extreme simplification: $\frac{\sin(Nx)\sin((N+1)x)\sin(Mx)\cos((M+1)x)}{\sin^2(x)}$ I am trying to simplify the function $$f_{N,M}(x)=\frac{\sin(Nx)\sin((N+1)x)\sin(Mx)\cos((M+1)x)}{\sin^2(x)},\qquad N,M\in\Bbb N$$
to the form $$f_{N,M}(x)=\sum_{k}\alpha_k^{(N,M)}\sin(2kx).$$ 
The reason I expect this to be possible is the $(N,M)=(3,5)$ case:
$$f_{3,5}(x)=-\frac12\sin 2x-\frac12\sin4x+\sin8x+\frac32\sin10x+\frac32\sin12x+\sin14x+\frac12\sin16x,$$
and the unexplained answer here.  The $(N,M)=(3,5)$ case, included in an answer to the linked question, is apparently due to the formula for the sine $\sin x=\frac1{2i}(u-1/u)$, where $u=e^{ix}$. I've tried to apply this to the general case at hand:
$$\begin{align}
f_{N,M}(x)&=\frac{\sin(Nx)\sin((N+1)x)\sin(Mx)\cos((M+1)x)}{\sin^2(x)}\\
&=\frac{2^{-1}(2i)^{-3}(u^{N}-u^{-N})(u^{N+1}-u^{-N-1})(u^M-u^{-M})(u^{M+1}+u^{-M-1})}{(2i)^{-2}(u-1/u)^2}\\
&=\frac1{4i}\frac{(u^{N}-u^{-N})(u^{N+1}-u^{-N-1})(u^M-u^{-M})(u^{M+1}+u^{-M-1})}{(u-1/u)^2}.
\end{align}$$
Then I defined the functions 
$$\begin{align}
a_j(x)&=x^j-\frac1{x^j}\\
r_j(x)&=x^j+\frac1{x^j}
\end{align}$$
so that $$4if_{N,M}(x)=\frac{a_N(u)a_{N+1}(u)a_M(u)r_{M+1}(u)}{a_1^2(u)}.$$
Letting the numerator be $\eta$, we note that 
$$r_N(u)r_{N+1}(u)=a_{2N+1}(u)-a_1(u)$$
as well as 
$$r_M(u)a_{M+1}(u)=r_{2M+1}(u)-r_1(u)$$
so that 
$$\eta=a_{2N+1}(u)r_{2M+1}(u)-a_{2N+1}(u)r_1(u)-a_1(u)r_{2M+1}(u)+a_1(u)r_1(u).$$
We see that $$a_p(u)r_q(u)=4i\cos(px)\sin(qx)=2i(s_{q+p}+s_{q-p}),\qquad s_K=\sin Kx$$
so that 
$$\frac{\eta}{2i}=s_{2M+2N+2}+s_{2M-2N}+s_{2N}+s_2-(s_{2N+2}+s_{2M+2}+s_{2M}).$$
This almost seems like something I'm looking for, except for the denominator of $4if_{N,M}$, namely $\sin^2x$, which I can't seem to get rid of. Could I have some help? Thanks.
