Unit 3 · Topic 3.2 · about 30 minutes · 7 parts · 14 scored checks, 2 of them multiple choice

Sine, Cosine, and Tangent

Built from College Board's course framework · Every answer key independently checked

The unit circle centered at the origin with an angle of 2.5 radians in standard position. The initial ray lies along the positive x-axis and the terminal ray turns counterclockwise into Quadrant II, meeting the circle at P, about (-0.80, 0.60). The arc from (1, 0) to P is highlighted and has length 2.5. A horizontal leg from the origin to (-0.80, 0) is labeled cos theta, about -0.80, and a vertical leg up to P is labeled sin theta, about 0.60.

Part 1 Think first

One wheel, rolled one radius

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A bike wheel has a radius of 30 centimeters. You roll it forward in a straight line for exactly 30 centimeters, without slipping. About how much of a full turn has the wheel made?

In short. Put an angle's vertex at the origin with one ray along the positive $x$ -axis, and the point $P$ where its other ray meets a circle centered at the origin gives the angle's sine, cosine and tangent.

The radian measure of an angle is the length of the arc it cuts off divided by the radius, so a full turn is $2 π$ radians. Positive angles turn counterclockwise, negative ones clockwise, and angles that share a terminal ray differ by a whole number of turns.
$\sin θ$ is the vertical displacement of $P$ divided by its distance from the origin, and $\cos θ$ is the horizontal displacement divided by that distance. On the unit circle, $P = (\cos θ, \sin θ)$ .
$\tan θ$ is the slope of the terminal ray, $\frac{y}{x} = \frac{\sin θ}{\cos θ}$ , and it is not defined when the ray is vertical.

Part 2 Learn it

Four short ideas, each checked as you go

2a. Measuring an angle with an arc

Put the vertex of an angle at the origin and lay one of its rays along the positive $x$ -axis. The angle is now in standard position. The other ray, the one that can point anywhere, is the terminal ray.

Draw a circle of radius $r$ centered at the origin. The angle cuts off an arc of that circle, and the radian measure of the angle is the arc length divided by the radius:

θ = \frac{arc length}{r}

On the unit circle, where $r = 1$ , an arc of length 1 gives an angle of exactly 1 radian. That is the wheel in Think first: rolling forward one radius turned it through one radian.

The size of the circle does not matter, because it is a ratio. On a circle of radius 5, an arc of length 5 is 1 radian and an arc of length 10 is 2 radians.

A full turn sweeps the whole circumference, $2 π r$ , so one revolution is $\frac{2 π r}{r} = 2 π$ radians, about 6.28. Half a turn is $π$ , about 3.14.

You may have measured angles in degrees, where a full turn is 360. In this course an angle with no unit is in radians, so never compare a radian measure with 90 or 360.

Quick check · Calculate

No calculator. A lawn sprinkler sits at the origin and sprays water 6 meters. It turns counterclockwise from the positive $x$ -axis, and the edge of the wet patch traces an arc 9 meters long. Through what angle, in radians, did the sprinkler turn?

Your answer

2b. Turning both ways, and coming back around

The sign of an angle says which way the terminal ray turned. A positive measure rotates counterclockwise from the positive $x$ -axis, and a negative measure rotates clockwise. So $\frac{π}{2}$ points the ray straight up the $y$ -axis, and $- \frac{π}{2}$ points it straight down.

The axes cut the plane into four quadrants, numbered counterclockwise from the upper right. For angles from 0 to $2 π$ the boundaries are $\frac{π}{2} \approx 1.57$ , $π \approx 3.14$ and $\frac{3 π}{2} \approx 4.71$ . An angle of 2.5 radians falls between 1.57 and 3.14, so its terminal ray is in Quadrant II.

Keep turning past $2 π$ and the ray comes back to where it started. Angles in standard position that share a terminal ray are coterminal, and they always differ by a whole number of revolutions, a multiple of $2 π$ . To find the coterminal angle from 0 up to $2 π$ , add or subtract $2 π$ until you land there:

$\frac{9 π}{4} - 2 π = \frac{π}{4}$ , so the ray of $\frac{9 π}{4}$ is in Quadrant I.
$- \frac{2 π}{3} + 2 π = \frac{4 π}{3}$ , so the ray of $- \frac{2 π}{3}$ is in Quadrant III.
$- 4 + 2 π \approx 2.28$ , so the ray of $- 4$ radians is in Quadrant II.

