I have the following expression that is part of a more complicated integral \begin{equation} \cos(\sqrt{x^2+2axy+y^2}) \end{equation}

For $a=\pm 1$ the expression reduces to sum and difference angle formulas.

I would like to express it, for any $-1\le a \le 1$, in terms of $\cos x$, $\sin x$, $\cos y$, $\sin y$, or more generally the sum of separable trigonometric products (e.g. trig($x$)trig($y$) ).

Expanding the cosine into a Taylor series, the square root disappears. One can then apply the binomial / multinomial theorem, and manipulate factorials, but it is unclear to me how to reduce the resulting series to sums and products of simple trig functions.

My final goal is to evaluate the average \begin{equation} \lim_{T\rightarrow\infty}\frac{1}{(2T)^2}\int^T_{-T}\int^T_{-T}\sin(x)\sin(y)\cos(\sqrt{x^2+2axy+y^2})\ dx\ dy \end{equation}

as a function of $a$.

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    $\begingroup$ Any reason to think it's actually possible? $\endgroup$ – Greg Martin Feb 14 '17 at 6:02
  • $\begingroup$ I do not necessarily know that it is possible, but the fact that the square root disappears in the series gave me hope. If it is impossible, can we prove that? Basically I am interested in the average of the expression above multiplied by sin(x) sin(y) (i.e. same way you would average sin squared to 1/2) . The average simplifies nicely if it can be reduced as I asked. If the averaging integral simply does not converge in the limit, is there a proof? $\endgroup$ – Omarco Feb 14 '17 at 19:07
  • $\begingroup$ It would be lovely if we could prove that no function of the form $h(\sqrt{x^2+y^2})$ could be written as a linear combination of functions of the form $f(x)g(y)$; and that might even be morally true; but it's not literally true—take $h(t)=e^{t^2}$. And I don't even know how to formulate a consistent general statement that would imply the impossibility of your question. $\endgroup$ – Greg Martin Feb 14 '17 at 20:14
  • $\begingroup$ As for the integral you say you want to average: it might be possible to integrate when written in polar coordinates. Specifically, change to polar coordinates, and then with $\theta$ fixed, your function looks something like $\rho \sin(a\rho) \sin(b\rho) \cos(c\rho)$; this function has an explicit antiderivative (integrate by parts to get rid of the $\rho$, and remember that products of $\sin$s and $\cos$s can be written as linear combinations of $\sin$s and $\cos$s, thanks to trig angle-addition formulas). $\endgroup$ – Greg Martin Feb 14 '17 at 20:21
  • $\begingroup$ So the integral simplifies greatly if I use the change of variables $u=\sqrt{\frac{1+a}{2}}(x+ y)$ and $v=\sqrt{\frac{1-a}{2}}(x- y)$. Then I am still left with something of the form $\int\int\big[\cos(\sqrt{\frac{2}{1-a}}v)-\cos(\sqrt{\frac{2}{1+a}}u)\big]\cos(\sqrt{u^2+v^2}) \ du \ dv$, which needs to be evaluated. $\endgroup$ – Omarco Feb 15 '17 at 9:31

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