Sine and cosine function values (unit circle)

AP Precalculus· AP Precalculus CED — Trigonometric and Polar Functions· 14 min read

1. Unit Circle Definition of Sine and Cosine★★☆☆☆⏱ 3 min

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The unit circle definition extends right-triangle trigonometry from acute angles to all real-number angles, and it is the foundation for all trigonometric concepts tested on the AP Precalculus exam. The unit circle is centered at the origin with radius 1, and follows the equation:

x^2 + y^2 = 1

Angles on the unit circle follow standard position convention: measured from the positive x-axis, with counterclockwise rotation as positive and clockwise rotation as negative. For any angle $θ$ , the terminal side intersects the unit circle at point $P (x, y)$ . By definition:

📘 Definition

Unit Circle Sine and Cosine

$cos θ, sin θ$

For the intersection point $(x, y)$ of the terminal side of $θ$ with the unit circle, $cos θ = x$ (the x-coordinate) and $sin θ = y$ (the y-coordinate). This definition holds for all real angles.

Example:

This matches the right triangle definition for acute angles: for a hypotenuse of length 1, $cos θ = \frac{adjacent}{hypotenuse} = x$ and $sin θ = \frac{opposite}{hypotenuse} = y$ .

📐 Worked Example

State the coordinates of the intersection of the terminal side of $θ = \frac{7 π}{6}$ with the unit circle, then give the values of $cos (\frac{7 π}{6})$ and $sin (\frac{7 π}{6})$ .

1
First, identify the quadrant: $\frac{7 π}{6}$ is between $π$ and $\frac{3 π}{2}$ , so it lies in Quadrant III, where both x and y coordinates are negative.
2
Calculate the reference angle, the acute angle between the terminal side and the x-axis:
3
$\frac{7\pi}{6} - \pi = \frac{\pi}{6}$
4
For $\frac{π}{6}$ in Quadrant I, the known unit circle intersection point is:
5
$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$
6
Apply the Quadrant III sign rule to get the final intersection point:
7
$\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)$
8
By the unit circle definition, $cos θ = x$ and $sin θ = y$ , so the final values are:
9
$\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \quad \sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$

Exam tip:

On multiple-choice questions, you can often eliminate two wrong options immediately just by checking the sign of sine and cosine based on quadrant, before doing any calculation to find the magnitude.

2. Reference Angles and Finding Exact Values★★★☆☆⏱ 4 min

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A reference angle is the acute angle that the terminal side of any angle makes with the x-axis, always between $0$ and $\frac{π}{2}$ . Due to the symmetry of the unit circle, the absolute value of sine and cosine for any angle is equal to the sine and cosine of its reference angle. Only the sign of the value changes, based on which quadrant the angle falls into.

To find the exact value of sine or cosine for any angle, follow these steps:

If the angle is negative or larger than $2 π$ , find a coterminal angle between $0$ and $2 π$ by adding or subtracting integer multiples of $2 π$ .
Identify the quadrant of the coterminal angle, to get the correct sign of the final value.
Calculate the reference angle $α$ using quadrant-specific rules: Q1: $α = θ$ ; Q2: $α = π - θ$ ; Q3: $α = θ - π$ ; Q4: $α = 2 π - θ$ .
Use the known value of sine/cosine for $α$ , and apply the correct sign from step 2.

📐 Worked Example

Find the exact value of $sin (- \frac{7 π}{4})$ .

1
Find a positive coterminal angle between $0$ and $2 π$ by adding $2 π = \frac{8 π}{4}$ to the negative angle:
2
$-\frac{7\pi}{4} + \frac{8\pi}{4} = \frac{\pi}{4}$
3
$\frac{π}{4}$ is in Quadrant I, where sine (the y-coordinate) is positive.
4
The reference angle for a Quadrant I angle is the angle itself, so $α = \frac{π}{4}$ . We know $sin (\frac{π}{4}) = \frac{2}{2}$ , so the final value is:
5
$\sin\left(-\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}$

Exam tip:

Always reduce radian fractions to their simplest form immediately. For example, rewrite $\frac{10 π}{8}$ as $\frac{5 π}{4}$ right away, to avoid miscounting quadrants or misidentifying common angles.