 A: Introducing the Chebyshev polynomials of the second kind:
\begin{equation}
U_n(\cos(x))=\frac{\sin\left( \left( n+1 \right)x \right)}{\sin x}
\end{equation} 
the terms of the summation can be written as
\begin{align}
f_{N,M}(x)&=\frac{\sin(Nx)\sin((N+1)x)\sin(Mx)\cos((M+1)x)}{\sin^2(x)}\\
&=\cos((M+1)x)\sin(x)U_{N-1}\left(  \cos(x)\right)U_N\left(  \cos(x)\right)U_{M-1}\left(  \cos(x)\right)
\end{align}
From the product decomposition:
\begin{equation}
U_n(z)U_m(z)=\sum_{p=0}^nU_{m-n+2p}(z)\quad\text{ for }m\ge n
\end{equation} 
we deduce
\begin{equation}
U_{N-1}\left(  \cos(x)\right)U_N\left(  \cos(x)\right)=\sum_{p=0}^{N-1}U_{2p+1}(\cos x)
\end{equation} 
To evaluate 
\begin{equation}
U_{N-1}\left(  \cos(x)\right)U_N\left(  \cos(x)\right)U_{M-1}\left(  \cos(x)\right)=\sum_{p=0}^{N-1}U_{2p+1}(\cos x)U_{M-1}\left(  \cos(x)\right)
\end{equation} 
using the above decomposition formula for the product, care must be taken to the relative values of the indices of the polynomials. As $1\le 2p+1\le 2N-1$, if $M\ge 2N$, we have $M-1\ge2p+1$ and thus
\begin{equation}
U_{N-1}\left(  \cos(x)\right)U_N\left(  \cos(x)\right)U_{M-1}\left(  \cos(x)\right)=\sum_{p=0}^{N-1}\sum_{q=0}^{2p+1}U_{M-2p+2q-2}\left(  \cos(x)\right)
\end{equation} 
which gives
\begin{align}
f_{N,M}(x)&=\cos((M+1)x)\sin(x)\sum_{p=0}^{N-1}\sum_{q=0}^{2p+1}U_{M-2p+2q-2}\left(  \cos(x)\right)\\
&=\sum_{p=0}^{N-1}\sum_{q=0}^{2p+1}\cos((M+1)x)\sin\left( \left( M-2p+2q-1 \right) x\right)\\
&=\frac{1}{2}\sum_{p=0}^{N-1}\sum_{q=0}^{2p+1}\left[
\sin\left( \left( 2M-2p+2q \right)x \right)-\sin\left( \left( 2p-2q+2 \right) x\right)
\right]
\end{align}
the proposed decomposition in terms of $\sin\left( 2kx \right)$ holds.
When $M<2N$, two contributions must be taken into account:
\begin{align}
U_{N-1}\left(  \cos(x)\right)U_N\left(  \cos(x)\right)U_{M-1}\left(  \cos(x)\right)&=\sum_{p=0}^{\lfloor M/2\rfloor-1}U_{2p+1}(\cos x)U_{M-1}\left(  \cos(x)\right)\\
&\,\quad+\sum_{p=\lfloor M/2\rfloor}^{N-1}U_{2p+1}(\cos x)U_{M-1}\left(  \cos(x)\right)\\
&=\sum_{p=0}^{\lfloor M/2-1\rfloor}\sum_{q=0}^{2p+1}U_{M-2p+2q-2}\left(  \cos(x)\right)\\
&\,\quad +\sum_{p=\lfloor M/2\rfloor}^{N-1}\sum_{q=0}^{M-1}U_{2p+2q-M+2}\left(  \cos(x)\right)
\end{align}
Proceeding as above, we find
\begin{align}
f_{N,M}(x)&=\sum_{p=0}^{\lfloor M/2-1\rfloor}\sum_{q=0}^{2p+1}\sin\left( \left(M-2p+2q-1  \right)x \right)\cos((M+1)x)\\
&\,\quad +\sum_{p=\lfloor M/2\rfloor}^{N-1}\sum_{q=0}^{M-1}\sin\left( \left( 2p+2q-M+3 \right)x \right)\cos((M+1)x)\\
&=\frac{1}{2}\sum_{p=0}^{\lfloor M/2-1\rfloor}\sum_{q=0}^{2p+1}\left[\sin\left( \left(2M-2p+2q  \right)x \right)-
\sin\left( \left(2p-2q+2  \right)x \right)
\right]\\
&\,\quad +\frac{1}{2}\sum_{p=\lfloor M/2\rfloor}^{N-1}\sum_{q=0}^{M-1}\left[\sin\left( \left( 2p+2q+4 \right)x \right)+\sin\left( \left( 2p+2q-2M+2 \right)x \right)
\right]
\end{align}
Here again, the proposed decomposition holds.
