Unit 3 · Topic 3.3 · about 30 minutes · 7 parts · 12 scored checks, 1 of them multiple choice

Sine and Cosine Function Values

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

A unit circle centered at the origin with sixteen points marked at every multiple of pi/6 and pi/4 from 0 to 2pi, each labeled with its angle and its exact coordinates. In Quadrant I the points are (square root of 3 over 2, 1/2) at pi/6, (square root of 2 over 2, square root of 2 over 2) at pi/4 and (1/2, square root of 3 over 2) at pi/3. The other quadrants repeat the same three sizes with a negative x-coordinate left of the y-axis and a negative y-coordinate below the x-axis. A shaded triangle at 5pi/6 and another at 7pi/4 show the same leg lengths turned into Quadrants II and IV.

Part 1 Think first

Where is the end of the sprinkler arm?

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

Predict first

A lawn sprinkler has a straight arm 10 meters long that starts out pointing due east. It turns counterclockwise through $\frac{π}{6}$ radians, one twelfth of a full turn. How far north of the sprinkler head is the tip of the arm now?

In short. When the terminal ray of an angle $θ$ in standard position meets a circle of radius $r$ centered at the origin, it meets it at the point $P = (r \cos θ, r \sin θ)$ .

Half of an equilateral triangle gives the exact values at $\frac{π}{6}$ and $\frac{π}{3}$ , and an isosceles right triangle gives them at $\frac{π}{4}$ : the sizes are $\frac{1}{2}$ , $\frac{\sqrt{2}}{2}$ and $\frac{\sqrt{3}}{2}$ .
Any multiple of $\frac{π}{6}$ or $\frac{π}{4}$ whose terminal ray is off the axes makes one of those triangles with the $x$ -axis. The quadrant sets the signs: $\cos θ < 0$ left of the $y$ -axis, $\sin θ < 0$ below the $x$ -axis.
On a circle of radius $r$ , find the unit circle values first, then multiply both by $r$ .

Part 2 Learn it

From two triangles to any circle

3a. Two triangles give the exact values

Put an angle of $\frac{π}{3}$ in standard position, and let its terminal ray meet the unit circle at $P$ . The origin, $P$ and the point $(1, 0)$ form a triangle with two sides of length 1 and an angle of $\frac{π}{3}$ between them. That makes it equilateral, every side 1.

Drop a segment from $P$ straight down to the $x$ -axis. It cuts the equilateral triangle in half, so it lands halfway to $(1, 0)$ , at $x = \frac{1}{2}$ . The Pythagorean theorem gives the height: $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ . So $\cos \frac{π}{3} = \frac{1}{2}$ and $\sin \frac{π}{3} = \frac{\sqrt{3}}{2}$ .

The angle $\frac{π}{6}$ uses the same half triangle lying on its long side. The ray is shallow, so $P$ is far across and only a little way up: $\cos \frac{π}{6} = \frac{\sqrt{3}}{2}$ and $\sin \frac{π}{6} = \frac{1}{2}$ .

For $\frac{π}{4}$ the ray splits the quadrant evenly, so the triangle is an isosceles right triangle and its legs are equal. With hypotenuse 1, each leg $a$ satisfies $a^{2} + a^{2} = 1$ , so $a = \frac{\sqrt{2}}{2}$ . Both $\cos \frac{π}{4}$ and $\sin \frac{π}{4}$ equal $\frac{\sqrt{2}}{2}$ .

A quick test when you forget which is which: a steeper ray means a higher point, so a bigger sine.

Quick check · Fill the statement

No calculator. Tap a blank, then tap a value.

$\cos \frac{π}{3} =$ , $\sin \frac{π}{3} =$ , and $\sin \frac{π}{4} =$ .

3b. Same sizes, new signs

Every multiple of $\frac{π}{6}$ or $\frac{π}{4}$ whose terminal ray is not on an axis makes one of those same triangles with the $x$ -axis. Only the direction changes.

