Show that $|\sin(0.1) - 0.1| \leq 0.001$ with the lagrange remainder 
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 ?

You also know that $\vert \sin x \vert \le \vert x \vert$. Hence
$$\vert \sin(c) \vert \frac{0.01}{2} \le 0.1 \frac{0.01}{2} = 0.0005 < 0.001$$
