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

MHB Math Helper

#### Fernando Revilla

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