Free AP Precalculus Trigonometric and Polar Functions

AP Precalculus · Trigonometric and Polar Functions · free lesson

Trigonometry and Polar Coordinates

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Polar coordinates

Rectangular coordinates $(x, y)$ locate a point by how far to go across and how far up. Polar coordinates $(r, \theta)$ locate it by a turn and a distance: from the origin (the pole), turn through the angle $\theta$ from the positive $x$ -axis (counterclockwise for $\theta > 0$ ), and go $r$ units along that ray.

The polar grid below has circles of radius 1, 2 and 3 about the pole and a ray every $\frac{\pi}{6}$ . The point $A$ is $\left(3, \frac{\pi}{3}\right)$ : turn $\frac{\pi}{3}$ , go out 3. The point $B$ is $\left(2, \frac{7\pi}{6}\right)$ .

From polar to rectangular, and back

The point $(r, \theta)$ with $r > 0$ is on the circle of radius $r$ at angle $\theta$ , so by the definitions of cosine and sine its rectangular coordinates are $r$ times those of the unit-circle point:

Rule.

Polar to rectangular: $x = r\cos\theta$ and $y = r\sin\theta$ .

Rectangular to polar: $r^{2} = x^{2} + y^{2}$ , and $\tan\theta = \frac{y}{x}$ (for $x \ne 0$ ), with $\theta$ chosen in the quadrant of the point.

For $A$ : $x = {3}\cos {\frac{\pi}{3}} = {\frac{3}{2}}$ and $y = {3}\sin {\frac{\pi}{3}} = {\frac{3\sqrt{3}}{2}}$ . For $B$ : $x = -\sqrt{3}$ and $y = -1$ .

Going back needs care with the angle. Take the point $({-3}, {3})$ . Its distance from the pole is $r = \sqrt{({-3})^2 + {3}^2} = {3\sqrt{2}}$ . The inverse tangent of $\frac{y}{x} = {-1}$ gives $-\frac{\pi}{4}$ , an angle in the fourth quadrant, but the point is in the second quadrant ( $x < 0$ , $y > 0$ ). The inverse tangent only gives angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ , so when $x < 0$ add $\pi$ : $\theta = {-\frac{\pi}{4}} + \pi = {\frac{3\pi}{4}}$ . The point is $\left({3\sqrt{2}}, {\frac{3\pi}{4}}\right)$ .

Warning.

The usual slip is to stop at $\tan^{-1}\frac{y}{x}$ . Check the quadrant every time: when $x < 0$ , the angle is $\tan^{-1}\frac{y}{x} + \pi$ .

Many names for one point, and negative r

Turning a full turn more or less brings you back to the same ray, so $(r, \theta)$ and $(r, \theta + 2\pi n)$ name the same point for every integer $n$ . A negative $r$ means: face along the ray of $\theta$ , then walk $\lvert r \rvert$ units backward, through the pole. So $(-r, \theta)$ lies on the ray opposite to $\theta$ , and $(-r, \theta + \pi)$ is the same point as $(r, \theta)$ .

Move the sliders. The blue ray is the direction $\theta$ ; the red segment goes from the pole to $P = (r, \theta)$ . When $r$ is negative, $P$ is on the other side of the pole from the blue ray. The rectangular coordinates are still $x = r\cos\theta$ and $y = r\sin\theta$ .

For example, $\left({-2}, {\frac{\pi}{4}}\right)$ is on the ray of $\frac{5\pi}{4}$ , in the third quadrant: $x = {-2}\cos {\frac{\pi}{4}} = {-\sqrt{2}}$ and $y = {-2}\sin {\frac{\pi}{4}} = {-\sqrt{2}}$ . The same point is $\left({2}, {\frac{5\pi}{4}}\right)$ and $\left({-2}, {\frac{9\pi}{4}}\right)$ .

