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Show that $|\sin(0.1) - 0.1| \leq 0.001$

I know that's a basic exercise on taylor polynomial but I have made a mistake somewhere that I don't find out. Anyway, here's my attempt :

Because the function $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \rightarrow \sin(x)$ is $1$ time derivable,

by the Taylor polynomial formula we find: \begin{equation*} \sin(x) = x + R^1_0 \sin(x) \end{equation*}

Therefore, \begin{equation*} |sin(0.1) - 0.1| = |R^1_0 \sin(0.1)| \end{equation*}

Because $f$ est 2 times derivable,

by the Lagrange remainder formula, $\exists c \in ]0, 0.1[$ such that \begin{align*} R^1_0 \sin(0.1) &= f^{(2)}(c) \frac{(0.1)^2}{2!} \\ &= - sin(c) \frac{0.01}{2} \\ |R^1_0 \sin(0.1)| &= |sin(c) \cdot 0.005| \end{align*}

Because $|sin(x)| \leq 1$, $\forall x \in \mathbb{R}$ \begin{align*} |R^1_0 \sin(0.1)| &\leq |0.005| \end{align*}

However, $0.005 > 0.001$ so I'm wondering where I did a mistake ?

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You also know that $\vert \sin x \vert \le \vert x \vert$. Hence

$$\vert \sin(c) \vert \frac{0.01}{2} \le 0.1 \frac{0.01}{2} = 0.0005 < 0.001$$

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