AP Precalculus Quiz: Sine Cosine And Tangent
20 questions · exam conditions
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Sine Cosine And TangentQuestion 1 of 20

A circle is centered at the origin and has a radius of 4. The terminal ray of an angle θ\theta intersects the circle at a point P in Quadrant III with a y-coordinate of −2-2. What is the value of cos⁡θ\cos\theta?

−32-\frac{\sqrt{3}}{2}
32\frac{\sqrt{3}}{2}
−12-\frac{1}{2}
12\frac{1}{2}
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AP Precalculus Quiz

AP Precalculus Quiz: Sine Cosine And Tangent

Practice Sine Cosine And Tangent in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sine Cosine And Tangent, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A circle is centered at the origin and has a radius of 4. The terminal ray of an angle θ\theta intersects the circle at a point P in Quadrant III with a y-coordinate of −2-2. What is the value of cos⁡θ\cos\theta?

  1. −32-\frac{\sqrt{3}}{2} (correct answer)
  2. 32\frac{\sqrt{3}}{2}
  3. −12-\frac{1}{2}
  4. 12\frac{1}{2}
Explanation: The equation of the circle is x2+y2=42=16x^2 + y^2 = 4^2 = 16. Given y=−2y=-2, we have x2+(−2)2=16x^2 + (-2)^2 = 16, so x2+4=16x^2 + 4 = 16, and x2=12x^2 = 12. Thus, x=±12=±23x = \pm\sqrt{12} = \pm 2\sqrt{3}. Since the point is in Quadrant III, the x-coordinate must be negative, so x=−23x = -2\sqrt{3}. The value of cos⁡θ\cos\theta is xr=−234=−32\frac{x}{r} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}.

Question 2

Let PP be the point where the terminal ray of an angle θ\theta in standard position intersects the unit circle. What are the coordinates of point PP?

  1. (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) (correct answer)
  2. (sin⁡θ,cos⁡θ)(\sin\theta, \cos\theta)
  3. (tan⁡θ,1)(\tan\theta, 1)
  4. (sec⁡θ,csc⁡θ)(\sec\theta, \csc\theta)
Explanation: By the definition of trigonometric functions on the unit circle, the x-coordinate of the point of intersection is defined as the cosine of the angle (x=cos⁡θx = \cos\theta), and the y-coordinate is defined as the sine of the angle (y=sin⁡θy = \sin\theta). Therefore, the coordinates of point PP are (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta).

Question 3

If the terminal ray of an angle θ\theta in standard position lies in Quadrant IV, which of the following statements must be true?

  1. sin⁡θ>0\sin\theta > 0 and cos⁡θ<0\cos\theta < 0
  2. sin⁡θ<0\sin\theta < 0 and cos⁡θ>0\cos\theta > 0 (correct answer)
  3. sin⁡θ<0\sin\theta < 0 and cos⁡θ<0\cos\theta < 0
  4. sin⁡θ>0\sin\theta > 0 and cos⁡θ>0\cos\theta > 0
Explanation: In Quadrant IV, the x-coordinates are positive and the y-coordinates are negative. On the unit circle, cos⁡θ\cos\theta corresponds to the x-coordinate and sin⁡θ\sin\theta corresponds to the y-coordinate. Therefore, in Quadrant IV, cos⁡θ>0\cos\theta > 0 and sin⁡θ<0\sin\theta < 0.

Question 4

The terminal ray of an angle θ\theta in standard position passes through the point (−8,15)(-8, 15). What is the value of cos⁡θ\cos\theta?

  1. −817-\frac{8}{17} (correct answer)
  2. 1517\frac{15}{17}
  3. −815-\frac{8}{15}
  4. 1715\frac{17}{15}
Explanation: The distance from the origin to the point (−8,15)(-8, 15) is the radius rr, calculated using the Pythagorean theorem: r=(−8)2+152=64+225=289=17r = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17. The cosine of the angle is the ratio of the x-coordinate to the radius, so cos⁡θ=xr=−817\cos\theta = \frac{x}{r} = -\frac{8}{17}.

Question 5

An angle in standard position has a measure of 11π4\frac{11\pi}{4} radians. This angle is coterminal with an angle that has which of the following radian measures?

  1. π4\frac{\pi}{4}
  2. 3π4\frac{3\pi}{4} (correct answer)
  3. 5π4\frac{5\pi}{4}
  4. 7π4\frac{7\pi}{4}
Explanation: To find a coterminal angle, we can add or subtract multiples of 2π2\pi. We have 11π4−2π=11π4−8π4=3π4\frac{11\pi}{4} - 2\pi = \frac{11\pi}{4} - \frac{8\pi}{4} = \frac{3\pi}{4}. Since 3π4\frac{3\pi}{4} is between 0 and 2π2\pi, it is the principal coterminal angle.

