Conditions for Moving Function Outside Sine Argument Are there multiple discrete functions $f[n]$ such that 
\begin{align}
\sin \left( f[n] x[n] \right) = f[n] \sin \left( x[n] \right)
\end{align}
I know that the above equation holds for $f[n] = 0$, $f[n] = 1$, $f[n] = u[n]$  (where $u[n] = 0$ for $n < 0$ and 1 otherwise) and $f[n] = \delta[n]$ (where $\delta[n] = 1$ for $n = 0$ and $0$ otherwise).  If there are others, are there any special conditions for $f[n]$ and $x[n]$ that allow for this type of "commuting"?  Is it basically just variations of functions that modulate between 0 and 1?


One has $\sin(f\cdot x)=f\cdot\sin(x)$ for every $x$ if and only if $f=-1$ or $0$ or $1$.

The if part is obvious. For the only if part, consider the limit of the identity $\sin(f\cdot x)=f\cdot\sin(x)$ when $x\to0$. Then $\sin(x)=x-x^3/6+o(x^3)$ hence 
$$
f\cdot\sin(x)=f\cdot x-f\cdot x^3/6+o(x^3).
$$
Furthermore $f\cdot x\to0$ hence 
$$
\sin(f\cdot x)=f\cdot x-f^3\cdot x^3/6+o(x^3).
$$
The expansion along powers of $x$ is unique hence $f/6=f^3/6$, that is, $f^3-f=0$, that is, $f\cdot(f-1)\cdot(f+1)=0$. Finally, $f=-1$ or $0$ or $1$.
