Find an upper bound for the magnitude of the error in the approximation $ \sin(x) \approx x$ and $ |x|< \frac{\pi}{20}$ 
Find an upper bound for the magnitude of the error in the approximation 
  $$ \sin(x) \approx x$$ and
  $$ |x|< \frac{\pi}{20}$$

First, $ |x|< \frac{\pi}{20}$ describes an open interval about $x_0=0$ [$0$ being the middle of the interval $(- \pi/20, \pi/20)$].
$$f(x) = T(x) + R(x) $$
$$f(x) =f(x_0) +f'(x_0)(x-x_0)+ \frac{1}{2} f''(c) (x-x_0)^2$$
Where $R(x) =\frac{1}{2} f''(c) (x-x_0)^2$ and $c$ is between $x$ and $0$ by definition. 
$$f(x) =\sin(0) + \cos(0) \cdot x + \frac{1}{2} (-\sin(c)) x^2$$
$$f(x) = x - \frac{1}{2} \sin(c) x^2$$
As $c$ is between $0$ and $x$, we have $ 0 \leq |c|<|x|$, and adding the property that $0 \leq |\sin c| \leq |c|$ $\forall c$, and the given that $|x|< \frac{\pi}{20}$ we have:
$$ |\sin c| \leq |c| < |x| < \frac{\pi}{20}$$
I dont see how to find the upper bound (without using integral) What is the process to narrow $R(x)$ to his bound?
greatly appreciated
 A: by second order Taylor expansion
$$\sin (x)=x-\frac {x^2}{2}\sin (c) $$
with $$|\sin (c)|\le |c|<|x|<\frac {\pi}{20}.$$
and $$\frac {x^2}{2}<\frac {\pi^2}{20.20.2} $$
hence
$$|\sin (x)-x|<\frac {\pi^3}{16000} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\verts{x} < {\pi \over 20}}$.

\begin{align}
\sin\pars{x} & =
\mrm{sgn}\pars{x}\int_{0}^{\verts{x}}\cos\pars{t}\,\dd t =
\mrm{sgn}\pars{x}\int_{0}^{\verts{x}}\cos\pars{\verts{x} - t}\,\dd t
\\[5mm] & =
\left.\vphantom{\Large A}\mrm{sgn}\pars{x}\cos\pars{\verts{x} - t}\,t
\,\right\vert_{\ t\ =\ 0}^{\ t\ =\ \verts{x}} -
\mrm{sgn}\pars{x}\int_{0}^{\verts{x}}t
\bracks{-\sin\pars{\verts{x} - t}\pars{-1}}\,\dd t
\\[5mm] & =
x - \mrm{sgn}\pars{x}\int_{0}^{\verts{x}}t\sin\pars{\verts{x} - t}\,\dd t
\\[1cm]
&\mbox{Then,}\quad
\verts{\int_{0}^{\verts{x}}t\sin\pars{\verts{x} - t}\,\dd t} <
\int_{0}^{\verts{x}}t\,\dd t = {x^{2} \over 2}
\bbx{< {\pi^{2} \over 800} \approx 1.2337 \times 10^{-2}}
\end{align}
A: With $f(x)=\sin x$ we have $f(0)=f''(0)=0$ and $f'(0)=-f'''(0)=1$ and $f'''(x)=-\cos x.$ Therefore there exists $c$ with $|c|\leq |x| $such that   $$f(x)=f(0)+xf'(0)+x^2f''(0)/2+x^3f'''(c)/6=$$ $$=x-x^3(\cos c)/6.$$ For $|x|<\pi /20$ we have $|x^3(\cos c)/6|\leq |x^3/6|<\pi^3/48000.$