Question 6

If cos⁡θ>0\cos\theta > 0 and tan⁡θ<0\tan\theta < 0, in which quadrant does the terminal ray of angle θ\theta lie?

  1. Quadrant I
  2. Quadrant II
  3. Quadrant III
  4. Quadrant IV (correct answer)
Explanation: The condition cos⁡θ>0\cos\theta > 0 means the x-coordinate is positive, which occurs in Quadrant I and Quadrant IV. The condition tan⁡θ<0\tan\theta < 0 means the ratio of the y-coordinate to the x-coordinate is negative, which occurs when their signs are different. This happens in Quadrant II (x<0, y>0) and Quadrant IV (x>0, y<0). The only quadrant that satisfies both conditions is Quadrant IV.

Question 7

Which of the following describes the radian measure of an angle in standard position?

  1. The length of the arc subtended by the angle on the unit circle. (correct answer)
  2. The radius of the circle divided by the length of the subtended arc.
  3. The x-coordinate of the point where the terminal ray intersects a circle.
  4. The number of degrees in the angle multiplied by π\pi.
Explanation: The radian measure of an angle is defined as the ratio of the length of the subtended arc (ss) to the radius of the circle (rr), i.e., θ=s/r\theta = s/r. For a unit circle, the radius r=1r=1, so the radian measure is equal to the arc length, θ=s\theta = s.

Question 8

The terminal ray of an angle θ\theta in standard position has a slope of 12\frac{1}{2}. What is the value of tan⁡θ\tan\theta?

  1. 12\frac{1}{2} (correct answer)
  2. −12-\frac{1}{2}
  3. 2
  4. -2
Explanation: The tangent of an angle in standard position is defined as the slope of its terminal ray. Since the slope of the terminal ray is given as 12\frac{1}{2}, the value of tan⁡θ\tan\theta is 12\frac{1}{2}.

Question 9

Let θ\theta be an angle such that cos⁡θ=−22\cos\theta = -\frac{\sqrt{2}}{2} and sin⁡θ=22\sin\theta = \frac{\sqrt{2}}{2}. What is a possible value for θ\theta in radians?

  1. π4\frac{\pi}{4}
  2. 3π4\frac{3\pi}{4} (correct answer)
  3. 5π4\frac{5\pi}{4}
  4. 7π4\frac{7\pi}{4}
Explanation: The condition cos⁡θ<0\cos\theta < 0 (negative x-coordinate) and sin⁡θ>0\sin\theta > 0 (positive y-coordinate) places the angle in Quadrant II. The reference angle for which cosine and sine have a magnitude of 22\frac{\sqrt{2}}{2} is π4\frac{\pi}{4}. The angle in Quadrant II with this reference angle is π−π4=3π4\pi - \frac{\pi}{4} = \frac{3\pi}{4}.

Question 10

Given that tan⁡θ=−3\tan\theta = -\sqrt{3} and 3π2<θ<2π\frac{3\pi}{2} < \theta < 2\pi, what is the value of cos⁡θ\cos\theta?

  1. 12\frac{1}{2} (correct answer)
  2. −12-\frac{1}{2}
  3. 32\frac{\sqrt{3}}{2}
  4. −32-\frac{\sqrt{3}}{2}
Explanation: The angle θ\theta is in Quadrant IV, where cosine is positive and sine is negative. The reference angle for which tan⁡θ′=3\tan\theta' = \sqrt{3} is θ′=π3\theta'=\frac{\pi}{3}. The angle in Quadrant IV with this reference angle is 2π−π3=5π32\pi - \frac{\pi}{3} = \frac{5\pi}{3}. We need to find cos⁡(5π3)\cos(\frac{5\pi}{3}). Since cosine is positive in Q4, cos⁡(5π3)=cos⁡(π3)=12\cos(\frac{5\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}.

Question 11

What is the value of cos⁡(−π)\cos(-\pi)?

  1. 1
  2. -1 (correct answer)
  3. 0
  4. Undefined
Explanation: The angle −π-\pi is coterminal with π\pi. The terminal ray for an angle of π\pi radians lies on the negative x-axis. The point where this ray intersects the unit circle is (−1,0)(-1, 0). The cosine of the angle is the x-coordinate of this point, which is -1.

