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?
 A: If $x[n]=pi/2$, $sin(f[n]*\pi/2)=f[n]$. Sine can only take on values between -1 and 1 so $f[n]\in [-1,1]$
\begin{align}
\sin \left( f[n] x[n] \right) - f[n] \sin \left( x[n] \right) =0
\end{align}
This equation has to hold for all functions $x[n]$ so let's assume $x[n]=\sqrt{2}$ so that we can find some functions $f[n]$ that may work. A plot of $f[n]$ vs $sin(f[n]\sqrt{2})-f[n]sin(\sqrt{2})$ for $f[n]\in [-1,1]$ shows that the only values that $f[n]$ can take are -1, 0 and 1. (plot)
$f[n]={-1, 0, 1}$
$\ $
Alternate (analytic) solution:
\begin{align}
&fsin(x)=sin(fx)\\
\\
\text{apply $\frac{d^2}{dx^2}$ to both sides}\\
&-fsin(x)=-f^2sin(fx)\\
\text{divide both sides by $-f$}\\
&sin(x)=fsin(fx)\\
\text{substitute into first equation}\\
&f^2sin(fx)=sin(fx)&\\
\text{this equation can be true iff $sin(fx)=0$ or $f^2=1$}\\
\\
&f=-1,0,1
\end{align}
A: 
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 $1$.