3. Finding Unknown Values with the Pythagorean Identity★★★☆☆⏱ 3 min

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The Pythagorean identity for sine and cosine is derived directly from the unit circle equation. Since $x^{2} + y^{2} = 1$ , and $x = cos θ$ , $y = sin θ$ , we get the identity that holds for all real angles $θ$ :

\cos^2\theta + \sin^2\theta = 1

This identity is a common exam tool to find an unknown sine or cosine value when you know the other value and the quadrant of $θ$ . The key step after solving for the squared value is to pick the correct sign based on the quadrant, since taking the square root gives both a positive and negative solution.

📐 Worked Example

Given that $cos θ = \frac{2}{3}$ and $θ$ is in Quadrant IV, find $sin θ$ .

1
Substitute the known value of $cos θ$ into the Pythagorean identity:
2
$\left(\frac{2}{3}\right)^2 + \sin^2\theta = 1$
3
Simplify and solve for $sin^{2} θ$ :
4
$\frac{4}{9} + \sin^2\theta = 1 \implies \sin^2\theta = 1 - \frac{4}{9} = \frac{5}{9}$
5
Take the square root of both sides to get two possible solutions:
6
$\sin\theta = \pm \frac{\sqrt{5}}{3}$
7
$θ$ is in Quadrant IV, where y-coordinates (and thus $sin θ$ ) are negative, so the final solution is:
8
$\sin\theta = -\frac{\sqrt{5}}{3}$

Exam tip:

Never skip writing the $\pm$ when taking the square root. Explicitly writing the sign option reminds you to select the correct sign based on quadrant, which is the most commonly missed point on this problem type.

4. AP-Style Practice Worked Examples★★★★☆⏱ 4 min

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📐 Worked Example

Which of the following is equal to $sin (\frac{17 π}{4})$ ?
A) $\frac{2}{2}$
B) $- \frac{2}{2}$
C) $\frac{1}{2}$
D) $- \frac{1}{2}$

1
Find a coterminal angle between $0$ and $2 π = \frac{8 π}{4}$ by subtracting $2 \cdot 2 π = \frac{16 π}{4}$ :
2
$\frac{17\pi}{4} - \frac{16\pi}{4} = \frac{\pi}{4}$
3
$\frac{π}{4}$ lies in Quadrant I, where sine is positive. The exact value of $sin (\frac{π}{4})$ is $\frac{2}{2}$ , so the correct answer is A.

📐 Worked Example

Consider angle $θ$ with terminal side passing through the point $(- 2, 3)$ on the coordinate plane.
(a) Find the length $r$ from the origin to the point.
(b) What are the exact values of $cos θ$ and $sin θ$ ?
(c) If $ϕ$ is coterminal with $θ$ and $0 \leq ϕ < 2 π$ , what quadrant is $ϕ$ in, and what is the reference angle for $ϕ$ ?

1
Part (a): Use the distance formula from the origin:
2
$r = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}}$
3
Part (b): For any point $(x, y)$ at distance $r$ from the origin, the unit circle intersection is $(\frac{x}{r}, \frac{y}{r})$ , so:
4
$\cos\theta = \frac{-2}{\sqrt{13}} = -\frac{2\sqrt{13}}{13}, \quad \sin\theta = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13}$
5
Part (c): The x-coordinate is negative and the y-coordinate is positive, so $ϕ$ is in Quadrant II. For a Quadrant II angle, the reference angle is $α = π - ϕ$ , so:
6
$\alpha = \pi - \arccos\left(\frac{2\sqrt{13}}{13}\right)$

📐 Worked Example

A Ferris wheel with radius 10 meters has its center 15 meters above the ground. A rider starts at the 3 o'clock position (same height as the center, right of the center). The wheel rotates counterclockwise by 210 degrees. What is the rider's height above the ground after this rotation, to the nearest tenth of a meter?

1
First convert 210 degrees to radians:
2
$210^\circ = \frac{7\pi}{6} \text{ radians}$
3
The vertical position of the rider relative to the center of the wheel is $r sin θ$ , where $r = 10$ meters. From unit circle values:
4
$\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$
5
Calculate relative vertical position, then add the center's height above ground:
6
$10 \cdot \left(-\frac{1}{2}\right) = -5 \implies 15 + (-5) = 10.0$
7
The rider's height after rotation is 10.0 meters.

