Answer
$2x-\dfrac{8x^3}{3!}$
Work Step by Step
The Taylor approximation for degree $n$ centered at point $a$ can be written as:
$P_n(x)=f(a)+f'(a)(x-a)+\dfrac{1}{2}f''(a)(x-a)^2+.......+\dfrac{1}{n!}f^n(a)(x-a)^n ~~~~........(1)$----
We have: $f(x)=\sin (2x)$; $n=3$ and $a=0$
and $f(0)=\sin (0)=0$
Further, $f'(x)=2 \cos (2x) \implies f'(0)=2\cos (0)=2\\ f''(x)=-4 \sin (2x) \implies f''(0)=0\\ f'''(x)=-8 \cos (2x) \implies f'''(0)=-8$
Now, plug these values in the equation (1) to obtain:
$P_3(x)=f($2x-\dfrac{8x^3}{3!}$0)+f'(0)x+\dfrac{1}{2!}f''(0)x^2+f'''(0)\dfrac{x^3}{3!} \\=0+2x+0-\dfrac{8x^3}{3!}\\=2x-\dfrac{8x^3}{3!}$