Complex numbers in polar form

The complex number $z = a + bi$ is drawn as the point $(a, b)$ of the complex plane: the real part across, the imaginary part up. Its distance from 0 is the modulus $r = \sqrt{a^2 + b^2}$ , and its angle $\theta$ is found exactly as for a point. Since $a = r\cos\theta$ and $b = r\sin\theta$ ,

Rule.

$z = a + bi = r\cos\theta + (r\sin\theta)\,i = r(\cos\theta + i\sin\theta)$ , the polar form of $z$ .

For $z = {-2 + 2\sqrt{3}i}$ : the modulus is $\sqrt{({-2})^2 + ({2\sqrt{3}})^2} = {4}$ , and the point is in the second quadrant, so $\theta = \tan^{-1}\frac{ {2\sqrt{3}} }{ {-2} } + \pi = {\frac{2\pi}{3}}$ . So $z = {4}\left(\cos {\frac{2\pi}{3}} + i\sin {\frac{2\pi}{3}}\right)$ . The imaginary part is the number $b = 2\sqrt{3}$ , not $bi$ : the point is $({-2}, {2\sqrt{3}})$ .

Worked examples

Each example is one kind of question the AP exam asks about polar coordinates. Every value in the working is found by the engine; angles are in radians.

Example.

1. Polar to rectangular. Find the rectangular coordinates of the points $P = \left(6, \frac{5\pi}{4}\right)$ and $Q = \left(4, -\frac{\pi}{6}\right)$ .

Use $x = r\cos\theta$ and $y = r\sin\theta$ with the exact values of the unit circle:

\begin{aligned} P: \quad x &= {6}\cos\frac{5\pi}{4} = {-3\sqrt{2}}, \quad y = {6}\sin\frac{5\pi}{4} = {-3\sqrt{2}} \\ Q: \quad x &= {4}\cos\left(-\frac{\pi}{6}\right) = {2\sqrt{3}}, \quad y = {4}\sin\left(-\frac{\pi}{6}\right) = {-2} \end{aligned}

Check: $P$ is in the third quadrant, where both coordinates are negative, and $Q$ , at a negative angle, is below the positive $x$ -axis.

Result.

$P = \left({-3\sqrt{2}}, {-3\sqrt{2}}\right)$ and $Q = \left({2\sqrt{3}}, {-2}\right)$ .

Example.

2. Rectangular to polar, with the quadrant. Give polar coordinates $(r, \theta)$ with $r > 0$ and $0 \le \theta < 2\pi$ for the point $(-2, -2\sqrt{3})$ .

Below, $r$ is the distance from the pole, $u = \tan^{-1}\frac{y}{x}$ is the inverse tangent's angle, and $t$ is the angle of the point:

\begin{aligned} r &= \sqrt{x^{2} + y^{2}} \\ &= \sqrt{(-2)^{2} + (-2\sqrt{3})^{2}} \\ &= 4 \\ u &= \tan^{-1}(\frac{y}{x}) \\ &= \tan^{-1}(\frac{-2\sqrt{3}}{-2}) \\ &= \frac{\pi}{3} \approx 1.047197551 \\ t &= u + \pi \\ &= \frac{\pi}{3} + \pi \\ &= \frac{4\pi}{3} \approx 4.188790205 \end{aligned}

The inverse tangent gives $u = \frac{\pi}{3}$ , in the first quadrant, but the point has $x < 0$ and $y < 0$ : it is in the third quadrant. So add $\pi$ , which gives $t = \frac{4\pi}{3}$ , already in $0 \le \theta < 2\pi$ .

Check: $x = {4}\cos {\frac{4\pi}{3}} = {-2}$ and $y = {4}\sin {\frac{4\pi}{3}} = {-2\sqrt{3}}$ .

Result.

$\left({4}, {\frac{4\pi}{3}}\right)$ . The answer $\left({4}, {\frac{\pi}{3}}\right)$ names the point $({2}, {2\sqrt{3}})$ , on the opposite side of the pole.

Example.

3. Other names for the same point. Which of $\left(-3, \frac{7\pi}{6}\right)$ , $\left(3, \frac{13\pi}{6}\right)$ and $\left(-3, \frac{\pi}{6}\right)$ name the same point as $\left(3, \frac{\pi}{6}\right)$ ?

