MHB Jason Z's question at Yahoo Answers (Maclaurin series cuestion)

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To find the coefficient of x^3 in the Maclaurin series for sin(x)cos(x), the function can be rewritten as (1/2)sin(2x). The Maclaurin series for sin(t) is t - (t^3)/6 + ..., which leads to sin(2x) being 2x - (8x^3)/6 + .... Therefore, the coefficient of x^3 in sin(x)cos(x) is calculated as (1/2) multiplied by (8/6), resulting in 2/3. This solution provides a clear method for determining the desired coefficient.
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Here is the question:

Find the coefficient of x^3 in the macluarin series for sinxcosx

show work please and thank you!

Here is a link to the question:

Maclaurin Series Question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Jason Z,

We have $\sin x\cos x=\dfrac{1}{2}\sin 2x$. On the other hand for all $t\in\mathbb{R}$, $\sin t= t-\dfrac{t^3}{3!}+\ldots$ so $$\sin x\cos x=\frac{1}{2}\left(2x-\frac{(2x)^3}{3!}+\ldots\right)\Rightarrow \mbox{coef. }(x^3)=\frac{1}{2}\cdot \frac{8}{6}=\frac{2}{3}$$
 
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