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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.

gen-ℤ ready to perish
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    It might depend on what "in terms of" means (though I am very much inclined to suggest that the answer is "no" regardless). For one thing, it is apparent that for any function $g:\Bbb R^4\to\Bbb R$, the map $g(\sin x,\cos x,\sin y,\cos y)$ is a bi-periodic function $\Bbb R^2\to\Bbb R$, while $\sin(xy)$ is not. –  Aug 06 '17 at 17:07
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    Half and double or $\sin kx$ are much more different from $xy$ Think to $e^{xy}$ or $\log(xy)$. No way. Forget it :) – Raffaele Aug 06 '17 at 17:10
  • Taylor series give nothing. Try Ryan series. – hamam_Abdallah Aug 06 '17 at 17:13
  • @Blue I think you are partially right. This question and its answers are more general, but one answer does explicitly say $$\sin(n x)=\sum_{d=0}^\infty \frac{d\cdot \cos\left(\frac{n-d}{2}\pi\right)\frac{2^dn}{n+d}{{\frac{n+d}2}\choose d}}{n}\cos^{d-1}(x)\sin(x) \ \ \ \ \ \forall |x|\leq\pi$$ However, these answers already have provided different formulae and insight, so I'm not sure. – gen-ℤ ready to perish Aug 06 '17 at 19:57

1 Answers1

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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 : $$\sin(xy)=\sum_{k=0}^\infty \frac{(-1)^k \left(\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n \left(\sin(x)\right)^{2n+1} }{(2n+1)n!} \right)^{2k+1}\left(\sum_{n=0}^\infty \frac{\left(\frac{1}{2} \right)_n \left(\sin(y)\right)^{2n+1} }{(2n+1)n!} \right)^{2k+1}}{(2k+1)!}$$

JJacquelin
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