Compare rectangular coordinates, each found by $x = r\cos\theta$ , $y = r\sin\theta$ :

Polar name	$x$	$y$
$\left(3, \frac{\pi}{6}\right)$	$\frac{3\sqrt{3}}{2}$	$\frac{3}{2}$
$\left(-3, \frac{7\pi}{6}\right)$	$\frac{3\sqrt{3}}{2}$	$\frac{3}{2}$
$\left(3, \frac{13\pi}{6}\right)$	$\frac{3\sqrt{3}}{2}$	$\frac{3}{2}$
$\left(-3, \frac{\pi}{6}\right)$	$-\frac{3\sqrt{3}}{2}$	$-\frac{3}{2}$

$\left(-3, \frac{7\pi}{6}\right)$ turns half a turn more and walks backward, which lands on the original point; $\left(3, \frac{13\pi}{6}\right)$ turns one full turn more. $\left(-3, \frac{\pi}{6}\right)$ walks backward along the same ray and lands on the opposite side of the pole.

Result.

The first two name the point $\left(3, \frac{\pi}{6}\right)$ ; the third does not.

Example.

4. A point with negative r. Plot $\left(-2, \frac{2\pi}{3}\right)$ and find its rectangular coordinates.

Face along the ray of $\frac{2\pi}{3}$ (dashed, in the second quadrant), then walk 2 units backward through the pole. The point is on the ray of $\frac{2\pi}{3} + \pi = {\frac{5\pi}{3}}$ , in the fourth quadrant.

\begin{aligned} x &= ({-2})\cos\frac{2\pi}{3} = {1}, \quad y = ({-2})\sin\frac{2\pi}{3} = {-\sqrt{3}} \end{aligned}

Result.

$\left(-2, \frac{2\pi}{3}\right)$ is the point $\left({1}, {-\sqrt{3}}\right)$ ; with a positive radius it is $\left({2}, {\frac{5\pi}{3}}\right)$ .

Example.

5. A complex number in polar form, and back. Write $z = 3 - 3i$ in polar form, and write $w = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$ as $a + bi$ .

$z$ is the point $({3}, {-3})$ of the complex plane, in the fourth quadrant. Its modulus is $\sqrt{ {3}^2 + ({-3})^2} = {3\sqrt{2}}$ , and since $a > 0$ the inverse tangent's angle is right as it is: $\theta = \tan^{-1}\frac{ {-3} }{ {3} } = {-\frac{\pi}{4}}$ (or $\frac{7\pi}{4}$ , the same ray).

\begin{aligned} z &= {3\sqrt{2}}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right) \\ w &= 2\cos\frac{5\pi}{6} + \left(2\sin\frac{5\pi}{6}\right)i = {-\sqrt{3} + i} \end{aligned}

Result.

$z = {3\sqrt{2}}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$ and $w = {-\sqrt{3} + i}$ .

Example.

6. A distance and a direction (calculator). A drone flies in a straight line from its launch point, 850 meters in the direction 2.4 radians counterclockwise from due east. How far east and how far north of the launch point is it, correct to three decimal places?

Put the launch point at the pole, east along the positive $x$ -axis and north along the positive $y$ -axis. The drone is at the polar point $(850, 2.4)$ . With the calculator in radian mode:

\begin{aligned} x &= 850\cos 2.4 = {-626.785} \\ y &= 850\sin 2.4 = {574.144} \end{aligned}

The angle 2.4 is between $\frac{\pi}{2}$ and $\pi$ , so the drone is north and west of the launch point: $x < 0$ means west. Check: $\sqrt{x^2 + y^2} = {850.000}$ meters.

Result.

About $626.785$ meters west (an $x$ -coordinate of $-626.785$ ) and $574.144$ meters north.

Practice

Answer AP-style questions on trigonometry and polar coordinates. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise trigonometry and polar coordinates with new numbers