Taylor Polynomial Question

Question

Can anyone help me with this question?

Calculate the Taylor polynomials $T_2(x)$ and $T_3(x)$ centered at $x=\frac{\pi}{6}$ for $f(x)=\sin(x)$.

Answer 1

So $f(x)=\sin(x)$, $f'(x)=\cos(x)$, $f''(x)=-\sin(x)$. Thus $f({\pi\over 6})=1/2$, $f'({\pi\over 6})=\sqrt{3}/2$, $f''({\pi\over 6})=-1/2$. The Taylor series begins $$T(x)={1\over 2}+{\sqrt{3}\over 2}\left(x-{\pi\over 6}\right)-{1\over 2}{1\over 2!}\left(x-{\pi\over 6}\right)^2$$