# Converting Coordinates from Rectangular to Polar

## Key Questions

• Let's look at the trig formulas SYR, CXR, TYX:

$\sin \theta = \frac{y}{r}$
$\cos \theta = \frac{x}{r}$
$\tan \theta = \frac{y}{x}$

Since we are given the Cartesian coordinates, we are given $x$ and $y$. For polar coordinates, we need to figure out $r$ and $\theta$. $r$ is easy, we just use Pythagorean:

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

To figure out $\theta$, I like to use cosine because the range of arccosine is in quadrants I and II and adjusting $\theta '$ is easier. So,

$\theta ' = {\cos}^{- 1} \frac{x}{r}$

If $y \ge 0$ then $\theta = \theta '$.
If $y < 0$ then $\theta = 2 \pi - \theta '$ (in radians) or $\theta = 360 - \theta '$ (in degrees).

Our final answer is $\left(r , \theta\right)$.

Let's look at a concrete example: Convert $\left(- 3 , 3 \sqrt{3}\right)$ to polar coordinates:

$r = \sqrt{{\left(- 3\right)}^{2} + {\left(3 \sqrt{3}\right)}^{2}} = \sqrt{36} = 6$
$\theta ' = {\cos}^{- 1} \left(\frac{- 3}{6}\right) = \frac{2 \pi}{3}$
$y < 0$ so, $\theta = 2 \pi - \frac{2 \pi}{3} = \frac{4 \pi}{3}$

So the polar coordinates are $\left(6 , \frac{4 \pi}{3}\right)$.

• I presume we're looking for a radius $r$ and angle $\theta$ such that $a + b i = r \left(\cos \theta + i \sin \theta\right)$.

Pythagoras theorem gives us $r = \sqrt{{a}^{2} + {b}^{2}}$.

Simple trigonometry gives us $\tan \theta = \frac{b}{a}$, so $\theta = \arctan \left(\frac{b}{a}\right)$.