two facts. First, for $|z|<1$ we have the expansion $$\frac{z\sin\phi_k}{1-2z\cos\phi_k+z^2}=\sum_{n=0}^\infty z^n\sin n\phi_k=\frac1{1-z^N}\sum_{n=0}^{N-1}z^n\sin n\phi_k$$ (the last expression is valid for "arbitrary" $z$). Second, for $n\in\mathbb{Z}$ we have $$\frac1N\sum_{k=0}^{N-1}\cos n\phi_k=\mathbf{1}_{N\,\mid\,n}:=\begin{cases}1,&N\mid n\\0,&N\not\mid n\end{cases}.$$

Thus, $$ 2(1-z^N)^2 S_N(z)=\frac2N\sum_{m,n,k=0}^{N-1}z^{m+n}\sin m\phi_k\sin n\phi_k \\=\frac1N\sum_{k,m,n=0}^{N-1}z^{m+n}\big(\cos(m-n)\phi_k-\cos(m+n)\phi_k\big) \\=\sum_{m,n=0}^{N-1}z^{m+n}(\mathbf{1}_{N\,\mid\,(m-n)}-\mathbf{1}_{N\,\mid\,(m+n)})=\sum_{n=1}^{N-1}z^{2n}-(N-1)z^N $$

$$\implies\bbox[5pt,border: 2pt solid black]{S_N(z)=\frac1{2(1-z^N)^2}\left(\frac{z^2-z^{2N}}{1-z^2}-(N-1)z^N\right).}$$

metamorphy
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