What is the Mauclarin series for $\sin x \cos x$? I would think that you could just multiply out the representation for $\sin x$ with the representation for $\cos x$, but that's apparently wrong.

If I start multiplying it out and differentiating I'll get some nasty derivatives. Does anyone have a better way?

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    $\begingroup$ You could multiply them out, but that's very tedious and hard, and would be very difficult to get the general nth term. I think the easiest way would be to realize you have $.5\sin 2x$ here, and use that instead. $\endgroup$ – davidlowryduda Nov 24 '13 at 9:51

You are not wrong to multiply out the two series expansions for $\cos x$ and $\sin x$, but it would be very annoying and hard to give the $n$th coefficient in this way.

However, if you recognize that $\cos x \sin x = \frac{1}{2} \sin (2x)$, then you can write down the Taylor expansion immediately from the expansion for $\sin x$.


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