Answer
Please see the figure attached.
The tangent vectors at $t = \frac{\pi }{3}$ and $t = \frac{{3\pi }}{4}$ and the corresponding points are listed in the table below:
$\begin{array}{*{20}{c}}
t&{{\bf{r}}\left( t \right)}&{{\rm{r}}'\left( t \right)}\\
{\frac{\pi }{3}}&{\left( {0.18,0.5} \right)}&{\left( {0.5,0.87} \right)}\\
{\frac{{3\pi }}{4}}&{\left( {1.65,1.71} \right)}&{\left( {1.71,0.71} \right)}
\end{array}$
Work Step by Step
Recall that the derivative vector ${\rm{r}}'\left( {{t_0}} \right)$ points in the direction tangent to the path traced by ${\bf{r}}\left( t \right)$ at $t = {t_0}$. For the curve ${\bf{r}}\left( t \right) = \left( {t - \sin t,1 - \cos t} \right)$ we get ${\rm{r}}'\left( t \right) = \left( {1 - \cos t,\sin t} \right)$. Thus, we obtain the tangent vectors at $t = \frac{\pi }{3}$ and $t = \frac{{3\pi }}{4}$ and the corresponding points listed in the table below:
$\begin{array}{*{20}{c}}
t&{{\bf{r}}\left( t \right)}&{{\rm{r}}'\left( t \right)}\\
{\frac{\pi }{3}}&{\left( {0.18,0.5} \right)}&{\left( {0.5,0.87} \right)}\\
{\frac{{3\pi }}{4}}&{\left( {1.65,1.71} \right)}&{\left( {1.71,0.71} \right)}
\end{array}$
We evaluate several points for the interval $0 \le t \le 2\pi $ and list them in the following table. Then we plot the points and join them to obtain the curve.
$\begin{array}{*{20}{c}}
t&{{\bf{r}}\left( t \right)}\\
0&{\left( {0,0} \right)}\\
{\frac{\pi }{4}}&{\left( {0.078,0.29} \right)}\\
{\frac{\pi }{2}}&{\left( {0.57,1} \right)}\\
{\frac{{3\pi }}{4}}&{\left( {1.65,1.71} \right)}\\
\pi &{\left( {3.14,2} \right)}\\
{\frac{{5\pi }}{4}}&{\left( {4.63,1.71} \right)}\\
{\frac{{3\pi }}{2}}&{\left( {5.71,1} \right)}\\
{\frac{{7\pi }}{4}}&{\left( {6.2,0.29} \right)}\\
{2\pi }&{\left( {6.28,0} \right)}
\end{array}$
