Given:
[tex](-4,\frac{7\pi}{6})[/tex][tex]\text{Here r=-4 and }\theta=\frac{7\pi}{6}\text{.}[/tex]
Recall the formula for rectangular coordinates
[tex]x=r\cos \theta[/tex][tex]y=r\sin \theta[/tex]
[tex]\text{Substitute r=-4 and }\theta=\frac{7\pi}{6}\text{ in x=rcos}\theta,\text{ we get}[/tex]
[tex]x=-4\cos (\frac{7\pi}{6})[/tex]
[tex]Use\cos (\frac{7\pi}{6})=\cos (\pi+\frac{\pi}{6})=-\cos (\frac{\pi}{6})=-\frac{\sqrt[]{3}}{2}\text{.}[/tex]
[tex]x=-4(\frac{-\sqrt[]{3}}{2})=2\sqrt[]{3}_{}[/tex]
[tex]\text{Substitute r=-4 and }\theta=\frac{7\pi}{6}\text{ in y=rsin}\theta,\text{ we get}[/tex]
[tex]y=-4\sin (\frac{7\pi}{6})[/tex]
[tex]Use\text{ sin}(\frac{7\pi}{6})=sin(\pi+\frac{\pi}{6})=-sin(\frac{\pi}{6})=-\frac{1}{2}\text{.}[/tex]
[tex]y=-4(-\frac{1}{2})=2[/tex]
Hence the rectangular coordinates are
[tex](x,y)=(2\sqrt[]{3},\text{ 2)}[/tex]
