Proof $|\sin(x) - x| \le \frac{1}{3.2}|x|^3$

Question

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?

Answer 1

Stop with the $x^3$ term:

$$\sin x - x = \frac{\cos \alpha}{3!} x^3$$

for some $\alpha \in (0, x)$. This form of Taylor's theorem is sometimes called the Lagrange form.

Hence

$$ \left| \sin x - x \right| = \frac{|\cos \alpha|}{3!} |x|^3$$

Now $|\cos\alpha| \leq 1$ and hence

$$ \left| \sin x - x \right| \leq \frac{1}{3!} |x|^3$$

Answer 2

$$\sin x - x = \int_{0}^{x}(\cos t-1)\, dt$$ 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 $\sin\frac{t}{2}\leq\frac{t}{2}.$