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