Answer
Please see the sketch attached.
Work Step by Step
We evaluate ${{\bf{r}}_1}\left( \theta \right) = \left( {\theta ,\cos \theta } \right)$ and ${{\bf{r}}_2}\left( \theta \right) = \left( {\cos \theta ,\theta } \right)$ for $\theta = 0,\frac{\pi }{4},\frac{\pi }{2},\frac{{3\pi }}{4},...,2\pi $ and list these points in the following tables:
$\begin{array}{*{20}{c}}
\theta &{{{\bf{r}}_1}\left( \theta \right) = \left( {\theta ,\cos \theta } \right)}\\
0&{\left( {0,1} \right)}\\
{\frac{\pi }{4}}&{\left( {\frac{\pi }{4},\frac{1}{{\sqrt 2 }}} \right)}\\
{\frac{\pi }{2}}&{\left( {\frac{\pi }{2},0} \right)}\\
{\frac{{3\pi }}{4}}&{\left( {\frac{{3\pi }}{4}, - \frac{1}{{\sqrt 2 }}} \right)}\\
\pi &{(\pi , - 1\} }\\
{\frac{{5\pi }}{4}}&{\left( {\frac{{5\pi }}{4}, - \frac{1}{{\sqrt 2 }}} \right)}\\
{\frac{{3\pi }}{2}}&{\left( {\frac{{3\pi }}{2},0} \right)}\\
{\frac{{7\pi }}{4}}&{\left( {\frac{{7\pi }}{4},\frac{1}{{\sqrt 2 }}} \right)}\\
{2\pi }&{\left( {2\pi ,1} \right)}
\end{array}\begin{array}{*{20}{c}}
{{{\bf{r}}_2}\left( \theta \right) = \left( {\cos \theta ,\theta } \right)}\\
{\left( {1,0} \right)}\\
{\left( {\frac{1}{{\sqrt 2 }},\frac{\pi }{4}} \right)}\\
{\left( {0,\frac{\pi }{2}} \right)}\\
{\left( { - \frac{1}{{\sqrt 2 }},\frac{{3\pi }}{4}} \right)}\\
{\left( { - 1,\pi } \right)}\\
{\left( { - \frac{1}{{\sqrt 2 }},\frac{{5\pi }}{4}} \right)}\\
{\left( {0,\frac{{3\pi }}{2}} \right)}\\
{\left( {\frac{1}{{\sqrt 2 }},\frac{{7\pi }}{4}} \right)}\\
{\left( {1,2\pi } \right)}
\end{array}$
Then, we plot the points in the $xy$-plane and join them by smooth curves to obtain the paths.
