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.
Fernando Revilla
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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 $$\sin x\cos x=\frac{1}{2}\left(2x-\frac{(2x)^3}{3!}+\ldots\right)\Rightarrow \mbox{coef. }(x^3)=\frac{1}{2}\cdot \frac{8}{6}=\frac{2}{3}$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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