Answer
$$T_3(x) =-\left(x-\frac{\pi}{2}\right)+\frac{1}{6}\left(x-\frac{\pi}{2}\right)^{3} $$
Work Step by Step
Given $$f(x) =\cos x,\ \ a=\pi /2$$
Since
\begin{align*}
f(x) &=\cos x\ \ \ \ \ \ \ \ \ \ \ \ f(\pi/2) =0\\
f'(x) &=-\sin x\ \ \ \ \ \ \ f'(\pi/2 )= -1 \\
f''(x) &=-\cos \ \ \ \ \ \ \ \ f''(\pi/2 )=0\\
f'''(x) &=\sin x \ \ \ \ \ \ \ \ \ f'''(\pi/2 )=1
\end{align*}
Then
\begin{align*}
T_3(x)&= \sum_{n=0}^{3}\frac{f^{(n)}(\pi/2)}{n!}(x-\pi/2)^n\\
& =-\left(x-\frac{\pi}{2}\right)+\frac{1}{6}\left(x-\frac{\pi}{2}\right)^{3}
\end{align*}
