AP Precalculus Quiz: Sine Cosine And Tangent

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.

How to use this quiz

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.

Question 1

1. $-\frac{\sqrt{3}}{2}$ (correct answer)
2. $\frac{\sqrt{3}}{2}$
3. $-\frac{1}{2}$
4. $\frac{1}{2}$

Explanation: The equation of the circle is $x^2 + y^2 = 4^2 = 16$. Given $y=-2$, we have $x^2 + (-2)^2 = 16$, so $x^2 + 4 = 16$, and $x^2 = 12$. Thus, $x = \pm\sqrt{12} = \pm 2\sqrt{3}$. Since the point is in Quadrant III, the x-coordinate must be negative, so $x = -2\sqrt{3}$. The value of $\cos\theta$ is $\frac{x}{r} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$.

Question 2

1. $(\cos\theta, \sin\theta)$ (correct answer)
2. $(\sin\theta, \cos\theta)$
3. $(\tan\theta, 1)$
4. $(\sec\theta, \csc\theta)$

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 ($x = \cos\theta$), and the y-coordinate is defined as the sine of the angle ($y = \sin\theta$). Therefore, the coordinates of point $P$ are $(\cos\theta, \sin\theta)$.

Question 3

1. $\sin\theta > 0$ and $\cos\theta < 0$
2. $\sin\theta < 0$ and $\cos\theta > 0$ (correct answer)
3. $\sin\theta < 0$ and $\cos\theta < 0$
4. $\sin\theta > 0$ and $\cos\theta > 0$

Explanation: In Quadrant IV, the x-coordinates are positive and the y-coordinates are negative. On the unit circle, $\cos\theta$ corresponds to the x-coordinate and $\sin\theta$ corresponds to the y-coordinate. Therefore, in Quadrant IV, $\cos\theta > 0$ and $\sin\theta < 0$.

Question 4

1. $-\frac{8}{17}$ (correct answer)
2. $\frac{15}{17}$
3. $-\frac{8}{15}$
4. $\frac{17}{15}$

Explanation: The distance from the origin to the point $(-8, 15)$ is the radius $r$, calculated using the Pythagorean theorem: $r = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17$. The cosine of the angle is the ratio of the x-coordinate to the radius, so $\cos\theta = \frac{x}{r} = -\frac{8}{17}$.

Question 5

1. $\frac{\pi}{4}$
2. $\frac{3\pi}{4}$ (correct answer)
3. $\frac{5\pi}{4}$
4. $\frac{7\pi}{4}$

Explanation: To find a coterminal angle, we can add or subtract multiples of $2\pi$. We have $\frac{11\pi}{4} - 2\pi = \frac{11\pi}{4} - \frac{8\pi}{4} = \frac{3\pi}{4}$. Since $\frac{3\pi}{4}$ is between 0 and $2\pi$, it is the principal coterminal angle.

Question 6

1. Quadrant I
2. Quadrant II
3. Quadrant III
4. Quadrant IV (correct answer)

Explanation: The condition $\cos\theta > 0$ means the x-coordinate is positive, which occurs in Quadrant I and Quadrant IV. The condition $\tan\theta < 0$ 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

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 $\pi$.

Explanation: The radian measure of an angle is defined as the ratio of the length of the subtended arc ($s$) to the radius of the circle ($r$), i.e., $\theta = s/r$. For a unit circle, the radius $r=1$, so the radian measure is equal to the arc length, $\theta = s$.

Question 8

1. $\frac{1}{2}$ (correct answer)
2. $-\frac{1}{2}$
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 $\frac{1}{2}$, the value of $\tan\theta$ is $\frac{1}{2}$.

Question 9

1. $\frac{\pi}{4}$
2. $\frac{3\pi}{4}$ (correct answer)
3. $\frac{5\pi}{4}$
4. $\frac{7\pi}{4}$

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

Question 10

1. $\frac{1}{2}$ (correct answer)
2. $-\frac{1}{2}$
3. $\frac{\sqrt{3}}{2}$
4. $-\frac{\sqrt{3}}{2}$

Explanation: The angle $\theta$ is in Quadrant IV, where cosine is positive and sine is negative. The reference angle for which $\tan\theta' = \sqrt{3}$ is $\theta'=\frac{\pi}{3}$. The angle in Quadrant IV with this reference angle is $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. We need to find $\cos(\frac{5\pi}{3})$. Since cosine is positive in Q4, $\cos(\frac{5\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.

