Answer
$p_{0}(x)=0 \\ p_{1}(x)=x \\p_{2}(x)=x \\p_{3}(x)=x-\dfrac{x^3}{6}$
Work Step by Step
Taylor polynomial of order $n$ for the function $f(x)$ at the point $k$ can be defined as:
$p_n(x)=f(k)+\dfrac{f'(k)}{1!}(x-k)+\dfrac{f''(k)}{2!}(x-k)^2+....+\dfrac{f^{n}(k)}{n!}(x-k)^n$
Here, $f(0)=0 ; f'(x)=\cos x \implies f'(0)=1; f''(x)=-\sin x \implies f''(0)=0; f'''(x)=-\cos x \implies f'''(0)=-1$
Thus, $p_{0}(x)=0 \\ p_{1}(x)=x \\p_{2}(x)=x \\p_{3}(x)=x-\dfrac{x^3}{6}$
