## Calculus (3rd Edition)

$\begin{array}{*{20}{c}} {Polar}&{Rectangular}\\ {\left( {r,\theta } \right)}&{\left( {x,y} \right)}\\ {\left( {1,\frac{\pi }{6}} \right)}&{\left( {\frac{1}{2}\sqrt 3 ,\frac{1}{2}} \right)}\\ {\left( {3,\frac{{5\pi }}{4}} \right)}&{\left( { - \frac{3}{2}\sqrt 2 , - \frac{3}{2}\sqrt 2 } \right)} \end{array}$
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: $\begin{array}{*{20}{c}} {Polar}&{Rectangular}\\ {\left( {r,\theta } \right)}&{\left( {x,y} \right)}\\ {\left( {1,\frac{\pi }{6}} \right)}&{\left( {\frac{1}{2}\sqrt 3 ,\frac{1}{2}} \right)}\\ {\left( {3,\frac{{5\pi }}{4}} \right)}&{\left( { - \frac{3}{2}\sqrt 2 , - \frac{3}{2}\sqrt 2 } \right)} \end{array}$