Answer
$\begin{array}{*{20}{c}}
{Polar}&{Rectangular}\\
{\left( {r,\theta } \right)}&{\left( {x,y} \right)}\\
{\left( {1,\frac{\pi }{6}} \right)}&{\left( {\frac{1}{2}\sqrt 3 ,\frac{1}{2}} \right)}\\
{\left( {3,\frac{{5\pi }}{4}} \right)}&{\left( { - \frac{3}{2}\sqrt 2 , - \frac{3}{2}\sqrt 2 } \right)}
\end{array}$
Work Step by Step
We have $\left( {r,\theta } \right) = \left( {1,\frac{\pi }{6}} \right)$.
Using the conversion formula from polar coordinates to rectangular coordinates given by
$x = r\cos \theta $, ${\ \ \ }$ $y = r\sin \theta $
we obtain
$x = \cos \frac{\pi }{6} = \frac{1}{2}\sqrt 3 $, ${\ \ \ }$ $y = \sin \frac{\pi }{6} = \frac{1}{2}$.
$\left( {x,y} \right) = \left( {\frac{1}{2}\sqrt 3 ,\frac{1}{2}} \right)$
Similarly for $\left( {r,\theta } \right) = \left( {3,\frac{{5\pi }}{4}} \right)$,
$x = 3\cos \frac{{5\pi }}{4} = - \frac{3}{2}\sqrt 2 $, ${\ \ }$ $y = 3\sin \frac{{5\pi }}{4} = - \frac{3}{2}\sqrt 2 $.
$\left( {x,y} \right) = \left( { - \frac{3}{2}\sqrt 2 , - \frac{3}{2}\sqrt 2 } \right)$
In summary:
$\begin{array}{*{20}{c}}
{Polar}&{Rectangular}\\
{\left( {r,\theta } \right)}&{\left( {x,y} \right)}\\
{\left( {1,\frac{\pi }{6}} \right)}&{\left( {\frac{1}{2}\sqrt 3 ,\frac{1}{2}} \right)}\\
{\left( {3,\frac{{5\pi }}{4}} \right)}&{\left( { - \frac{3}{2}\sqrt 2 , - \frac{3}{2}\sqrt 2 } \right)}
\end{array}$
