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. $\sin\theta$ 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 2

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 3

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 4

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 5

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 6

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, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-5/13}{-12/13} = \frac{5}{12}$.

Question 7

1. $6.5\tan(40^\circ)\approx 5.5\text{ m}$
2. $\dfrac{6.5}{\tan(40^\circ)}\approx 7.7\text{ m}$ (correct answer)
3. $\dfrac{6.5}{\tan(50^\circ)}\approx 5.5\text{ m}$
4. $\dfrac{6.5}{\tan(40)}\approx 1.7\text{ m}$

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 sun's angle of elevation creates a right triangle where the flagpole is the opposite side and the shadow is the adjacent side to the angle. Choice B is correct because tan(40°) = opposite/adjacent = 6.5/shadow length, so shadow length = 6.5/tan(40°) ≈ 7.7 m. Choice A incorrectly multiplies instead of dividing, which would give the height if we knew the shadow length, not the shadow length from the height. To help students: Emphasize the relationship between the sun's angle and the resulting shadow, practice setting up the tangent ratio correctly, and reinforce when to multiply versus divide. Watch for: confusion about which quantity is known versus unknown, and errors in algebraic manipulation of the tangent equation.

Question 8

1. $6.9\text{ m}$
2. $20.8\text{ m}$ (correct answer)
3. $24.0\text{ m}$
4. $10.4\text{ m}$

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, a 12.0 m light pole casts a shadow when the sun's angle of elevation is 30°, requiring the tangent function to find shadow length. Choice B is correct because it applies the tangent function correctly: tan(30°) = opposite/adjacent = 12.0/shadow length, so shadow length = 12.0/tan(30°) ≈ 12.0/0.5774 ≈ 20.8 m. Choice A is incorrect because it appears to multiply 12.0 by tan(30°) instead of dividing, giving a much smaller value. To help students: Emphasize drawing clear diagrams with the sun's rays, vertical pole, and horizontal shadow, practice setting up ratios correctly, and remember that lower sun angles create longer shadows. Watch for: confusion about which way to set up the division, and forgetting that tan(30°) = 1/√3 ≈ 0.577.

Question 9

1. $2.1\text{ m}$
2. $6.4\text{ m}$ (correct answer)
3. $17.5\text{ m}$
4. $5.6\text{ m}$

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, a ladder leaning at a 70° angle with the ground must reach a 6.0 m height, and we need to find the ladder length using the sine function. Choice B is correct because it applies the sine function correctly: sin(70°) = opposite/hypotenuse = 6.0/ladder length, so ladder length = 6.0/sin(70°) ≈ 6.0/0.9397 ≈ 6.4 m. Choice A is incorrect because it appears to multiply 6.0 by sin(70°) instead of dividing, giving a much smaller value. To help students: Emphasize identifying which side is the hypotenuse (the ladder), practice recognizing when to use sine versus other functions, and always check that answers make physical sense. Watch for: confusion about which angle to use and mixing up multiplication versus division when solving for the unknown.

Question 10

1. $250\tan(12^\circ)\approx 53\text{ m}$ (correct answer)
2. $250\cos(12^\circ)\approx 244\text{ m}$
3. $250\tan(78^\circ)\approx 1176\text{ m}$
4. $250\tan(12)\approx 160\text{ m}$

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 depression from a ship to an object below creates a right triangle where the horizontal distance is adjacent and the depth is opposite to the angle. Choice A is correct because it applies the tangent function correctly: depth = horizontal distance × tan(angle of depression) = 250 × tan(12°) ≈ 53 m. Choice D is incorrect because it uses 12 radians instead of 12 degrees, which would give an entirely different result. To help students: Emphasize that angles of depression are measured from the horizontal downward, practice drawing diagrams to visualize the problem setup, and reinforce the importance of using degree mode on calculators. Watch for: confusion between angles of elevation and depression, and mixing up degree and radian measures.

Question 11

1. $50\cos(40^\circ)\approx 38.3\text{ m/s}$
2. $50\sin(40^\circ)\approx 32.1\text{ m/s}$ (correct answer)
3. $50\sin(50^\circ)\approx 38.3\text{ m/s}$
4. $50\sin(40)\approx 37.3\text{ m/s}$

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 real-world applications involving heights, distances, and angles. In this scenario, the projectile's velocity can be decomposed into components, where the vertical component is opposite to the launch angle. Choice B is correct because the vertical speed component equals the total speed multiplied by sin(40°): vertical speed = 50 × sin(40°) ≈ 32.1 m/s. Choice A incorrectly uses cosine, which would give the horizontal component instead of the vertical component. To help students: Use vector diagrams showing velocity decomposition, emphasize that sine gives the vertical (opposite) component while cosine gives the horizontal (adjacent) component, and practice identifying which component is requested. Watch for: confusion between horizontal and vertical components, and mixing up sine and cosine for different velocity components.

Question 12

1. $15.3\text{ m/s}$ (correct answer)
2. $12.9\text{ m/s}$
3. $30.6\text{ m/s}$
4. $17.3\text{ m/s}$

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 real-world applications involving velocity components in projectile motion. In this scenario, a projectile launched at 20 m/s at a 40° angle requires finding the horizontal velocity component using the cosine function. Choice A is correct because it applies the cosine function correctly: horizontal component = initial velocity × cos(angle) = 20 × cos(40°) ≈ 20 × 0.7660 ≈ 15.3 m/s. Choice B is incorrect because it appears to use sine instead of cosine, giving the vertical component rather than horizontal. To help students: Emphasize understanding velocity vector decomposition, practice identifying when to use cosine (for horizontal) versus sine (for vertical) components, and reinforce the connection to right triangle trigonometry. Watch for: confusion between horizontal and vertical components, and mixing up sine and cosine functions in projectile problems.

Question 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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 $\frac{1}{2}$.