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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $$\sin x\cos x=\frac{1}{2}\left(2x-\frac{(2x)^3}{3!}+\ldots\right)\Rightarrow \mbox{coef. }(x^3)=\frac{1}{2}\cdot \frac{8}{6}=\frac{2}{3}$$