Trigonometry and Polar Coordinates

Part 1 · Hook

Why this matters

An air traffic radar does not report a plane's

x

and

y

. It reports how far away the plane is and in which direction. Distance and direction are a second way to name every point in the plane, and for anything that spins or sweeps, it is the natural one.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\cos\frac{2\pi}{3}$ ?

$-\frac{1}{2}$
$\frac{1}{2}$
$-\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{2}$

Show the answer

$\frac{2\pi}{3}$ is in Quadrant II with reference angle $\frac{\pi}{3}$ , so its cosine is $-\frac{1}{2}$ .

Correct: $-\frac{1}{2}$ :
$\frac{1}{2}$ :
$-\frac{\sqrt{3}}{2}$ :
$\frac{\sqrt{3}}{2}$ :

2. What is $\tan^{-1}(-1)$ ?

$-\frac{\pi}{4}$
$\frac{3\pi}{4}$
$\frac{7\pi}{4}$
$-1$

Show the answer

The angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with tangent $-1$ is $-\frac{\pi}{4}$ .

Correct: $-\frac{\pi}{4}$ :
$\frac{3\pi}{4}$ :
$\frac{7\pi}{4}$ :
$-1$ :

Part 4 · See it

See it first

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. — A polar grid: circles of constant distance $r$ and rays of constant angle $\theta$ . P is $\left(3, \frac{\pi}{6}\right)$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Pick a distance $r$ and an angle $\theta$ the point $(r, \theta)$ : turn to $\theta$ , then go $r$ units out from the pole
Use the circle of radius $r$ $x = r\cos\theta$ and $y = r\sin\theta$
Go back from $(x, y)$ $r = \sqrt{x^2 + y^2}$ and $\tan\theta = \frac{y}{x}$ , with the quadrant checked
Add $2\pi$ to $\theta$ , or negate $r$ and add $\pi$ the same point gets another name

Part 6 · Key ideas

Key ideas

$(r, \theta)$ means: face the direction $\theta$ , then move $r$ units. A negative $r$ moves backward through the pole.
$x = r\cos\theta$ , $y = r\sin\theta$ and $r^2 = x^2 + y^2$ convert between the two systems.
$\tan^{-1}\frac{y}{x}$ gives an angle in Quadrant I or IV; if the point is in Quadrant II or III, add $\pi$ .
A complex number $a + bi$ is the point $(a, b)$ , so it has a polar form $r(\cos\theta + i\sin\theta)$ .

Part 7 · Misconception

A common mistake

The wrong idea: The polar angle of $(-3, 3)$ is $\tan^{-1}\left(\frac{3}{-3}\right) = -\frac{\pi}{4}$ .

What actually happens: $\tan^{-1}$ only returns angles in Quadrants I and IV. $(-3, 3)$ is in Quadrant II, so the angle is $-\frac{\pi}{4} + \pi = \frac{3\pi}{4}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Diagram

Four points on a polar grid

The circles have radius $1$ , $2$ , $3$ and $4$ , and the rays are $\frac{\pi}{6}$ apart, measured counterclockwise from the polar axis (the positive $x$ -axis).

1. Which polar coordinates $(r, \theta)$ name point Q in the diagram?

$\left(2, \frac{5\pi}{6}\right)$
$\left(\frac{5\pi}{6}, 2\right)$
$\left(2, \frac{\pi}{6}\right)$
$\left(-2, \frac{5\pi}{6}\right)$

Show the answer

Q is on the circle of radius $2$ , on the ray $\frac{5\pi}{6}$ (five steps of $\frac{\pi}{6}$ from the polar axis). Polar coordinates list the distance first, then the angle: $\left(2, \frac{5\pi}{6}\right)$ .

