Answer
\begin{aligned}
T_{2}(x) &= 1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^{2}\\
T_3(x)&=1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^{2}
\end{aligned}
Work Step by Step
Since
\begin{array}{ll}
{f(x)=\sin x,} & {f\left(\frac{\pi}{2}\right)=1} \\
{f^{\prime}(x)=\cos x,} & {f^{\prime}\left(\frac{\pi}{2}\right)=0} \\
{f^{\prime \prime}(x)=-\sin x,} & {f^{\prime \prime}\left(\frac{\pi}{2}\right)=-1} \\
{f^{\prime \prime \prime}(x)=-\cos x,} & {f^{\prime \prime \prime}\left(\frac{\pi}{2}\right)=0}
\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} \\
&=1+(0)\left(x-\frac{\pi}{2}\right)-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^{2} \\
&=1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^{2}
\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} \\
&=1+(0)\left(x-\frac{\pi}{2}\right)-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^{2}+(0)\left(x-\frac{\pi}{2}\right)^{3} \\
&=1-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^{2}
\end{aligned}
