Answer
$$T_2=x,\ \ T_3= x-\frac{1}{6} x^{3}$$
Work Step by Step
Since
\begin{array}{ll}
{f(x)=\sin x,} & {f(0)=0} \\
{f^{\prime}(x)=\cos x,} & {f^{\prime}(0)=1} \\
{f^{\prime \prime}(x)=-\sin x,} & {f^{\prime \prime}(0)=0} \\
{f^{\prime \prime \prime}(x)=-\cos x,} & {f^{\prime \prime \prime}(0)=-1}
\end{array}
Then
\begin{aligned}
T_{2}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2} \\
&=0+(x-0)+(0)(x-0)^{2} \\
&=x
\end{aligned}
and
\begin{aligned}
T_{3}(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3} \\
&=0+(0)(x-0)+(x-0)^{2}-\frac{1}{6}(x-0)^{3} \\
&=x-\frac{1}{6} x^{3}
\end{aligned}
