Science:Math Exam Resources/Courses/MATH100/December 2016/Question 13 (a)/Solution 2

We know (or can calculate) that the Taylor expansion of $\sin (x)$ centred at $x=0$ is

$\sin (x)=x-\frac{x^3}{3!}+\cdots .$

Hence the Taylor expansion of $f(x)=(x+1)\sin (x)$ centred at $x=0$ is

$(x+1)\sin (x)=(x+1)(x-\frac{x^3}{3!}+\cdots )=x^2-\frac{x^4}{3!}+x-\frac{x^3}{3!}+\cdots .$

Discarding the terms of degree higher than 3, we obtain the third degree Taylor polynomial

$T_3(x)=x+x^2-\frac{x^3}{6}.$