Take $\frac{5 π}{6}$ . It is $\frac{π}{6}$ short of $π$ , so its ray sits $\frac{π}{6}$ above the negative $x$ -axis. The triangle is the $\frac{π}{6}$ triangle turned into Quadrant II, with legs $\frac{\sqrt{3}}{2}$ across and $\frac{1}{2}$ up. Across is now to the left, so $P = (- \frac{\sqrt{3}}{2}, \frac{1}{2})$ , which means $\cos \frac{5 π}{6} = - \frac{\sqrt{3}}{2}$ and $\sin \frac{5 π}{6} = \frac{1}{2}$ .

Ask two questions about any of these angles. First, how far is the ray from the $x$ -axis: $\frac{π}{6}$ , $\frac{π}{4}$ or $\frac{π}{3}$ ? That picks the triangle and the sizes. The angle $\frac{7 π}{4}$ is $\frac{π}{4}$ short of $2 π$ , and $\frac{4 π}{3}$ is $\frac{π}{3}$ past $π$ . Second, which quadrant? That picks the signs.

Signs of the coordinates of $P$ by quadrant
Quadrant	$\cos θ$ , the $x$ -coordinate	$\sin θ$ , the $y$ -coordinate
I	positive	positive
II	negative	positive
III	negative	negative
IV	positive	negative

Quick check · Sort it

No calculator. Each card is an angle, in radians, in standard position. Tap a card, then the signs of its cosine and sine.

$\cos θ > 0$ and $\sin θ > 0$

$\cos θ < 0$ and $\sin θ > 0$

$\cos θ < 0$ and $\sin θ < 0$

$\cos θ > 0$ and $\sin θ < 0$

3c. Circles bigger than 1

Back to the sprinkler. Its arm is 10 meters long, so the tip moves on a circle of radius 10 centered at the sprinkler head, not on the unit circle.

Draw the $\frac{π}{6}$ triangle again with a hypotenuse of 10 instead of 1. It is the same shape, so every side is 10 times as long: the legs are $10 \cdot \frac{\sqrt{3}}{2} = 5 \sqrt{3}$ across and $10 \cdot \frac{1}{2} = 5$ up. With east as the positive $x$ -axis and north as the positive $y$ -axis, the tip is at $(5 \sqrt{3}, 5)$ , about 8.66 meters east and exactly 5 meters north.

That works for every angle and every radius. If the terminal ray of $θ$ meets a circle of radius $r$ centered at the origin at the point $P$ , then

P = (r \cos θ, r \sin θ)

Find $\cos θ$ and $\sin θ$ as if the circle had radius 1, signs included, then multiply both by $r$ . The 5.2 meters from Think first, $10 \cdot \frac{π}{6}$ , is how far the tip traveled along the circle. It does not tell you where the tip ended up.

Quick check · Calculate

No calculator. A circle of radius 12 is centered at the origin. The terminal ray of an angle of $\frac{2 π}{3}$ radians in standard position meets the circle at $P$ . What is the $x$ -coordinate of $P$ ?

Your answer

Part 3 Watch it done

A point on a circle of radius 8

Watch each step, then do the last one yourself.

A circle of radius 8 is centered at the origin. The terminal ray of an angle of $\frac{7 π}{6}$ radians in standard position meets the circle at $P$ . Find the exact coordinates of $P$ .

Step 1 · Place the ray
$\frac{7 π}{6} = π + \frac{π}{6}$ , so the ray goes $\frac{π}{6}$ past the negative $x$ -axis, into Quadrant III. The triangle it makes with the $x$ -axis has an angle of $\frac{π}{6}$ at the origin.
Step 2 · Read the triangle
On the unit circle, the $\frac{π}{6}$ triangle has a horizontal leg of $\frac{\sqrt{3}}{2}$ and a vertical leg of $\frac{1}{2}$ .
Why this step?
Why is the horizontal leg the long one here, and not the vertical one?
The ray is only $\frac{π}{6}$ away from the $x$ -axis, a shallow angle, so the point is much farther across than it is down. A steep ray, $\frac{π}{3}$ from the $x$ -axis, would make the vertical leg the long one.
Step 3 · Attach the signs
In Quadrant III the point is left of the $y$ -axis and below the $x$ -axis, so both coordinates are negative and $\cos \frac{7 π}{6} = - \frac{\sqrt{3}}{2}$ .
Step 4 · Scale by the radius
The $x$ -coordinate of $P$ is $8 \cos \frac{7 π}{6} = 8 \cdot (- \frac{\sqrt{3}}{2}) = - 4 \sqrt{3}$ .
Your step · Calculate

