Let $N_x,N_y>1$. Let $1\le l,l'\le N_x$ and $1\le k,k'\le N_y$. If $l\ne l'$ or $k\ne k'$, prove
$$\sum\limits_{n=1}^{N_x}\sum\limits_{m=1}^{N_y}\sin\left(l\cdot \frac{n\pi}{N_x+1}\right)\sin\left(l'\cdot \frac{n\pi}{N_x+1}\right) \sin\left(k\cdot \frac{m\pi}{N_y+1}\right)\sin\left(k'\cdot \frac{m\pi}{N_y+1}\right)=0.$$
I've been working on this for hours and still can't figure it out. I'm sure there must be an identity that makes this clear. If I break this down into cases I have $l=l'$ and $k\ne k'$, $l\ne l'$ and $k= k'$, $l\ne l'$ and $k\ne k'$ to consider. Do to the symmetry of the summation I assume that once I show the first case I will clearly see the others.
So assume $l=l'$ and $k\ne k'$. Then
\begin{align*} &\sin\left(l\cdot \frac{n\pi}{N_x+1}\right)\sin\left(l'\cdot \frac{n\pi}{N_x+1}\right) \sin\left(k\cdot \frac{m\pi}{N_y+1}\right)\sin\left(k'\cdot \frac{m\pi}{N_y+1}\right)\\[1em] &=\frac{1}{2}\left[\cos\left([l-l']\frac{n\pi}{N_x+1}\right)-\cos\left([l+l']\frac{n\pi}{N_x+1}\right)\right] \sin\left(k\cdot \frac{m\pi}{N_y+1}\right)\sin\left(k'\cdot \frac{m\pi}{N_y+1}\right)\\[1em] &=\frac{1}{2}\left[1-\cos\left(2l\frac{n\pi}{N_x+1}\right)\right] \sin\left(k\cdot \frac{m\pi}{N_y+1}\right)\sin\left(k'\cdot \frac{m\pi}{N_y+1}\right) \end{align*}
Under the assumption I am moving in the right direction, I don't see where to go now. Any help will be greatly appreciated!