Conversion between Cartesian and Polar Coordinates in Plane

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Theorem

Let $S$ be the plane.

Let a Cartesian plane $\CC$ be applied to $S$.

Let a polar coordinate plane $\PP$ be superimposed upon $\CC$ such that:

$(1): \quad$ The origin of $\CC$ coincides with the pole of $\PP$.
$(2): \quad$ The $x$-axis of $\CC$ coincides with the polar axis of $\PP$.


Let $p$ be a point in $S$.

Let $p$ be specified as $p = \polar {r, \theta}$ expressed in the polar coordinates of $\PP$.

Then $p$ is expressed as $\tuple {r \cos \theta, r \sin \theta}$ in $\CC$.


Contrariwise, let $p$ be expressed as $\tuple {x, y}$ in the cartesian coordinates of $\CC$.

Then $p$ is expressed as:

$p = \polar {\sqrt {x^2 + y^2}, \arctan \dfrac y x + \pi \sqbrk {x < 0 \text{ or } y < 0} + \pi \sqbrk {x > 0 \text{ and } y < 0} }$

where:

$\sqbrk {\, \cdot \,}$ is Iverson's convention.
$\arctan$ denotes the arctangent function.


Proof

Let $P$ be a point in the plane expressed:

in Cartesian coordinates as $\tuple {x, y}$
in polar coordinates as $\polar {r, \theta}$.


Cartesian-polar-conversion.png


As specified, we identify:

the origins of both coordinate systems with a distinguished point $O$
the $x$-axis of $C$ with the polar axis of $P$.

Let a perpendicular $PM$ be dropped from $P$ to the $x$-axis.

The triangle $OMP$ is a right triangle:

whose hypotenuse is $OP$, whose length is $r$
whose legs are $OM$ and $MP$
whose angle $POM$ is $\theta$.

By definition of sine and cosine

$x = r \cos \theta$
$y = r \sin \theta$

The result follows.

$\blacksquare$


Examples

Example: $\polar {4, \dfrac \pi 3}$

The point $P$ defined in polar coordinates as:

$P = \polar {4, \dfrac \pi 3}$

can be expressed in the corresponding Cartesian coordinates as:

$P = \tuple {2, 2 \sqrt 3}$


Example: $\polar {-2, -\dfrac \pi 4}$

The point $P$ defined in polar coordinates as:

$P = \polar {-2, -\dfrac \pi 4}$

can be expressed in the corresponding Cartesian coordinates as:

$P = \tuple {\sqrt 2, -\sqrt 2}$


Sources