# What is the Cartesian form of ( -2 , (13pi)/6 ) ?

Jan 19, 2017

I got:

$\left(\sqrt{3} , 1\right)$

assuming $r > 0$. If $r < 0$, then we would necessarily have $r = - \sqrt{{x}^{2} + {y}^{2}}$ to keep the equations below self-consistent, and then we would have $x = - \sqrt{3}$ and $y = - 1$.

Recall:

$x = r \cos \theta$
$y = r \sin \theta$
$r = \sqrt{{x}^{2} + {y}^{2}}$

The coordinates you are given are in polar, and are the $r$ and $\theta$ coordinates. Thus, you need to use the expression for $r$ to calculate $x$ or $y$, and the expression for $\theta$ to get the other coordinate.

However... $r$ is only positive or $0$. You cannot have a negative radius, as it is physically impossible. So, there's a typo in the problem. It should be $\left(2 , \frac{13 \pi}{6}\right)$.

$2 = \sqrt{{x}^{2} + {y}^{2}} \implies 4 = {x}^{2} + {y}^{2}$

$y = r \sin \left(\theta\right) = 2 \sin \left(\frac{13 \pi}{6}\right)$

$\implies {y}^{2} = 4 {\sin}^{2} \left(\frac{13 \pi}{6}\right) = 4 {\sin}^{2} \left({390}^{\circ}\right) = 4 {\sin}^{2} \left({30}^{\circ}\right) = 1$

$\implies \textcolor{g r e e n}{y = 1}$

(to keep consistent with the equations above, $y > 0$, since $\cos \theta > 0$ and $r > 0$.)

Therefore, we can solve for $x$:

$\implies {x}^{2} = 4 - {y}^{2} = 4 - 1 = 3$

$\implies \textcolor{g r e e n}{x = \sqrt{3}}$

(to keep consistent with the equations above, $x y > 0$, since $\cos \theta > 0$ and $r > 0$.)

So, the cartesian coordinates are:

$\textcolor{b l u e}{\left(x , y\right) = \left(\sqrt{3} , 1\right)}$

To make sure it worked...

$r = \sqrt{3 + 1} = 2$ color(blue)(sqrt"")

$\theta = \arccos \left(\frac{x}{r}\right) = \arcsin \left(\frac{y}{r}\right)$

$= \arccos \left(\frac{\sqrt{3}}{2}\right)$

As $\cos \theta = \frac{\sqrt{3}}{2}$ when $\theta = {30}^{\circ}$, $\arccos \left(\frac{\sqrt{3}}{2}\right) = \theta = {30}^{\circ}$, which is coterminal with ${390}^{\circ}$. color(blue)(sqrt"")