Proof $|\sin(x) - x| \le \frac{1}{3.2}|x|^3$ So, by Taylor polynomial centered at $0$ we have:
$$\sin(x) = x-\frac{x^3}{3!}+\sin^4(x_o)\frac{x^4}{4!}$$
Where $\sin^4(x_0) = \sin(x_o)$ is the fourth derivative of sine in a point $x_0\in [0,x]$.
Then we have:
$$\sin(x)-x = -\frac{x^3}{3!}+\sin(x_0)\frac{x^4}{4!}$$
I thought about proving the error $\sin(x_0)\frac{x^4}{4!}$ is always positive, but for $x>\pi$ this is not the case. 
Someone has a hint?

$$\sin x - x = \int_{0}^{x}(\cos t-1)\, dt$$
hence, for any $x>0$:
$$|\sin x - x| = \int_{0}^{x}2\sin^2\frac{t}{2}\,dt\leq \int_{0}^{x}\frac{t^2}{2}\,dt=\frac{x^3}{6}$$
since $\sin\frac{t}{2}\leq\frac{t}{2}.$
