Evaluate cos(7π)/(6). [1 point]

If sinθ = (1)/(2) and θ is in Quadrant II, what is cosθ? [1 point]

College Board AP 2023 (style)-style practice: Section I, Part A (multiple choice, no calculator). If cosθ = (3)/(5) and θ is in Quadrant IV, what is sinθ? (A) (4)/(5) (B) -(4)/(5) (C) (3)/(5) (D) -(3)/(5)

The correct answer is (B), -(4)/(5). The Pythagorean identity gives sin^2θ = 1 - cos^2θ = 1 - (9)/(25) = (16)/(25), so sinθ = (4)/(5). In Quadrant IV the y-coordinate is negative, so sinθ = -(4)/(5).

United StatesPrecalculusSyllabus dot point

How do sine and cosine values move around the unit circle, and how do symmetry and the Pythagorean identity connect them?

Topic 3.3 Sine and Cosine Function Values: determine sine and cosine values using the unit circle, reference angles, symmetry and the Pythagorean identity.

A focused answer to AP Precalculus Topic 3.3, covering how sine and cosine values are generated around the unit circle, reference angles, even-odd symmetry, coterminal angles, and the Pythagorean identity.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

As an angle sweeps around the unit circle, $\cos\theta$ traces the $x$ -coordinate and $\sin\theta$ traces the $y$ -coordinate of the terminal point. Any angle reduces to a reference angle (the acute angle to the $x$ -axis), whose known sine and cosine give the size of the values; the quadrant then fixes the signs. Cosine is even, so $\cos(-\theta) = \cos\theta$ , and sine is odd, so $\sin(-\theta) = -\sin\theta$ , because negating the angle reflects the point across the $x$ -axis. Angles that differ by a full turn ( $2\pi$ ) are coterminal and share the same values. The Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ comes straight from $x^2 + y^2 = 1$ , and lets you find sine from cosine (or the reverse) once the quadrant settles the sign.

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What this topic is asking
Reference angles and quadrant signs
Coterminal angles
Even-odd symmetry
The Pythagorean identity
Putting it together
Try this

What this topic is asking

The College Board (Topic 3.3) wants you to generate sine and cosine values around the whole unit circle, not just for acute angles. You use reference angles to reduce any angle to a known one, the quadrant to fix the sign, the symmetry of the circle (even cosine, odd sine), coterminal angles for inputs beyond one revolution, and the Pythagorean identity to find one value from the other.

Reference angles and quadrant signs

So evaluating any angle is a two-step routine: find the reference angle for the size, then read the quadrant for the sign. This works for every angle, including negative ones and those past $2\pi$ .

Coterminal angles

To evaluate a large angle, subtract multiples of $2\pi$ until the result lies in $[0, 2\pi)$ , then proceed with the reference angle.

Even-odd symmetry

The point at angle $-\theta$ is the reflection of the point at angle $\theta$ across the $x$ -axis, sending $(x, y)$ to $(x, -y)$ . The $x$ -coordinate is unchanged and the $y$ -coordinate flips, so cosine is even ( $\cos(-\theta) = \cos\theta$ ) and sine is odd ( $\sin(-\theta) = -\sin\theta$ ). These symmetries let you evaluate negative angles instantly.

The Pythagorean identity

Finding sine from cosine

Given $\cos\theta = -\frac{5}{13}$ with $\theta$ in Quadrant III, find $\sin\theta$ and $\tan\theta$ .

step 1 Apply the Pythagorean identity

$\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$ .

step 2 Take the square root

$\sin\theta = \pm\frac{12}{13}$ .

step 3 Choose the sign from the quadrant

In Quadrant III the $y$ -coordinate is negative, so $\sin\theta = -\frac{12}{13}$ .

step 4 Find the tangent

$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-12/13}{-5/13} = \frac{12}{5}$ , positive as expected in Quadrant III.

Putting it together

A point worth stating once is that the Pythagorean identity gives the magnitude but never the sign; the quadrant must always decide the sign separately. Students who skip the quadrant check get the right number with the wrong sign, which on the unit circle is a completely different point. Combining the reference angle (for size) with the quadrant (for sign) and the identity (to swap between sine and cosine) handles every value question in this topic, and it is the engine behind the graphs in Topic 3.4 and the identities in Topic 3.12.

Try this

Q1. Evaluate $\cos\frac{7\pi}{6}$ . [1 point]

Cue. Reference angle $\frac{\pi}{6}$ , Quadrant III (cosine negative): $\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$ .

Q2. If $\sin\theta = \frac{1}{2}$ and $\theta$ is in Quadrant II, what is $\cos\theta$ ? [1 point]

Cue. $\cos^2\theta = 1 - \frac{1}{4} = \frac{3}{4}$ , and Quadrant II makes cosine negative: $\cos\theta = -\frac{\sqrt{3}}{2}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). If

\cos\theta = \frac{3}{5}

and

\theta

is in Quadrant IV, what is

\sin\theta

? (A)

\frac{4}{5}

(B)

-\frac{4}{5}

(C)

\frac{3}{5}

(D)

-\frac{3}{5}

Show worked answer →

The correct answer is (B), $-\frac{4}{5}$ .

The Pythagorean identity gives $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$ , so $\sin\theta = \pm\frac{4}{5}$ . In Quadrant IV the $y$ -coordinate is negative, so $\sin\theta = -\frac{4}{5}$ .

AP 2024 (style)3 marksSection II (free response, no calculator). (a) Using the unit circle, explain why

\cos(-\theta) = \cos\theta

and

\sin(-\theta) = -\sin\theta

. (b) Use these to evaluate

\sin\left(-\frac{\pi}{6}\right)

and

\cos\left(-\frac{\pi}{6}\right)

Show worked answer →

A 3-point question on symmetry of sine and cosine.

(a) Symmetry (2 points): the angle $-\theta$ reflects the terminal point across the $x$ -axis, sending $(x, y)$ to $(x, -y)$ . The $x$ -coordinate (cosine) is unchanged, so $\cos(-\theta) = \cos\theta$ (cosine is even); the $y$ -coordinate (sine) flips sign, so $\sin(-\theta) = -\sin\theta$ (sine is odd).
(b) Values (1 point): $\cos\left(-\frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{6} = -\frac{1}{2}$ .

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)