Adding $π$ is only half a turn. It points the ray the opposite way, so it never gives a coterminal angle.

Quick check · Match

No calculator. Each angle is in standard position, in radians. Tap an angle, then the angle from 0 up to $2 π$ that shares its terminal ray.

2c. Sine and cosine come from the point P

The terminal ray of $θ$ meets the circle of radius $r$ at a point $P = (x, y)$ . Its horizontal displacement from the $y$ -axis is $x$ , and its vertical displacement from the $x$ -axis is $y$ . Both carry a sign: left of the $y$ -axis, $x$ is negative, and under the $x$ -axis, $y$ is negative. The distance from the origin to $P$ is the radius, $r$ .

Sine and cosine are ratios of those distances:

\sin θ = \frac{y}{r} \cos θ = \frac{x}{r}

Say the terminal ray meets the circle of radius 5 at $P = (- 3, 4)$ . Then $\sin θ = \frac{4}{5} = 0.8$ and $\cos θ = \frac{- 3}{5} = - 0.6$ .

Follow the same ray out to the circle of radius 10 and you reach $(- 6, 8)$ . Both displacements doubled and so did the distance, so neither ratio changed. Sine and cosine belong to the angle, not to whichever circle you drew.

On the unit circle $r = 1$ , and the division disappears. The sine of the angle is the $y$ -coordinate of $P$ , and the cosine is the $x$ -coordinate. That is why the unit circle is the picture people draw.

Quick check · Multiple choice

No calculator. The terminal ray of an angle $θ$ in standard position meets the circle of radius 25 centered at the origin at $P = (- 24, 7)$ . Which is true?

2d. Tangent is the slope of the terminal ray

The terminal ray starts at the origin and passes through $P = (x, y)$ . Its slope, rise over run from the origin to $P$ , is the tangent of the angle:

\tan θ = \frac{y}{x}

Divide the top and bottom by $r$ and you get $\frac{y / r}{x / r}$ , which is $\frac{\sin θ}{\cos θ}$ . Any point on the ray gives the same slope, so use whichever point you have. For the ray through $(- 3, 4)$ , $\tan θ = \frac{4}{- 3} \approx - 1.33$ .

When the terminal ray lies along the $y$ -axis, $x = 0$ and the ray is vertical. A vertical line has no slope, so $\tan θ$ is not defined at $\frac{π}{2}$ , at $\frac{3 π}{2}$ , or at any angle coterminal with them. Along the $x$ -axis the ray is horizontal and $\tan θ = 0$ .

The signs come straight from the coordinates. In Quadrant I everything is positive. In Quadrant II, $x < 0 < y$ , so only sine is positive. In Quadrant III both coordinates are negative, so sine and cosine are negative, and tangent, a negative divided by a negative, is positive. In Quadrant IV only cosine is positive.

Quick check · Fill the statement

No calculator. Each angle $θ$ is in standard position. Tap a blank, then tap an answer.

When the terminal ray of $θ$ is in Quadrant III, $\cos θ$ is and $\tan θ$ is . When the terminal ray lies along the positive $y$ -axis, $\cos θ$ is and $\tan θ$ is .

Part 3 Watch it done

Three values from one point

Watch each step, then do the last one yourself.

The terminal ray of an angle $θ$ in standard position meets the circle of radius 13 centered at the origin at $P = (- 5, - 12)$ . Name the quadrant of the terminal ray, and find $\sin θ$ , $\cos θ$ and $\tan θ$ .

Step 1 · Place the ray
Both coordinates of $P$ are negative, so $P$ is left of the $y$ -axis and under the $x$ -axis. The terminal ray is in Quadrant III, so one measure of $θ$ is between $π$ and $\frac{3 π}{2}$ . Any angle coterminal with it has the same ray.
Step 2 · Sine and cosine as ratios
The distance from the origin to $P$ is the radius, 13. You can confirm it: $\sqrt{(- 5)^{2} + (- 12)^{2}} = \sqrt{169} = 13$ . The vertical displacement is $- 12$ , so $\sin θ = - \frac{12}{13}$ . The horizontal displacement is $- 5$ , so $\cos θ = - \frac{5}{13}$ . Both are negative, as Quadrant III says they should be.
Why this step?
Why divide by 13, instead of reading $\sin θ = - 12$ straight off the point?
Because $P$ is on a circle of radius 13, not the unit circle. Sine is the vertical displacement divided by the distance from the origin, and only when that distance is 1 does the division disappear. A sine of $- 12$ is impossible: on the unit circle no point is 12 units from the $x$ -axis.
Step 3 · Set up the tangent
The tangent is the slope of the terminal ray. The ray runs from the origin to $P$ , so its slope is $\frac{- 12 - 0}{- 5 - 0}$ . The same number comes from $\frac{\sin θ}{\cos θ}$ , because the 13s cancel.
Your step · Calculate

No calculator. Finish it. What is $\tan θ$ ? Give a decimal.