5. Common Pitfalls

Wrong move:

Leaves the value positive regardless of quadrant after finding the correct magnitude, e.g., $cos (\frac{5 π}{4}) = \frac{2}{2}$ instead of $- \frac{2}{2}$

Why:

Students stop after recalling the common angle value and forget to apply the quadrant sign rule.

Correct move:

After finding the magnitude, explicitly state the quadrant and assign the correct sign before writing your final answer.

Wrong move:

Misapplies the reference angle formula for Quadrant IV, using the Quadrant III rule, e.g., reference angle for $\frac{5 π}{3}$ calculated as $\frac{5 π}{3} - π = \frac{2 π}{3}$ instead of $\frac{π}{3}$

Why:

Students mix up the order of subtraction for Q3 vs Q4.

Correct move:

For any coterminal angle in $[0, 2 π)$ , first write down which quadrant it is in, then use the matching reference angle formula for that quadrant.

Wrong move:

Uses the sign of the known value to pick the sign of the unknown value, e.g., assuming $sin θ$ is positive if $cos θ$ is positive

Why:

Students forget that sine and cosine have opposite signs in Quadrants II and IV.

Correct move:

Always assign the sign of the unknown function based solely on the given quadrant of $θ$ , not the sign of the known function.

Wrong move:

Calculates a reference angle for an angle outside $[0, 2 π)$ without first finding a coterminal angle in the range

Why:

Students try to subtract $π$ directly from a large or negative angle, leading to an incorrect reference angle.

Correct move:

For any angle outside $[0, 2 π)$ , first add or subtract multiples of $2 π$ to get a coterminal angle in the correct range before calculating the reference angle.

Wrong move:

Swaps sine and cosine, writing $sin θ = x$ and $cos θ = y$ for unit circle point $(x, y)$

Why:

Students misremember the order when memorizing the definition.

Correct move:

Use the mnemonic 'cos(x), sin(y)' to always recall that cosine maps to the x-coordinate and sine maps to the y-coordinate.

6. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Unit Circle Definition	$cos θ = x, sin θ = y$	$(x, y)$ = intersection of terminal side with unit circle $x^{2} + y^{2} = 1$ , counterclockwise = positive angle
Pythagorean Identity	$cos^{2} θ + sin^{2} θ = 1$	True for all real angles, derived directly from unit circle equation
Coterminal Angles	$θ \pm 2 π k, k \in Z$	All coterminal angles have identical sine and cosine values
Reference Angle (Q1)	$α = θ$	$0 < θ < π /2$ , $∣ sin θ ∣ = sin α, ∣ cos θ ∣ = cos α$
Reference Angle (Q2)	$α = π - θ$	$π /2 < θ < π$ , $∣ sin θ ∣ = sin α, ∣ cos θ ∣ = cos α$
Reference Angle (Q3)	$α = θ - π$	$π < θ < 3 π /2$ , $∣ sin θ ∣ = sin α, ∣ cos θ ∣ = cos α$
Reference Angle (Q4)	$α = 2 π - θ$	$3 π /2 < θ < 2 π$ , $∣ sin θ ∣ = sin α, ∣ cos θ ∣ = cos α$
Quadrant Sign Mnemonic	All Students Take Calculus	Q1 = All positive, Q2 = Sine positive, Q3 = Tangent positive, Q4 = Cosine positive

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
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2023 · AP Precalculus
FRQ find unknown cosine value

What's Next

This topic is the foundational building block for all remaining trigonometric and polar topics in Unit 3 of AP Precalculus. Immediately next, you will use unit circle sine and cosine values to graph sine and cosine functions, identify their amplitude, period, and phase shift, and model periodic real-world phenomena like seasonal temperature variation or tidal motion. Without mastering exact unit circle values and sign rules, you will not be able to correctly evaluate trigonometric functions at key points, find intercepts and extrema of trig graphs, or convert between rectangular and polar coordinates later in the unit. This topic also underpins future work with trigonometric identities and inverse trigonometric functions, which are tested heavily on the AP Precalculus exam.