Question 12

A projectile is launched at 30 m/s at a 25° angle above horizontal. Using the information provided in the passage, what vertical component of velocity results (nearest tenth) using sin⁡\sin?​

  1. 12.7 m/s12.7\text{ m/s} (correct answer)
  2. 27.2 m/s27.2\text{ m/s}
  3. 63.9 m/s63.9\text{ m/s}
  4. 11.7 m/s11.7\text{ m/s}
Explanation: This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for decomposing velocity vectors in projectile motion problems. In this scenario, a projectile launched at 30 m/s at a 25° angle requires finding the vertical velocity component using the sine function. Choice A is correct because it applies the sine function correctly: vertical component = initial velocity × sin(angle) = 30 × sin(25°) ≈ 30 × 0.4226 ≈ 12.7 m/s. Choice D is incorrect because it appears to use a slightly different calculation or rounding, possibly from calculator error or using the wrong function. To help students: Emphasize understanding velocity vector decomposition, practice using sine for vertical components and cosine for horizontal, and reinforce the physical meaning of these components. Watch for: confusion between sine and cosine usage, and errors in calculator degree mode settings.

Question 13

A circle is centered at the origin with radius rr. If the terminal ray of an angle θ\theta in standard position intersects the circle at point (x,y)(x, y), which expression represents sin⁡θ\sin\theta?

  1. yr\frac{y}{r} (correct answer)
  2. xr\frac{x}{r}
  3. yx\frac{y}{x}
  4. yy
Explanation: For any circle of radius rr centered at the origin, the sine of an angle θ\theta whose terminal ray passes through the point (x,y)(x,y) on the circle is defined as the ratio of the y-coordinate to the radius, sin⁡θ=yr\sin\theta = \frac{y}{r}. This is a generalization of the unit circle definition where r=1r=1.

Question 14

An angle in standard position subtends an arc of length 5π5\pi on a circle of radius 1515. What is the radian measure of this angle?

  1. π3\frac{\pi}{3} (correct answer)
  2. 3
  3. 75π75\pi
  4. 13\frac{1}{3}
Explanation: The radian measure θ\theta is the ratio of the arc length ss to the radius rr. Here, s=5πs = 5\pi and r=15r = 15. So, θ=sr=5π15=π3\theta = \frac{s}{r} = \frac{5\pi}{15} = \frac{\pi}{3} radians.

Question 15

A student is 25 m from a tower and measures a 48° angle of elevation to its top. Based on the scenario described, what is the tower's height (nearest tenth)?

  1. 25tan⁡(48∘)≈27.8 m25\tan(48^\circ)\approx 27.8\text{ m} (correct answer)
  2. 25sin⁡(48∘)≈18.6 m25\sin(48^\circ)\approx 18.6\text{ m}
  3. 25tan⁡(42∘)≈22.5 m25\tan(42^\circ)\approx 22.5\text{ m}
  4. 25tan⁡(48)≈28.9 m25\tan(48)\approx 28.9\text{ m}
Explanation: This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, the angle of elevation from the student to the tower top creates a right triangle where the distance is adjacent and the height is opposite to the angle. Choice A is correct because it applies the tangent function correctly: height = distance × tan(angle) = 25 × tan(48°) ≈ 27.8 m. Choice D is incorrect because it uses 48 radians instead of 48 degrees, which would result in a different calculation entirely. To help students: Draw clear diagrams showing the angle of elevation, identify the known and unknown sides relative to the angle, and ensure calculator mode matches the angle units given. Watch for: confusion between degrees and radians, and errors in identifying which trigonometric function relates the given information.

Question 16

As an angle θ\theta increases from π\pi to 3π2\frac{3\pi}{2}, what is the behavior of sin⁡θ\sin\theta and cos⁡θ\cos\theta?

  1. sin⁡θ\sin\theta decreases from 0 to -1, and cos⁡θ\cos\theta increases from -1 to 0. (correct answer)
  2. sin⁡θ\sin\theta increases from -1 to 0, and cos⁡θ\cos\theta decreases from 0 to -1.
  3. Both sin⁡θ\sin\theta and cos⁡θ\cos\theta decrease over the interval.
  4. Both sin⁡θ\sin\theta and cos⁡θ\cos\theta increase over the interval.
Explanation: This interval represents Quadrant III. At θ=π\theta=\pi, the point on the unit circle is (−1,0)(-1, 0), so cos⁡π=−1\cos\pi=-1 and sin⁡π=0\sin\pi=0. At θ=3π2\theta=\frac{3\pi}{2}, the point is (0,−1)(0, -1), so cos⁡(3π2)=0\cos(\frac{3\pi}{2})=0 and sin⁡(3π2)=−1\sin(\frac{3\pi}{2})=-1. As θ\theta goes from π\pi to 3π2\frac{3\pi}{2}, the y-coordinate (sin⁡θ\sin\theta) goes from 0 to -1 (a decrease), and the x-coordinate (cos⁡θ\cos\theta) goes from -1 to 0 (an increase).

