# Sine and cosine function values (unit circle)

> AP Precalculus · Trigonometric and Polar Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u3-sine-and-cosine-function-values/

This guide covers the unit circle definition of sine and cosine, reference angle calculation, quadrant sign rules, finding exact values for common angles, and using the Pythagorean identity for unknown trigonometric values.

**Prerequisites:** Radian and degree angle conversion; Coordinate plane point plotting; Basic right triangle trigonometry

## Learning objectives

- Define sine and cosine using the unit circle definition
- Find exact sine and cosine values for common angles using reference angles
- Use the Pythagorean identity to find unknown trigonometric values
- Apply quadrant sign rules to get correct final values

## Unit Circle Definition of Sine and Cosine

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 $\theta$, the terminal side intersects the unit circle at point $P(x,y)$. By definition:

**Unit Circle Sine and Cosine** — For the intersection point $(x,y)$ of the terminal side of $\theta$ with the unit circle, $\cos\theta = x$ (the x-coordinate) and $\sin\theta = y$ (the y-coordinate). This definition holds for all real angles.

*Notation:* $\cos\theta, \sin\theta$

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

> **mnemonic**
>
> "cos(x), sin(y)" — remember that cosine always maps to the x-coordinate and sine maps to the y-coordinate to avoid swapping the two.

**Worked example:** State the coordinates of the intersection of the terminal side of $\theta = \frac{7\pi}{6}$ with the unit circle, then give the values of $\cos\left(\frac{7\pi}{6}\right)$ and $\sin\left(\frac{7\pi}{6}\right)$.

1. First, identify the quadrant: $\frac{7\pi}{6}$ is between $\pi$ and $\frac{3\pi}{2}$, 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 $\frac{\pi}{6}$ 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, $\cos\theta = x$ and $\sin\theta = y$, 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.

*Calculator:* forbidden

## Reference Angles and Finding Exact Values

A reference angle is the acute angle that the terminal side of any angle makes with the x-axis, always between $0$ and $\frac{\pi}{2}$. 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:

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

**Worked example:** Find the exact value of $\sin\left(-\frac{7\pi}{4}\right)$.

1. Find a positive coterminal angle between $0$ and $2\pi$ by adding $2\pi = \frac{8\pi}{4}$ to the negative angle:
2. $$-\frac{7\pi}{4} + \frac{8\pi}{4} = \frac{\pi}{4}$$
3. $\frac{\pi}{4}$ 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 $\alpha = \frac{\pi}{4}$. We know $\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$, 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 $\frac{10\pi}{8}$ as $\frac{5\pi}{4}$ right away, to avoid miscounting quadrants or misidentifying common angles.

*Calculator:* forbidden

## Finding Unknown Values with the Pythagorean Identity

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

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

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

1. Substitute the known value of $\cos\theta$ into the Pythagorean identity:
2. $$\left(\frac{2}{3}\right)^2 + \sin^2\theta = 1$$
3. Simplify and solve for $\sin^2\theta$:
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. $\theta$ is in Quadrant IV, where y-coordinates (and thus $\sin\theta$) 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.

*Calculator:* forbidden

## AP-Style Practice Worked Examples

**Worked example:** Which of the following is equal to $\sin\left(\frac{17\pi}{4}\right)$?<br>A) $\frac{\sqrt{2}}{2}$<br>B) $-\frac{\sqrt{2}}{2}$<br>C) $\frac{1}{2}$<br>D) $-\frac{1}{2}$

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

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

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 $\left(\frac{x}{r}, \frac{y}{r}\right)$, 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 $\phi$ is in Quadrant II. For a Quadrant II angle, the reference angle is $\alpha = \pi - \phi$, 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 $r\sin\theta$, 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Leaves the value positive regardless of quadrant after finding the correct magnitude, e.g., $\cos\left(\frac{5\pi}{4}\right) = \frac{\sqrt{2}}{2}$ instead of $-\frac{\sqrt{2}}{2}$
  - Why it fails: Students stop after recalling the common angle value and forget to apply the quadrant sign rule.
  - Correct: After finding the magnitude, explicitly state the quadrant and assign the correct sign before writing your final answer.
- **Wrong:** Misapplies the reference angle formula for Quadrant IV, using the Quadrant III rule, e.g., reference angle for $\frac{5\pi}{3}$ calculated as $\frac{5\pi}{3} - \pi = \frac{2\pi}{3}$ instead of $\frac{\pi}{3}$
  - Why it fails: Students mix up the order of subtraction for Q3 vs Q4.
  - Correct: For any coterminal angle in $[0, 2\pi)$, first write down which quadrant it is in, then use the matching reference angle formula for that quadrant.
- **Wrong:** Uses the sign of the known value to pick the sign of the unknown value, e.g., assuming $\sin\theta$ is positive if $\cos\theta$ is positive
  - Why it fails: Students forget that sine and cosine have opposite signs in Quadrants II and IV.
  - Correct: Always assign the sign of the unknown function based solely on the given quadrant of $\theta$, not the sign of the known function.
- **Wrong:** Calculates a reference angle for an angle outside $[0, 2\pi)$ without first finding a coterminal angle in the range
  - Why it fails: Students try to subtract $\pi$ directly from a large or negative angle, leading to an incorrect reference angle.
  - Correct: For any angle outside $[0, 2\pi)$, first add or subtract multiples of $2\pi$ to get a coterminal angle in the correct range before calculating the reference angle.
- **Wrong:** Swaps sine and cosine, writing $\sin\theta = x$ and $\cos\theta = y$ for unit circle point $(x,y)$
  - Why it fails: Students misremember the order when memorizing the definition.
  - Correct: Use the mnemonic 'cos(x), sin(y)' to always recall that cosine maps to the x-coordinate and sine maps to the y-coordinate.

## Cheatsheet

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

## 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.

- [Pythagorean and reciprocal trigonometric identities](https://www.owlsprep.com/study/ap-precalculus-u3-trigonometric-identities/)
- [Inverse sine and inverse cosine functions](https://www.owlsprep.com/study/ap-precalculus-u3-inverse-trigonometric-functions/)
- [Sine and cosine function graphs](https://www.owlsprep.com/study/ap-precalculus-u3-sine-and-cosine-function-graphs/)

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