Can $\sin(xy)$ be written in terms of trigonometric functions of only $x$ or $y$? Can $\sin(xy)$ be written in terms of trigonometric functions of only $x$ or $y$? I am tempted to say yes, because the double- and half-angle formulae exist, and these would be special cases of $\sin(xy)$.
I first looked at the Taylor series,
$$\sin(xy) = \sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}y^{2n+1}}{(2n+1)!} = xy-\frac{x^3y^3}{6}+\frac{x^5y^5}{120}-\cdots$$
but as far as I know, neither $x$ nor $y$ can come out of the sum since they are not constants.
 A: If you think at $\sin(xy)$ be expressed as a combination of a FINITE number of  sums and powers of $\sin(x)$ and $\sin(y)$ , the answer is certainly no.
With infinite series, one can imagine a formal answer (more a joke than serious) 
Starting from :
$$\sin(xy)=\sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}y^{2k+1}}{(2k+1)!}$$
$$a=\sin(x)\quad\to\quad x=\sin^{-1}(a)=\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n a^{2n+1}}{(2n+1)n!}=\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n \sin^{2n+1}(x)}{(2n+1)n!}$$
And similarly 
$$y=\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n \sin^{2n+1}(y)}{(2n+1)n!}$$
$\left(\frac{1}{2} \right)_n$ is the Pochhammer symbol $=\frac{\Gamma(n+\frac{1}{2})}{\Gamma(\frac{1}{2})}$.
Finally, the monster :
$$\sin(xy)=\sum_{k=0}^\infty \frac{(-1)^k \left(\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n \left(\sin(x)\right)^{2n+1} }{(2n+1)n!} \right)^{2k+1}\left(\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n \left(\sin(y)\right)^{2n+1} }{(2n+1)n!} \right)^{2k+1}}{(2k+1)!}$$
