AP PreCalculus Practice Quiz: Sine and Cosine Function Values

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

A circle centered at the origin has a radius of 5. What are the coordinates of the point where the terminal ray of an angle θ = π/2 intersects the circle?

A)(5, 0)B)(0, 5)C)(-5, 0)D)(0, -5)

All Questions (9)

A circle centered at the origin has a radius of 5. What are the coordinates of the point where the terminal ray of an angle θ = π/2 intersects the circle?

A) (5, 0)

B) (0, 5)

C) (-5, 0)

D) (0, -5)

Correct Answer: B

The coordinates of a point on a circle of radius r are given by the formula (r cos θ, r sin θ). In this case, r = 5 and θ = π/2. We know that cos(π/2) = 0 and sin(π/2) = 1. Therefore, the coordinates are (5 * 0, 5 * 1), which simplifies to (0, 5).

A point P is on a circle of radius 10 centered at the origin. The point corresponds to an angle of θ = π/4 in standard position. What are the exact coordinates of P?

A) (10, 10)

B) (5, 5)

C) (5√2, 5√2)

D) (10√2, 10√2)

Correct Answer: C

The coordinates are given by (r cos θ, r sin θ). With r = 10 and θ = π/4, we use the known values from an isosceles right triangle: cos(π/4) = √2/2 and sin(π/4) = √2/2. The coordinates are (10 * (√2/2), 10 * (√2/2)), which simplifies to (5√2, 5√2).

The terminal ray of an angle θ intersects a circle of radius 4 at the point (-2√3, 2). Which of the following could be the value of θ?

A) π/6

B) 2π/3

C) 5π/6

D) 7π/6

Correct Answer: C

The coordinates are given by (r cos θ, r sin θ). So, 4 cos θ = -2√3 and 4 sin θ = 2. This gives cos θ = -√3/2 and sin θ = 1/2. An angle whose cosine is negative and sine is positive must be in Quadrant II. Based on the geometry of an equilateral triangle, the reference angle with cos(θ_ref) = √3/2 is π/6. The angle in Quadrant II with this reference angle is θ = π - π/6 = 5π/6.

Point P(x, y) is the intersection of the terminal ray of an angle θ and a circle of radius r centered at the origin. Which of the following expressions represents the value of cos θ?

A) x/r

B) y/r

C) r/x

D) r/y

Correct Answer: A

The coordinates of point P are defined as (r cos θ, r sin θ). Therefore, the x-coordinate is x = r cos θ. Solving for cos θ gives cos θ = x/r.

Using the geometry of an isosceles right triangle, determine the exact coordinates of the point P on a circle of radius 6 corresponding to an angle of 5π/4.

A) (3√2, 3√2)

B) (-3√2, 3√2)

C) (3√2, -3√2)

D) (-3√2, -3√2)

Correct Answer: D

The angle 5π/4 is in Quadrant III, where both cosine and sine are negative. The reference angle is 5π/4 - π = π/4. From the geometry of an isosceles right triangle (45-45-90), we know cos(π/4) = √2/2 and sin(π/4) = √2/2. The coordinates are (r cos θ, r sin θ). With r=6, cos(5π/4) = -cos(π/4) = -√2/2 and sin(5π/4) = -sin(π/4) = -√2/2. Thus, the coordinates are (6 * -√2/2, 6 * -√2/2) = (-3√2, -3√2).

The geometry of an equilateral triangle can be used to find the exact coordinates for a point on a circle of radius 20 corresponding to an angle of 11π/6. What are these coordinates?

A) (10√3, -10)

B) (-10, 10√3)

C) (10, -10√3)

D) (-10√3, -10)

Correct Answer: A

The angle 11π/6 is in Quadrant IV, where cosine is positive and sine is negative. The reference angle is 2π - 11π/6 = π/6. Using the geometry of a 30-60-90 triangle (derived from an equilateral triangle), cos(π/6) = √3/2 and sin(π/6) = 1/2. The coordinates are (r cos θ, r sin θ). With r=20, cos(11π/6) = cos(π/6) = √3/2 and sin(11π/6) = -sin(π/6) = -1/2. The coordinates are (20 * √3/2, 20 * -1/2) = (10√3, -10).

An angle θ in standard position has its terminal ray passing through a vertex of an isosceles right triangle. The triangle is placed in the first quadrant with one vertex at the origin and one leg along the positive x-axis. The hypotenuse of the triangle has a length of 7. What are the coordinates of the point where the terminal ray intersects a circle of radius 7 centered at the origin?

A) (7/2, 7√3/2)

B) (7√3/2, 7/2)

C) (7√2/2, 7√2/2)

D) (7, 7)

Correct Answer: C

An isosceles right triangle in the first quadrant with a vertex at the origin and a leg on the x-axis has its hypotenuse along the line y=x, which corresponds to an angle of θ = π/4. The point of intersection with a circle of radius r=7 is given by (r cos θ, r sin θ). For θ = π/4, the coordinates are (7 cos(π/4), 7 sin(π/4)) = (7 * √2/2, 7 * √2/2) = (7√2/2, 7√2/2).

A point P is located on a circle of radius `r` centered at the origin. The line segment from the origin to P makes an angle `θ` with the positive x-axis. Which of the following represents the coordinates of P?

A) (r sin θ, r cos θ)

B) (r cos θ, r sin θ)

C) (cos θ, sin θ)

D) (r, θ)

Correct Answer: B

This question tests the fundamental definition provided in the content. The x-coordinate of a point on a circle of radius r is given by r cos θ, and the y-coordinate is given by r sin θ. Therefore, the coordinates of point P are (r cos θ, r sin θ).

A point P on a circle of radius 2 is determined by the angle θ = -π/3. Using the geometry of an equilateral triangle, find the exact coordinates of P.

A) (1, -√3)

B) (-1, √3)

C) (√3, -1)

D) (-√3, 1)

Correct Answer: A

The angle θ = -π/3 is coterminal with 5π/3 and is in Quadrant IV. The reference angle is π/3. From the geometry of a 30-60-90 triangle (derived from an equilateral triangle), we know cos(π/3) = 1/2 and sin(π/3) = √3/2. Since P is in Quadrant IV, its x-coordinate (cosine) is positive and its y-coordinate (sine) is negative. The coordinates are (r cos θ, r sin θ) = (2 * cos(-π/3), 2 * sin(-π/3)) = (2 * (1/2), 2 * (-√3/2)) = (1, -√3).