# What is the Cartesian form of (4,(23pi )/12)?

Mar 31, 2018

$\left(x , y\right) \to \left(\sqrt{6} + \sqrt{2} , \sqrt{2} - \sqrt{6}\right)$

#### Explanation:

As seen in the image, we see that the $x$ coordinate is $r \cos \theta$, and the $y$ is $r \sin \theta$ obtained by simple trig

$r = 4 , \theta = \frac{23 \pi}{12}$

$\implies x = 4 \cos \left(\frac{23 \pi}{12}\right) = \sqrt{6} + \sqrt{2}$

$\implies y = 4 \sin \left(\frac{23 \pi}{12}\right) = \sqrt{2} - \sqrt{6}$

$\left(r , \theta\right) \to \left(4 , \frac{23 \pi}{12}\right)$

color(blue)((x,y) -> (sqrt(6) +sqrt(2) , sqrt(2) - sqrt(6) )