No calculator. Finish it. On the circle of radius 8 centered at the origin, what is the $y$ -coordinate of the point $P$ where the terminal ray of $\frac{7 π}{6}$ radians meets the circle?

Your answer

Part 4 Try it

Predict the point, then move it

The lab draws a circle of radius $r$ centered at the origin and the point $P$ where the terminal ray of $θ$ meets it. It starts with $r = 4$ and $θ = 0$ , so $P = (4, 0)$ . Use it to practise first, then answer the two questions below on paper, without the lab.

Lab

Unit Circle

Keep the Values menu on Exact. Set $r$ to 3 and type 2pi/3, then say the coordinates of $P$ before you look. Do the same with $r$ at 2 and 7pi/4. Watch which leg of the triangle changes sign as $P$ moves into a new quadrant.

Open the full Unit Circle lab

Try it · Fill the statement

No calculator. A circle of radius 4 is centered at the origin, and the terminal ray of an angle of $\frac{3 π}{4}$ radians in standard position meets it at $P$ . Tap a blank, then tap a value.

The $x$ -coordinate of $P$ is and the $y$ -coordinate of $P$ is .

Try it · Calculate

No calculator. A circle of radius 2.5 is centered at the origin, and the terminal ray of an angle of $\frac{5 π}{3}$ radians in standard position meets it at $P$ . What is the $x$ -coordinate of $P$ ? Give a decimal.

Your answer

Part 5 Explain it

Write it the way the exam asks

Nothing you type leaves this page.

Free response is 37.5% of the exam score. This one is shaped like Question 3, Modeling a Periodic Context.

Free-response practice · Question 3 style

No calculator. A Ferris wheel has a radius of 20 feet, and its hub is 30 feet above the ground. Place the origin at the hub, with the positive $x$ -axis pointing horizontally to the right and the positive $y$ -axis pointing straight up. A car starts level with the hub on the right, at the point $(20, 0)$ , and the wheel turns counterclockwise. (i) The wheel turns through $\frac{3 π}{4}$ radians. Determine the exact coordinates of the car relative to the hub. (ii) Determine the exact height of the car above the ground at that moment. (iii) Later, the car has turned through $\frac{4 π}{3}$ radians from its start. Explain how you can tell, without finding its height, that the car is below the hub.

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

Model answer

(i) The car is on a circle of radius 20 centered at the origin, so it is at $(20 \cos \frac{3 π}{4}, 20 \sin \frac{3 π}{4})$ . The ray of $\frac{3 π}{4}$ is $\frac{π}{4}$ from the negative $x$ -axis, in Quadrant II, so $\cos \frac{3 π}{4} = - \frac{\sqrt{2}}{2}$ and $\sin \frac{3 π}{4} = \frac{\sqrt{2}}{2}$ . The car is at $(- 10 \sqrt{2}, 10 \sqrt{2})$ . (ii) The car is $10 \sqrt{2}$ feet above the hub, so its height is $30 + 10 \sqrt{2}$ feet. (iii) The ray of $\frac{4 π}{3}$ is $\frac{π}{3}$ past the negative $x$ -axis, in Quadrant III, where $\sin θ < 0$ . So the car's vertical coordinate, $20 \sin \frac{4 π}{3}$ , is negative, and the car is below the hub.

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

Writes the car's position as $(20 \cos θ, 20 \sin θ)$ with $θ = \frac{3 π}{4}$ .
Gives $(- 10 \sqrt{2}, 10 \sqrt{2})$ , with the negative sign on the first coordinate.
Gives the height as $30 + 10 \sqrt{2}$ feet.
Explains that $\frac{4 π}{3}$ is in Quadrant III, where the sine is negative, so the vertical coordinate is negative.

