Answer
$A_{0}(x)=0 \\ A_{1}(x)=x \\A_{2}(x)=x \\A_{3}(x)=x-\dfrac{x^3}{6}$
Work Step by Step
The Taylor polynomial is defined for the function $f(x)$ at the point $a$ of order $n$ as:
$A_n(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
Now, we get $f(0)=0; \\f'(x)=\cos x \implies f'(0)=1 \\ f''(x)=-\sin x \implies f''(0)=0 \\f'''(x)=-\cos x \implies f'''(0)=-1$
This implies that $A_{0}(x)=0=0 \\ A_{1}(x)=0+x=x \\A_{2}(x)=0+x+(0)x^2=x \\A_{3}(x)=0+x+(0)x^2-(\dfrac{1}{6})x^3=x-(\dfrac{1}{6})^3$
