AP®MathsCollege BoardPrecalculusRevision NotesTrigonometric & Polar FunctionsPolar Coordinates & Polar FunctionsThe Polar Coordinate System

The Polar Coordinate System (College Board AP® Precalculus): Revision Note

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 20 April 2026

Polar coordinates

What are polar coordinates?

The polar coordinate system is an alternative to the rectangular (Cartesian) coordinate system for locating points in the plane
It is built on a grid of
- circles centered at the origin
- and lines passing through the origin
The positive $x$ -axis is called the polar axis

Concentric circles with radial lines form a polar graph, labelled from 1 to 5, with polar axis marked. Grid lines provide a polar coordinate system. — **A polar coordinate grid**

A point in the polar coordinate system is described by an ordered pair $(r, θ)$
- $r$ is the radius
  - This is a signed value related to the distance of the point from the origin
    - It can be positive or negative (see the next section below for details)
  - The numbers on the above grid (1, 2, 3, 4, 5) correspond to distances $r$ from the origin
- $θ$ is the angle in standard position whose terminal ray includes the point
  - Remember that angles are usually measured in radians in AP® Precalculus
Values of $θ$ for the radial lines on the grid can vary just as the same angle in standard position can be represented by different values
- The radial lines on the grid above divide the circle into 24 parts
  - so each line is $\frac{2 π}{24} = \frac{π}{12}$ radians away from the ones on either side
- Starting from zero at the polar axis, the $θ$ values
  - increase counter-clockwise
    - $0, \frac{π}{12}, \frac{2 π}{12} = \frac{π}{6}, \frac{3 π}{12} = \frac{π}{4},$ etc.
    - up to $2 π$ (which is in the same position as 0)
  - and decrease clockwise
    - $0, - \frac{π}{12}, - \frac{2 π}{12} = - \frac{π}{6}, - \frac{3 π}{12} = - \frac{π}{4},$ etc.
    - down to $- 2 π$ (which is in the same position as 0)

Polar coordinates grid with values of θ increasing counter-clockwise from 0 to 2π. — **Values of θ increasing counter-clockwise from 0 to 2π**

Polar coordinates grid with values of θ decreasing clockwise from 0 to -2π — **Values of θ decreasing clockwise from 0 to -2π**

These angular values can also go beyond $2 π$ or $- 2 π$
- E.g. continuing to go counter-clockwise from $2 π$
  - the $\frac{π}{12}$ line becomes $\frac{π}{12} + 2 π = \frac{25 π}{12}$
  - the $\frac{π}{6}$ line becomes $\frac{π}{6} + 2 π = \frac{13 π}{6}$
  - etc.
- Or continuing to go clockwise from $- 2 π$
  - the $- \frac{π}{12}$ line becomes $- \frac{π}{12} - 2 π = - \frac{25 π}{12}$
  - the $- \frac{π}{6}$ line becomes $- \frac{π}{6} - 2 π = - \frac{13 π}{6}$
  - etc.

How do I interpret the radius value r?

For a point with polar coordinates $(r, θ)$
- Draw the angle $θ$ in standard position
  - This gives the terminal ray
- If $r > 0$ , the point lies on the terminal ray
  - at distance $r$ from the origin
- If $r < 0$ , the point lies on the opposite ray (the ray pointing in the reverse direction)
  - at distance $| r |$ from the origin
- If $r = 0$ , the point is at the origin, regardless of the value of $θ$
Because of its behavior, $r$ is called a signed radius value
- It behaves like a coordinate on a number line laid along the terminal ray
- where the positive direction is the direction of the terminal ray itself

Examples of points in (r,θ) form plotted on a polar coordinate grid. The labelled points are (2, π/3), (-3, π/12) and (4, -π/4). — **Examples of points (r,θ) plotted on a polar coordinate grid**

Can the same point have more than one set of polar coordinates?

Yes, unlike with rectangular coordinates, a single point in the plane can be represented by many different polar coordinate pairs
This happens for two reasons
- Adding full revolutions to the angle
  - the coordinates $(r, θ)$ and $(r, θ + 2 π)$ represent the same point
  - More generally, $(r, θ + 2 π k)$ for any integer $k$ gives the same point
- Using a negative radius
  - the coordinates $(r, θ)$ and $(- r, θ + π)$ represent the same point
  - Rotating the terminal ray by an extra $π$ reverses its direction
    - and making $r$ negative reverses it back
E.g. the point with polar coordinates $(2, \frac{π}{3})$ can also be written as
- $(2, \frac{π}{3} + 2 π) = (2, \frac{7 π}{3})$
- or as $(- 2, \frac{π}{3} + π) = (- 2, \frac{4 π}{3})$
- or in infinitely many other ways

How do I convert from polar coordinates to rectangular coordinates?

