Find the rectangular coordinates of each point whose polar coordinates are:

(a) $\left( 4,\dfrac{\pi }{3}\right)$

(b) $\left( -2,\dfrac{3\pi }{4}\right)$

(c) $\left(-3,-\dfrac{5\pi }{6}\right)$

Solution (a) We use the equations $x=r\cos \theta$ and $y=r\sin \theta$ with $r=4$ and $\theta =\dfrac{\pi }{3}.$ $$x=4\cos \dfrac{\pi }{3}=4\left( \dfrac{1}{2}\right) =2 \qquad \hbox{and} \qquad y=4\sin \dfrac{\pi }{3}=4\left( \dfrac{\sqrt{3}}{2}\right) =2\sqrt{3}$$

The rectangular coordinates are $(2,2\sqrt{3}).$

(b) We use the equations $x=r\cos \theta$ and $y=r\sin \theta$ with $r=-2$ and $\theta =\dfrac{3\pi }{4}.$ $$x=-2\cos \dfrac{3\pi }{4}=-2\left(-\dfrac{\sqrt{2}}{2}\right) =\sqrt{2} \quad\hbox{and}\quad y=-2\sin \dfrac{3\pi }{4}=-2\left( \dfrac{\sqrt{2}}{2}\right) =-\sqrt{2}$$

The rectangular coordinates are $( \sqrt{2},-\sqrt{2}) .$

(c) We use the equations $x=r\cos \theta$ and $y=r\sin \theta$ with $r=-3$ and $\theta =-\dfrac{5\pi }{6}.$ $$\begin{eqnarray*} x&=&-3\cos \left( -\dfrac{5\pi }{6}\right) =-3\left( -\dfrac{\sqrt{3}}{2} \right) =\dfrac{3\sqrt{3}}{2}\;\hbox{}\; \\[4pt] y&=&-3\sin \left( -\dfrac{5\pi }{6}\right) =-3\left( -\dfrac{1}{2}\right) =\dfrac{3}{2} \end{eqnarray*}$$

Figure 34

The rectangular coordinates are $\left( \dfrac{3\sqrt{3}}{2},\dfrac{3}{2}\right)$.