What will be a reason why $\sin(xy)$ and $\sin(x/y)$ have no identities in form of separated functions of $x$ and $y$? What I am looking for is, for example, $\sin(xy)=f(x)g(y)$, where $f$ and $y$ are some functions for $x$ and $y$.
It is "trivial" to see the answer is no, but not really able to come up with a reasonable reason.
 A: $$\sin (x y) = f(x)g(y) = \sin(xy + 2\pi) = \sin(x ( y + 2\pi /x)) = f(x) g(y + 2\pi /x).$$ If this should hold for all $x \neq 0$, then $g(y) = g(y + 2 \pi /x)$, so $g$ has arbitrary small period, implying that it is constant. Then $\sin(xy) = C f(x)$ which can not be true.
A: If $\sin(xy)=f(x)g(y)$ for all $x,y$, then for any $n \in \mathbb N$ and any real numbers $x_1,\dots,x_n,y_1,\dots,y_n$, the $n \times n$ matrix
$$
\begin{bmatrix}
\sin(x_1y_1) & \sin(x_1y_2) &\dots &\sin(x_1y_n)
\\
\sin(x_2y_1) & \sin(x_2y_2) &\dots &\sin(x_2y_n)
\\
\vdots & \vdots & \ddots & \vdots
\\
\sin(x_ny_1) & \sin(x_ny_2) &\dots &\sin(x_ny_n)
\end{bmatrix}
$$
would have rank (at most) $1$.  But in fact, that is (almost) never true.  For example, the $2 \times 2$ matrix with $x_1=\pi/2,x_2=\pi/3,y_1 = 1,y_2=2$ is
$$
\begin{bmatrix}
1 & 0
\\
\frac{\sqrt{3}}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}
$$
with rank $2$.
A: I guess there is a simpler reason: if we had such $f$, $g$ then 
$f(x) g(\frac{\pi/2}{x})$ would not be zero for every $x\ne 0$, while $f(x) g(\frac{2\pi}{x})$ would be zero, clearly not possible. 