Question 11

1. 1
2. -1 (correct answer)
3. 0
4. Undefined

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

Question 12

1. $12.7\text{ m/s}$ (correct answer)
2. $27.2\text{ m/s}$
3. $63.9\text{ m/s}$
4. $11.7\text{ m/s}$

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

1. $\frac{y}{r}$ (correct answer)
2. $\frac{x}{r}$
3. $\frac{y}{x}$
4. $y$

Explanation: For any circle of radius $r$ centered at the origin, the sine of an angle $\theta$ whose terminal ray passes through the point $(x,y)$ on the circle is defined as the ratio of the y-coordinate to the radius, $\sin\theta = \frac{y}{r}$. This is a generalization of the unit circle definition where $r=1$.

Question 14

1. $\frac{\pi}{3}$ (correct answer)
2. 3
3. $75\pi$
4. $\frac{1}{3}$

Explanation: The radian measure $\theta$ is the ratio of the arc length $s$ to the radius $r$. Here, $s = 5\pi$ and $r = 15$. So, $\theta = \frac{s}{r} = \frac{5\pi}{15} = \frac{\pi}{3}$ radians.

Question 15

1. $25\tan(48^\circ)\approx 27.8\text{ m}$ (correct answer)
2. $25\sin(48^\circ)\approx 18.6\text{ m}$
3. $25\tan(42^\circ)\approx 22.5\text{ m}$
4. $25\tan(48)\approx 28.9\text{ m}$

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

1. $\sin\theta$ decreases from 0 to -1, and $\cos\theta$ increases from -1 to 0. (correct answer)
2. $\sin\theta$ increases from -1 to 0, and $\cos\theta$ decreases from 0 to -1.
3. Both $\sin\theta$ and $\cos\theta$ decrease over the interval.
4. Both $\sin\theta$ and $\cos\theta$ increase over the interval.

Explanation: This interval represents Quadrant III. At $\theta=\pi$, the point on the unit circle is $(-1, 0)$, so $\cos\pi=-1$ and $\sin\pi=0$. At $\theta=\frac{3\pi}{2}$, the point is $(0, -1)$, so $\cos(\frac{3\pi}{2})=0$ and $\sin(\frac{3\pi}{2})=-1$. As $\theta$ goes from $\pi$ to $\frac{3\pi}{2}$, the y-coordinate ($\sin\theta$) goes from 0 to -1 (a decrease), and the x-coordinate ($\cos\theta$) goes from -1 to 0 (an increase).

Question 17

1. A clockwise rotation of $300^\circ$. (correct answer)
2. A counterclockwise rotation of $300^\circ$.
3. A clockwise rotation of $60^\circ$.
4. A counterclockwise rotation of $60^\circ$.

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 $300^\circ$. Therefore, an angle of $-300^\circ$ is formed by a clockwise rotation of $300^\circ$.

Question 18

1. $4.2\sin(60^\circ)\approx 3.6\text{ m}$
2. $\dfrac{4.2}{\cos(60^\circ)}=8.4\text{ m}$
3. $\dfrac{4.2}{\sin(60^\circ)}\approx 4.8\text{ m}$ (correct answer)
4. $\dfrac{4.2}{\sin(30^\circ)}=8.4\text{ m}$

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

1. $\frac{\sqrt{2}}{2}$
2. $-\frac{\sqrt{2}}{2}$ (correct answer)
3. 1
4. -1

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

Question 20

1. $\frac{5}{12}$ (correct answer)
2. $-\frac{5}{12}$
3. $\frac{12}{5}$
4. $-\frac{12}{5}$

Explanation: The angle $\theta$ is in Quadrant III, where both sine and cosine are negative. Using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$, we have $(-\frac{5}{13})^2 + \cos^2\theta = 1$, which gives $\frac{25}{169} + \cos^2\theta = 1$. Thus, $\cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, so $\cos\theta = \pm\frac{12}{13}$. Since $\theta$ is in Quadrant III, $\cos\theta = -\frac{12}{13}$. Then, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-5/13}{-12/13} = \frac{5}{12}$.