Your answer

Part 4 Try it

Aim the camera

A security camera sits at the origin and starts pointing along the positive $x$ -axis. Its controller takes turn commands in radians: a positive number turns it counterclockwise and a negative number turns it clockwise. Where the camera ends up pointing is the terminal ray of the command.

For each command in the sort, decide the quadrant first, using the boundaries 1.57, 3.14, 4.71 and 6.28. Then type the angle into the lab (for example 2.5, -2 or 11pi/4) and read the Terminal ray box to check. The last box shows the coterminal angle from 0 up to $2 π$ .

Lab

Unit Circle

The lab is set to radians and decimals on the unit circle. Before you type each angle, say which quadrant the ray will land in and which of $\cos θ$ , $\sin θ$ and $\tan θ$ will be negative. Then check.

Open the full Unit Circle lab

Try it · Sort it

No calculator. A camera at the origin starts pointing along the positive $x$ -axis. Each card is one turn command in radians, positive for counterclockwise and negative for clockwise. Tap a card, then the quadrant the camera ends up pointing into.

Quadrant I

Quadrant II

Quadrant III

Quadrant IV

Try it · Calculate

No calculator. A camera at the origin, pointing along the positive $x$ -axis, is told to turn $- \frac{17 π}{4}$ radians, where negative means clockwise. A single counterclockwise turn of $\frac{k π}{4}$ radians, with $0 \leq \frac{k π}{4} < 2 π$ , leaves it pointing the same way. What is $k$ ?

Your answer

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page. Check your answer against the model and each scoring point.

Free response is 37.5% of the exam score. This one is shaped like Question 4, Symbolic Manipulations.

Free-response practice · Question 4 style

No calculator. The terminal ray of an angle $θ$ in standard position meets the circle of radius 17 centered at the origin at the point $P = (- 8, 15)$ . (A) Find the exact values of $\sin θ$ , $\cos θ$ and $\tan θ$ . (B) Let $α = θ - 6 π$ . Find the value of $\sin α$ , and explain your reasoning. (C) An angle $ϕ$ in standard position has the same tangent as $θ$ , but its terminal ray is in Quadrant IV. Find the point where the terminal ray of $ϕ$ meets the circle of radius 17 centered at the origin, and justify your answer.

Your answer stays on this page. It is never sent or saved.

Model answer

(A) The distance from the origin to $P$ is 17. So $\sin θ = \frac{15}{17}$ , the vertical displacement over that distance, and $\cos θ = - \frac{8}{17}$ , the horizontal displacement over that distance. $\tan θ$ is the slope of the terminal ray, $\frac{15}{- 8} = - \frac{15}{8}$ . (B) $\sin α = \frac{15}{17}$ . Since $α$ and $θ$ differ by $6 π$ , three full revolutions, they share a terminal ray. That ray meets the circle at the same point $P$ , so the sines are equal. (C) The point is $(8, - 15)$ . The terminal ray of $ϕ$ must have slope $- \frac{15}{8}$ , so it lies on the same line through the origin as $P$ , pointing the other way, into Quadrant IV. The point $(8, - 15)$ is on that line, since $\frac{- 15}{8} = - \frac{15}{8}$ , and it is 17 from the origin, since $8^{2} + 15^{2} = 289 = 17^{2}$ .

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Finds $\sin θ = \frac{15}{17}$ and $\cos θ = - \frac{8}{17}$ , dividing each displacement by 17.
Finds $\tan θ = - \frac{15}{8}$ as the slope of the terminal ray, or as $\frac{\sin θ}{\cos θ}$ .
Gives $\sin α = \frac{15}{17}$ and explains that $α$ and $θ$ differ by whole revolutions, so they share a terminal ray.
Finds $(8, - 15)$ and justifies it with the slope $- \frac{15}{8}$ and the distance 17 from the origin.

