Trigonometry and Polar Coordinates — AP® Precalculus Notes

Rectangular coordinates say how far to go right and how far to go up. Polar coordinates say how far to go and in which direction. Both name the same points; each is easier for some jobs. This page sets up the polar system, converts in both directions, and uses it to write complex numbers in polar form.

The polar grid

Start at a point called the pole (the origin) and draw a ray to the right, the polar axis (the positive $x$ -axis). The point $(r, \theta)$ is found in two moves:

Turn to the angle $\theta$ , measured counterclockwise from the polar axis, as for an angle in standard position.
Move $r$ units along that ray. If $r$ is negative, move $|r|$ units in the opposite direction, backward through the pole.

The grid has circles (all points with the same $r$ ) and rays (all points with the same $\theta$ ), as in Figure 1.

A polar grid with circles of radius 1 to 4 around the pole and rays every pi/6 radians. Point P is at distance 3 on the ray pi/6, Q at distance 2 on the ray 5 pi/6, R at distance 4 on the ray 3 pi/2 (straight below the pole) and S at distance 1 on the ray pi. — **Figure 1.** P is $\left(3, \frac{\pi}{6}\right)$ , Q is $\left(2, \frac{5\pi}{6}\right)$ , R is $\left(4, \frac{3\pi}{2}\right)$ and S is $(1, \pi)$ . LevlPrep original diagram.

One point, many names

In rectangular coordinates every point has one name. In polar coordinates it has infinitely many:

Adding a full turn changes nothing: $(r, \theta)$ and $(r, \theta + 2\pi)$ are the same point.
Reversing direction twice changes nothing: $(r, \theta)$ and $(-r, \theta + \pi)$ are the same point.
The pole is $(0, \theta)$ for every $\theta$ .

For example, Q $= \left(2, \frac{5\pi}{6}\right)$ $= \left(2, -\frac{7\pi}{6}\right)$ $= \left(-2, \frac{11\pi}{6}\right)$ .

Converting between the systems

The point $(r, \theta)$ lies on a circle of radius $|r|$ , so the circle facts from Topic 3.3 give the conversions.

Direction	Formulas	Watch out for
Polar to rectangular	$x = r\cos\theta$ , $y = r\sin\theta$	a negative $r$ flips both signs
Rectangular to polar	$r = \sqrt{x^2 + y^2}$ , $\tan\theta = \frac{y}{x}$	the quadrant of the angle
Equations	replace $x$ , $y$ and $x^2 + y^2$ with $r\cos\theta$ , $r\sin\theta$ and $r^2$	divide only by quantities that are not $0$

Worked example 1. Convert $\left(-4, \frac{5\pi}{6}\right)$ to rectangular coordinates.

$x$ $= -4\cos\frac{5\pi}{6}$ $= -4\cdot\left(-\frac{\sqrt{3}}{2}\right)$ $= 2\sqrt{3}$ .

$y$ $= -4\sin\frac{5\pi}{6}$ $= -4\cdot\frac{1}{2}$ $= -2$ .

The point is $\left(2\sqrt{3}, -2\right)$ , in Quadrant IV: the ray $\frac{5\pi}{6}$ points into Quadrant II, and the negative $r$ sends the point the other way.

Worked example 2. Convert $(-2, -2)$ to polar coordinates with $r > 0$ and $0 \le \theta < 2\pi$ .

$r = \sqrt{4 + 4} = 2\sqrt{2}$ .

$\tan\theta = \frac{-2}{-2} = 1$ . The inverse gives $\tan^{-1} 1 = \frac{\pi}{4}$ , in Quadrant I. But $(-2, -2)$ is in Quadrant III, so add $\pi$ : $\theta = \frac{5\pi}{4}$ .

The point is $\left(2\sqrt{2}, \frac{5\pi}{4}\right)$ . Check: $2\sqrt{2}\cos\frac{5\pi}{4}$ $= 2\sqrt{2}\cdot\left(-\frac{\sqrt{2}}{2}\right)$ $= -2$ .

Equations convert the same way. The circle $x^2 + y^2 = 25$ becomes $r^2 = 25$ , or just $r = 5$ . The line $y = 3$ becomes $r\sin\theta = 3$ , so $r = \frac{3}{\sin\theta} = 3\csc\theta$ .

Complex numbers in polar form

The complex plane plots $a + bi$ as the point $(a, b)$ : real part across, imaginary part up. Any point has polar coordinates, so

$a + bi = r(\cos\theta + i\sin\theta), \qquad r = \sqrt{a^2 + b^2}, \quad \tan\theta = \frac{b}{a}$

This is the polar form of a complex number. $r$ is its distance from $0$ .

Worked example 3. Write $z = \sqrt{3} - i$ in polar form.

The point is $\left(\sqrt{3}, -1\right)$ , in Quadrant IV. $r = \sqrt{3 + 1} = 2$ . $\tan\theta = -\frac{1}{\sqrt{3}}$ , reference angle $\frac{\pi}{6}$ , so $\theta = \frac{11\pi}{6}$ (or $-\frac{\pi}{6}$ ).

$z$ $= 2\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right)$ .

Common slips

Writing $(\theta, r)$ instead of $(r, \theta)$ .
Trusting $\tan^{-1}\frac{y}{x}$ in Quadrants II and III. Draw the point and add $\pi$ when it is on the left.
Reporting $r^2$ instead of $r$ .