Question 17

An angle measure of −300∘-300^\circ corresponds to a rotation from the positive x-axis. Which of the following describes this rotation?

  1. A clockwise rotation of 300∘300^\circ. (correct answer)
  2. A counterclockwise rotation of 300∘300^\circ.
  3. A clockwise rotation of 60∘60^\circ.
  4. A counterclockwise rotation of 60∘60^\circ.
Explanation: By convention, a negative angle measure indicates a clockwise rotation from the initial side (the positive x-axis). The magnitude of the rotation is 300∘300^\circ. Therefore, an angle of −300∘-300^\circ is formed by a clockwise rotation of 300∘300^\circ.

Question 18

A ladder must reach 4.2 m up a wall while making a 60° angle with the ground. Based on the scenario described, what ladder length is needed (nearest tenth)?

  1. 4.2sin⁡(60∘)≈3.6 m4.2\sin(60^\circ)\approx 3.6\text{ m}
  2. 4.2cos⁡(60∘)=8.4 m\dfrac{4.2}{\cos(60^\circ)}=8.4\text{ m}
  3. 4.2sin⁡(60∘)≈4.8 m\dfrac{4.2}{\sin(60^\circ)}\approx 4.8\text{ m} (correct answer)
  4. 4.2sin⁡(30∘)=8.4 m\dfrac{4.2}{\sin(30^\circ)}=8.4\text{ m}
Explanation: This question tests AP Precalculus skills involving trigonometric functions (sine, cosine, and tangent) in applied contexts. Trigonometric functions relate angles to side lengths in right triangles, useful for real-world applications involving heights, distances, and angles. In this scenario, the ladder forms a right triangle where the ladder is the hypotenuse, the wall height (4.2 m) is opposite to the ground angle (60°), and we need to find the ladder length. Choice C is correct because sin(60°) = opposite/hypotenuse = 4.2/ladder length, so ladder length = 4.2/sin(60°) ≈ 4.8 m. Choice B incorrectly uses cosine, which would relate the adjacent side (ground distance) to the hypotenuse, not the opposite side (wall height). To help students: Draw the triangle clearly labeling all parts, identify which trigonometric function relates the known side to the unknown side, and practice solving for the hypotenuse. Watch for: confusion between sine and cosine based on which sides are given, and errors in algebraic manipulation.

Question 19

The terminal ray of an angle θ\theta in standard position lies on the line y=−xy = -x, with x>0x>0. What is the value of sin⁡θ\sin\theta?

  1. 22\frac{\sqrt{2}}{2}
  2. −22-\frac{\sqrt{2}}{2} (correct answer)
  3. 1
  4. -1
Explanation: The line y=−xy=-x with x>0x>0 lies in Quadrant IV. This line makes a 45∘45^\circ angle with the negative y-axis and the positive x-axis. The angle in standard position is 315∘315^\circ or 7π4\frac{7\pi}{4} radians. A point on this ray could be (1,−1)(1, -1). The distance to the origin is r=12+(−1)2=2r = \sqrt{1^2 + (-1)^2} = \sqrt{2}. Therefore, sin⁡θ=yr=−12=−22\sin\theta = \frac{y}{r} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}.

Question 20

Given that sin⁡θ=−513\sin\theta = -\frac{5}{13} and π<θ<3π2\pi < \theta < \frac{3\pi}{2}, what is the value of tan⁡θ\tan\theta?

  1. 512\frac{5}{12} (correct answer)
  2. −512-\frac{5}{12}
  3. 125\frac{12}{5}
  4. −125-\frac{12}{5}
Explanation: The angle θ\theta is in Quadrant III, where both sine and cosine are negative. Using the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, we have (−513)2+cos⁡2θ=1(-\frac{5}{13})^2 + \cos^2\theta = 1, which gives 25169+cos⁡2θ=1\frac{25}{169} + \cos^2\theta = 1. Thus, cos⁡2θ=1−25169=144169\cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}, so cos⁡θ=±1213\cos\theta = \pm\frac{12}{13}. Since θ\theta is in Quadrant III, cos⁡θ=−1213\cos\theta = -\frac{12}{13}. Then, tan⁡θ=sin⁡θcos⁡θ=−5/13−12/13=512\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-5/13}{-12/13} = \frac{5}{12}.