Part 6 Check yourself

Four rays on one circle

The three case questions use the same table. Answer them from it.

The table lists four angles in standard position, measured in radians and rounded to three decimal places. For each angle it gives the point where the terminal ray meets the circle of radius 5 centered at the origin.

Four angles and the points on the circle of radius 5
Angle	Measure (radians)	Point where the terminal ray meets the circle
$α$	0.927	$(3, 4)$
$β$	2.498	$(- 4, 3)$
$γ$	4.069	$(- 3, - 4)$
$δ$	$- 0.644$	$(4, - 3)$

Case question 1 · Calculate

No calculator. The terminal ray of angle $γ$ meets the circle of radius 5 centered at the origin at $(- 3, - 4)$ . What is $\cos γ$ ? Give a decimal.

Your answer

Case question 2 · Select all

No calculator. The terminal rays of four angles meet the circle of radius 5 centered at the origin at the points given. Which two angles have a negative tangent? Pick two.

Select 2 answers, then check.

Case question 3 · Multiple choice

No calculator. Angle $δ$ in standard position has a measure of $- 0.644$ radians. Which of these is also the measure of an angle that shares its terminal ray?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds the angle in standard position that cuts off an arc of length 10 on the circle of radius 4 centered at the origin, then decides the sign of its cosine. Tap the first line that goes wrong.

Check yourself · Decide and justify

No calculator. The terminal ray of an angle $θ$ in standard position meets the unit circle at $(0.6, - 0.8)$ and meets the circle of radius 3 centered at the origin at $(1.8, - 2.4)$ . Which value is $\sin θ$ ?

Which reason supports your answer?

Part 7 Keep it

Bring back earlier ideas

Pulling an older idea back up is what makes it last.

From 2.8 Inverse Functions · Match

No calculator. Tap a function, then its inverse.

From 1.11 Equivalent Representations of Polynomial and Rational Expressions · Calculate

No calculator. Divide $x^{2} + 5 x + 9$ by $x + 2$ using long division. What is the remainder?

Your answer

In three lines

An angle in standard position has its vertex at the origin and one ray on the positive $x$ -axis; its radian measure is the arc length over the radius, and $2 π$ is one turn.
Positive angles turn counterclockwise and negative ones clockwise, and angles that share a terminal ray differ by a multiple of $2 π$ .
Where the terminal ray meets a circle of radius $r$ at $(x, y)$ : $\sin θ = \frac{y}{r}$ , $\cos θ = \frac{x}{r}$ , and $\tan θ = \frac{y}{x}$ , the slope of the ray.

Vocabulary

Standard position: An angle whose vertex is at the origin and one of whose rays lies along the positive $x$ -axis.
Terminal ray: The other ray of an angle in standard position, besides the one on the positive $x$ -axis; its direction gives the angle's sine, cosine and tangent.
Radian measure: The length of the arc an angle subtends on a circle centered at the origin, divided by that circle's radius, so one full revolution is $2 π$ radians.
Unit circle: The circle of radius 1 centered at the origin, on which an arc of length 1 subtends an angle of 1 radian.
Sine of an angle: The vertical displacement of $P$ from the $x$ -axis divided by the distance from the origin to $P$ ; on the unit circle, the $y$ -coordinate of $P$ .
Cosine of an angle: The horizontal displacement of $P$ from the $y$ -axis divided by the distance from the origin to $P$ ; on the unit circle, the $x$ -coordinate of $P$ .
Tangent of an angle: The slope of the terminal ray, when it exists, which equals $\frac{\sin θ}{\cos θ}$ .

Course alignment, for teachers

AP Precalculus topic 3.2, Unit 3: Trigonometric and Polar Functions.

Skill 2.A: Identify information from graphical, numerical, analytical, and verbal representations to answer a question or construct a model, with and without technology.
Skill 3.A: Describe the characteristics of a function with varying levels of precision, depending on the function representation and available mathematical tools.

This lesson maps all 5 essential knowledge statements of the topic to the part that teaches each one: 5 are checked by a scored question. The map is in the lesson file.

One wheel, rolled one radius

Four short ideas, each checked as you go

2a. Measuring an angle with an arc

2b. Turning both ways, and coming back around

2c. Sine and cosine come from the point P

2d. Tangent is the slope of the terminal ray

Three values from one point

Aim the camera

Unit Circle

Write it the way the exam asks

Four rays on one circle

Bring back earlier ideas

Vocabulary

Labs that use this topic