Correct: $\left(2, \frac{5\pi}{6}\right)$ : Right: distance $2$ , angle $\frac{5\pi}{6}$ .
$\left(\frac{5\pi}{6}, 2\right)$ : The order is $(r, \theta)$ : distance first, then angle.
$\left(2, \frac{\pi}{6}\right)$ : The ray at $\frac{\pi}{6}$ is in Quadrant I; Q is in Quadrant II.
$\left(-2, \frac{5\pi}{6}\right)$ : A negative $r$ points the opposite way, into Quadrant IV.

2. What are the rectangular coordinates of point P in the diagram?

$\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$
$\left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
$\left(3, \frac{\pi}{6}\right)$

Show the answer

P is $\left(3, \frac{\pi}{6}\right)$ in polar coordinates. $x$ $= r\cos\theta$ $= 3\cdot\frac{\sqrt{3}}{2}$ $= \frac{3\sqrt{3}}{2}$ and $y = r\sin\theta = 3\cdot\frac{1}{2} = \frac{3}{2}$ .

Correct: $\left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right)$ : Right: $x = 3\cos\frac{\pi}{6}$ , $y = 3\sin\frac{\pi}{6}$ .
$\left(\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$ : This swaps sine and cosine; $x$ uses cosine.
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ : This is the point on the unit circle; multiply by $r = 3$ .
$\left(3, \frac{\pi}{6}\right)$ : That is the polar form of P, not its rectangular coordinates.

Table

Four points in rectangular coordinates

Four points named by their rectangular coordinates (x, y)
Point	$x$	$y$
U	$-3$	$3$
V	$0$	$5$
W	$1$	$-\sqrt{3}$
Z	$-4$	$0$

3. Point U in the table has rectangular coordinates $(-3, 3)$ . Find $r > 0$ for U in polar coordinates. Give an exact value.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$r$ $= \sqrt{x^2 + y^2}$ $= \sqrt{9 + 9}$ $= \sqrt{18}$ $= 3\sqrt{2}$ .

Answer: $3\,\sqrt{2}$ ≈ 4.243

4. For point U, $(-3, 3)$ , which angle $\theta$ in $[0, 2\pi)$ goes with $r = 3\sqrt{2}$ ?

$\frac{3\pi}{4}$
$-\frac{\pi}{4}$
$\frac{7\pi}{4}$
$\frac{\pi}{4}$

Show the answer

$\tan\theta$ $= \frac{y}{x}$ $= \frac{3}{-3}$ $= -1$ , and $\tan^{-1}(-1) = -\frac{\pi}{4}$ . But U is in Quadrant II ( $x < 0$ , $y > 0$ ), and $\tan^{-1}$ only returns angles in Quadrants I and IV. Add $\pi$ : $\theta = \frac{3\pi}{4}$ .

Correct: $\frac{3\pi}{4}$ : Right: Quadrant II, reference angle $\frac{\pi}{4}$ .
$-\frac{\pi}{4}$ : That is $\tan^{-1}(-1)$ , which points into Quadrant IV, and it is not in $[0, 2\pi)$ .
$\frac{7\pi}{4}$ : $\frac{7\pi}{4}$ points into Quadrant IV; with $r > 0$ that names $(3, -3)$ .
$\frac{\pi}{4}$ : $\frac{\pi}{4}$ points into Quadrant I, where both coordinates are positive.

5. The point $(5, 2)$ is in Quadrant I. Find its polar angle $\theta$ , in radians, correct to three decimal places.

Calculator allowed. Give the answer correct to three decimal places, in radians.

Show the answer

$\tan\theta = \frac{2}{5}$ , and the point is in Quadrant I, so $\theta$ $= \tan^{-1}(0.4)$ $\approx 0.380506$ (calculator in radian mode).

Answer: 0.381 radians

Part 9 · Summary

Summary

A point in polar coordinates

(r, \theta)

is

r

units from the pole in the direction

\theta

. Convert with

x = r\cos\theta

,

y = r\sin\theta

,

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

and

\tan\theta = \frac{y}{x}

, checking the quadrant. Every point has many polar names, and a complex number

a + bi

can be written as

r(\cos\theta + i\sin\theta)

.

Part 10 · Up next

What comes next

Part 11 · Connections