In which quadrants is tanθ positive? [1 point]

College Board AP 2024 (style)-style practice: Section II (free response, no calculator). Let θ = (5π)/(6). (a) Give the unit-circle coordinates of the point at angle θ. (b) State sinθ, cosθ and tanθ, and explain the sign of each using the quadrant.

A 3-point question on unit-circle evaluation. (a) Coordinates (1 point): (5π)/(6) is in Quadrant II with reference angle (π)/(6), so the point is (-(sqrt(3))/(2), (1)/(2)). (b) Values and signs (2 points): cosθ = -(sqrt(3))/(2) (negative, since x 0), and tanθ = (sinθ)/(cosθ) = (1/2)/(-sqrt(3)/2) = -(1)/(sqrt(3)) (negative, since sine and cosine have opposite signs).

United StatesPrecalculusSyllabus dot point

How are sine, cosine and tangent defined on the unit circle, and how do they relate to right-triangle ratios?

Topic 3.2 Sine, Cosine, and Tangent: define sine, cosine and tangent using the unit circle and right-triangle ratios, and evaluate them at key angles.

A focused answer to AP Precalculus Topic 3.2, covering the unit-circle definitions of sine, cosine and tangent, their link to right-triangle ratios, radian measure, and evaluating them at the special angles.

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

On the unit circle (radius $1$ , centered at the origin), an angle $\theta$ measured counterclockwise from the positive $x$ -axis has a terminal point at $(\cos\theta, \sin\theta)$ . So cosine is the $x$ -coordinate and sine is the $y$ -coordinate of that point, and tangent is their ratio, $\tan\theta = \frac{\sin\theta}{\cos\theta}$ . This matches the right-triangle ratios (sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, tangent is opposite over adjacent) because the unit circle gives a hypotenuse of $1$ . Angles are measured in radians, where a full circle is $2\pi$ . The signs of sine and cosine follow the quadrant of the terminal point, and the special angles $0$ , $\frac{\pi}{6}$ , $\frac{\pi}{4}$ , $\frac{\pi}{3}$ , $\frac{\pi}{2}$ have exact values worth memorizing.

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What this topic is asking
The unit-circle definition
Radian measure
The right-triangle connection
Signs by quadrant
Try this

What this topic is asking

The College Board (Topic 3.2) wants you to define sine, cosine and tangent using the unit circle, connect them to right-triangle ratios, and evaluate them at the standard angles. An angle in standard position has its terminal ray meet the unit circle at a point whose coordinates are the cosine and sine of the angle; tangent is their ratio.

The unit-circle definition

Because the radius is $1$ , the coordinates themselves are the cosine and sine, with no division needed. This is why the unit circle is the natural home for the trigonometric functions.

Radian measure

AP Precalculus works primarily in radians because they make the function graphs and rates of change behave cleanly. A reference angle (the acute angle to the $x$ -axis) plus the quadrant fixes the exact value and sign of any trig function.

The right-triangle connection

For an acute angle, drop a vertical from the unit-circle point to the $x$ -axis. This forms a right triangle with hypotenuse $1$ , horizontal leg $\cos\theta$ and vertical leg $\sin\theta$ . So $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ reduce to the coordinates exactly, and $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ is the slope of the terminal ray.

Evaluating at a special angle

Find $\sin\frac{2\pi}{3}$ , $\cos\frac{2\pi}{3}$ and $\tan\frac{2\pi}{3}$ exactly.

step 1 Locate the angle

$\frac{2\pi}{3} = 120^\circ$ lies in Quadrant II, where $x < 0$ and $y > 0$ .

step 2 Find the reference angle

The reference angle to the $x$ -axis is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ (that is $60^\circ$ ).

step 3 Use the known values for the reference angle

For $\frac{\pi}{3}$ : cosine is $\frac{1}{2}$ and sine is $\frac{\sqrt{3}}{2}$ . Apply the Quadrant II signs: cosine negative, sine positive. So $\cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ .

step 4 Compute the tangent

$\tan\frac{2\pi}{3} = \frac{\sin}{\cos} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$ .

Signs by quadrant

In Quadrant I both coordinates are positive, so all three functions are positive. In Quadrant II only sine is positive ( $y > 0$ , $x < 0$ ). In Quadrant III both coordinates are negative, so tangent (their ratio) is positive while sine and cosine are negative. In Quadrant IV only cosine is positive. Recovering the sign from the quadrant, rather than memorizing a table, is the reliable method.

A distinction worth stating once is that the input to these functions is an angle, while the output is a pure ratio (a coordinate), with no units. This is why $\sin\frac{\pi}{2} = 1$ exactly: the terminal point at a quarter turn is $(0, 1)$ , so its $y$ -coordinate is $1$ . Reading values straight off the circle, rather than from a calculator, is the no-calculator skill the exam rewards.

Try this

Q1. What is $\cos\pi$ ? [1 point]

Cue. The terminal point at $\pi$ is $(-1, 0)$ , so $\cos\pi = -1$ .

Q2. In which quadrants is $\tan\theta$ positive? [1 point]

Cue. Quadrants I and III, where sine and cosine share the same sign.

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). On the unit circle, the terminal ray of angle

\theta

meets the circle at the point

\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)

. What is

\tan\theta

? (A)

-\sqrt{3}

(B)

-\frac{1}{\sqrt{3}}

(C)

\frac{1}{\sqrt{3}}

(D)

\sqrt{3}

Show worked answer →

The correct answer is (A), $-\sqrt{3}$ .

On the unit circle the point is $(\cos\theta, \sin\theta)$ , so $\cos\theta = -\frac{1}{2}$ and $\sin\theta = \frac{\sqrt{3}}{2}$ . Then $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$ .

AP 2024 (style)3 marksSection II (free response, no calculator). Let

\theta = \frac{5\pi}{6}

. (a) Give the unit-circle coordinates of the point at angle

\theta

. (b) State

\sin\theta

\cos\theta

and

\tan\theta

, and explain the sign of each using the quadrant.

Show worked answer →

A 3-point question on unit-circle evaluation.

(a) Coordinates (1 point): $\frac{5\pi}{6}$ is in Quadrant II with reference angle $\frac{\pi}{6}$ , so the point is $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ .
(b) Values and signs (2 points): $\cos\theta = -\frac{\sqrt{3}}{2}$ (negative, since $x < 0$ in Quadrant II), $\sin\theta = \frac{1}{2}$ (positive, since $y > 0$ ), and $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}}$ (negative, since sine and cosine have opposite signs).

Related dot points

Sources & how we know this

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