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$$\begin{eqnarray}&&|y \cdot \cos(xy)|\cdot \frac{|x|}{|\sin(xy)|} \\ &= &\frac{|x| \cdot |y \cdot \cos(xy)|}{|\sin(xy)|}\\ &= &\frac{|xy \cdot \cos(xy)|}{|\sin(xy)|}\\ &= &\left|\frac{xy \cdot \cos(xy)}{\sin(xy)}\right|\\ &= &\left|\frac{xy}{\tan(xy)}\right|\end{eqnarray}$$

I need to simplfy it somehow because it's used in a very big task which might get more complicated if I keep it unsimplified. I hope it's fine like that?

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    $\begingroup$ Yes, your simplification looks fine. But then again we do not know if will be useful in your computation, since we do not know what you are computing. $\endgroup$
    – Marko Karbevski
    Jun 12, 2017 at 20:10

2 Answers 2

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It is ''fine'' but with some attention. The starting function is not defined for $xy=k\pi$, the final function is not defined also for $xy=\frac{\pi}{2}+k\pi$, so this case require a special consideration.

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  • $\begingroup$ Hmm do you think it's better to keep it like that, unsimplified? I shall be as precise as possible. Edit: $x,y \neq 0$ $\endgroup$
    – cnmesr
    Jun 12, 2017 at 20:16
  • $\begingroup$ No, the simplification works well, but you have to consider separately the case $xy= \frac{\pi}{2}+k \pi$. In this case the value of the function is $0$ because $\cos xy=0$ and consider that the condition $x,y \ne 0$ is not sufficient to avoid problems with the final function. $\endgroup$
    – Emilio Novati
    Jun 12, 2017 at 20:21
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If you're worried about a calculation taking too long and if $|xy|$ is small, you can use the Laurant series for $\cot t$ centered at $t=0$. Your last expression is

$$= |xy||\cot xy | = |xy|\left| \frac{1}{xy}-\frac{xy}{3}-\frac{(xy)^3}{45} - \frac{2(xy)^5}{945} - \cdots \right| = \left| 1-\frac{(xy)^2}{3} - \frac{(xy)^4}{45} - \cdots \right|. $$

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