Part 6 Check yourself

Cars on a Ferris wheel

The two case questions use the same wheel. Answer them from it.

A Ferris wheel has a radius of 20 meters, and its hub is 22 meters above the ground. Put the origin at the hub, with the positive $x$ -axis pointing horizontally to the right and the positive $y$ -axis pointing straight up, in meters. Each car's position is given by an angle in standard position: the angle, in radians, of the ray from the hub to that car.

Cars on the Ferris wheel
Car	Angle $θ$ (radians)
A	$\frac{π}{3}$
B	$\frac{2 π}{3}$
C	$\frac{7 π}{6}$
D	$\frac{7 π}{4}$

Case question 1 · Match

No calculator. Tap a car on the Ferris wheel of radius 20 meters, then its coordinates relative to the hub, in meters.

Case question 2 · Calculate

No calculator. On the Ferris wheel of radius 20 meters whose hub is 22 meters above the ground, car C is at the angle $\frac{7 π}{6}$ radians. How high above the ground is car C, in meters?

Your answer

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

Check yourself · Spot the error

No calculator. A student finds the point $P$ where the terminal ray of $\frac{5 π}{6}$ radians meets a circle of radius 6 centered at the origin. Tap the first line that goes wrong.

Check yourself · Multiple choice

No calculator. For which angle $θ$ , in radians, is $\cos θ = - \frac{\sqrt{3}}{2}$ and $\sin θ = \frac{1}{2}$ ?

Part 7 Keep it

Bring back earlier ideas

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

From 2.13 Exponential and Logarithmic Equations and Inequalities · Match

No calculator. Tap each expression, then the expression equal to it for every $x$ .

From 1.13 Function Model Selection and Assumption Articulation · Decide and justify

No calculator. A beach stand records its daily lemonade sales at different temperatures. Sales rise as it gets warmer, peak at about 82 degrees, then fall about as fast on the hottest days, when people stay home. Which model type fits?

Which reason supports your answer?

In three lines

Half an equilateral triangle gives $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ at $\frac{π}{6}$ and $\frac{π}{3}$ ; an isosceles right triangle gives $\frac{\sqrt{2}}{2}$ twice at $\frac{π}{4}$ .
Off the axes, every multiple of $\frac{π}{6}$ or $\frac{π}{4}$ reuses one of those triangles, and the quadrant sets the signs.
On a circle of radius $r$ centered at the origin, the terminal ray of $θ$ meets the circle at $(r \cos θ, r \sin θ)$ .

Vocabulary

Quadrant: One of the four regions the axes cut the plane into, numbered I to IV counterclockwise from the upper right; the quadrant of a terminal ray sets the signs of its sine and cosine.
Standard position: An angle with its vertex at the origin and one ray along the positive $x$ -axis; positive angles turn counterclockwise.
Terminal ray: The ray of an angle in standard position that is not on the positive $x$ -axis; it meets a circle centered at the origin at the point $P = (r \cos θ, r \sin θ)$ .
Equilateral triangle: A triangle with three equal sides, whose half gives the exact values $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ at $\frac{π}{6}$ and $\frac{π}{3}$ .
Isosceles right triangle: A right triangle with two equal legs, each $\frac{\sqrt{2}}{2}$ when the hypotenuse is 1, which gives the exact values at $\frac{π}{4}$ .

Course alignment, for teachers

AP Precalculus topic 3.3, 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.B: Apply numerical results in a given mathematical or applied context.

This lesson maps all 2 essential knowledge statements of the topic to the part that teaches each one: 2 are checked by a scored question. The map is in the lesson file.

Where is the end of the sprinkler arm?

From two triangles to any circle

3a. Two triangles give the exact values

3b. Same sizes, new signs

3c. Circles bigger than 1

A point on a circle of radius 8

Predict the point, then move it

Unit Circle

Write it the way the exam asks

Cars on a Ferris wheel

Bring back earlier ideas

Vocabulary

Labs that use this topic