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?
Best Answer
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 $\sin x$.