Given a point with polar coordinates $(r, θ)$ , the rectangular coordinates $(x, y)$ can be found using
- $x = r \cos θ$
- $y = r \sin θ$
These formulas come directly from the right-triangle definitions of sine and cosine
- and they work for all values of $r$ and $θ$ (including negative $r$ )
E.g. to convert $(4, \frac{2 π}{3})$ to rectangular coordinates
- $x = 4 \cos \frac{2 π}{3} = 4 \cdot (- \frac{1}{2}) = - 2$
- $y = 4 \sin \frac{2 π}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2 \sqrt{3}$
- So the rectangular coordinates are $(- 2, 2 \sqrt{3})$

How do I convert from rectangular coordinates to polar coordinates?

Given a point with rectangular coordinates $(x, y)$ , the polar coordinates $(r, θ)$ can be found using:
- $r = \sqrt{x^{2} + y^{2}}$
- $θ = \arctan (\frac{y}{x})$ for $x > 0$
- $θ = \arctan (\frac{y}{x}) + π$ for $x < 0$
  - The adjustment by $π$ when $x < 0$ is needed because the $\arctan$ function only returns angles in the interval $(- \frac{π}{2}, \frac{π}{2})$
    - This range corresponds to terminal rays in the right half-plane (Quadrants I and IV)
    - For points in the left half-plane (Quadrants II and III), adding $π$ rotates the terminal ray to the correct side
E.g. to convert $(- 3, 3)$ to polar coordinates:
- $r = \sqrt{(- 3)^{2} + 3^{2}} = \sqrt{18} = 3 \sqrt{2}$
- Since $x < 0$ , $θ = \arctan (\frac{3}{- 3}) + π = \arctan (- 1) + π = - \frac{π}{4} + π = \frac{3 π}{4}$
- So one set of polar coordinates is $(3 \sqrt{2}, \frac{3 π}{4})$

Examiner Tips and Tricks

When converting from rectangular to polar, always check which quadrant the point is in before picking your angle.

A common error is to apply $\arctan (\frac{y}{x})$ without the $+ π$ adjustment when $x$ is negative, which puts the point in the wrong half-plane

When a question asks for one valid set of polar coordinates, any equivalent representation is usually acceptable.

But if the question specifies a range like $0 \leq θ < 2 π$ , make sure your answer falls in that range

Worked Example

A point $P$ in the plane has rectangular coordinates $(- 2 \sqrt{3}, 2)$ .

(a) Find the value of $r$ , where $r > 0$ , and the value of $θ$ , where $0 \leq θ < 2 π$ , for a polar coordinate representation $(r, θ)$ of point $P$ .

Answer:

Use the conversion formulas from rectangular to polar

$r = \sqrt{(- 2 \sqrt{3})^{2} + 2^{2}} = \sqrt{12 + 4} = \sqrt{16} = 4$

Because the $x$ -coordinate is negative and the $y$ -coordinate is positive, point $P$ lies in Quadrant II

so use the $x < 0$ version of the angle formula:

$θ = \arctan (\frac{2}{- 2 \sqrt{3}}) + π = \arctan (- \frac{1}{\sqrt{3}}) + π = - \frac{π}{6} + π = \frac{5 π}{6}$

So the polar coordinates are

$(4, \frac{5 π}{6})$

(b) Give a second polar coordinate representation $(r, θ)$ of point $P$ , different from the answer to part (a), with $r > 0$ and $θ$ any real number. Justify why your answer represents the same point.

Answer:

Adding $2 π$ to the angle gives another valid representation with the same (positive) radius

$(4, \frac{5 π}{6} + 2 π) = (4, \frac{17 π}{6})$

Adding or subtracting any other multiple of $2 π$ would also give an equivalent point

$(4, \frac{17 π}{6})$ represents the same point because adding $2 π$ to the angle corresponds to a full revolution, which returns the terminal ray to its original position. With the same $r$ and the same terminal ray, the point is unchanged.

(c) A second point $Q$ has polar coordinates $(5, \frac{7 π}{6})$ . Find the rectangular coordinates of $Q$ , giving exact values.

Answer:

Use the conversion formulas from polar to rectangular

$x = r \cos θ = 5 \cos \frac{7 π}{6} = 5 \cdot (- \frac{\sqrt{3}}{2}) = - \frac{5 \sqrt{3}}{2}$

$y = r \sin θ = 5 \sin \frac{7 π}{6} = 5 \cdot (- \frac{1}{2}) = - \frac{5}{2}$

So the rectangular coordinates of $Q$ are

$(- \frac{5 \sqrt{3}}{2}, - \frac{5}{2})$

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The Polar Coordinate System (College Board AP® Precalculus): Revision Note

Polar coordinates

What are polar coordinates?

How do I interpret the radius value r?

Can the same point have more than one set of polar coordinates?

How do I convert from polar coordinates to rectangular coordinates?

How do I convert from rectangular coordinates to polar coordinates?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis