Sine, Cosine, and Tangent - AP Precalculus

1. An angle measure of $-300^\circ$ corresponds to a rotation from the positive x-axis. Which of the following describes this rotation?
   Answer: A clockwise rotation of $300^\circ$. 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 $300^\circ$.
2. 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)?
   Answer: $\dfrac{4.2}{\sin(60^\circ)}\approx 4.8\text{ m}$ 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.
3. The terminal ray of an angle $\theta$ in standard position has a slope of $\frac{1}{2}$. What is the value of $\tan\theta$?
   Answer: $\frac{1}{2}$ 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 $\frac{1}{2}$.
4. A 12.0 m light pole casts a shadow when the sun's angle of elevation is 30°. Based on the scenario described, what is the shadow length (nearest tenth meter) using $\tan$?​
   Answer: $20.8\text{ m}$ 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.
5. For an angle $\theta$ in standard position, its terminal ray intersects the unit circle at point $(a, b)$. Which of the following expressions is always equal to 1?
   Answer: $a^2 + b^2$ The equation of the unit circle is $x^2 + y^2 = 1$. Since the point $(a, b)$ is on the unit circle, its coordinates must satisfy the equation. Therefore, $a^2 + b^2 = 1$. This is equivalent to the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$, since $a=\cos\theta$ and $b=\sin\theta$.
6. The terminal ray of an angle $\theta$ in standard position passes through the point $(-8, 15)$. What is the value of $\cos\theta$?
   Answer: $-\frac{8}{17}$ 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 $\cos\theta = \frac{x}{r} = -\frac{8}{17}$.
7. A worker needs a ladder to reach a 6.0 m height, leaning at a 70° angle with the ground. Based on the scenario described, what ladder length is required (nearest tenth) using $\sin$?​
   Answer: $6.4\text{ m}$ 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.
8. 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 $\sin$?​
   Answer: $12.7\text{ m/s}$ 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.
9. An angle in standard position has a measure of $\frac{11\pi}{4}$ radians. This angle is coterminal with an angle that has which of the following radian measures?
   Answer: $\frac{3\pi}{4}$ 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.
10. As an angle $\theta$ increases from $\pi$ to $\frac{3\pi}{2}$, what is the behavior of $\sin\theta$ and $\cos\theta$?
    Answer: $\sin\theta$ decreases from 0 to -1, and $\cos\theta$ increases from -1 to 0. 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).
11. Given that $\tan\theta = -\sqrt{3}$ and $\frac{3\pi}{2} < \theta < 2\pi$, what is the value of $\cos\theta$?
    Answer: $\frac{1}{2}$ 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, $\cos(\frac{5\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$.
12. Let $\theta$ be an angle such that $\cos\theta = -\frac{\sqrt{2}}{2}$ and $\sin\theta = \frac{\sqrt{2}}{2}$. What is a possible value for $\theta$ in radians?
    Answer: $\frac{3\pi}{4}$ 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 $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
13. Which of the following describes the radian measure of an angle in standard position?
    Answer: The length of the arc subtended by the angle on the unit circle. 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, $\theta = s$.
14. A circle is centered at the origin with radius $r$. If the terminal ray of an angle $\theta$ in standard position intersects the circle at point $(x, y)$, which expression represents $\sin\theta$?
    Answer: $\frac{y}{r}$ 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 $r=1$.
15. A projectile is launched at 20 m/s at a 40° angle above horizontal on level ground. Based on the scenario described, what horizontal component of velocity results (nearest tenth) using $\cos$?​
    Answer: $15.3\text{ m/s}$ 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.
16. 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)?
    Answer: $25\tan(48^\circ)\approx 27.8\text{ m}$ 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.
17. What is the value of $\sin(\frac{5\pi}{6})$?
    Answer: $\frac{1}{2}$ The angle $\frac{5\pi}{6}$ is in Quadrant II. The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$. In Quadrant II, the sine function is positive. Therefore, $\sin(\frac{5\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$.
18. If $\cos\theta > 0$ and $\tan\theta < 0$, in which quadrant does the terminal ray of angle $\theta$ lie?
    Answer: Quadrant IV 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.
19. The terminal ray of an angle $\theta$ in standard position lies on the line $y = -x$, with $x>0$. What is the value of $\sin\theta$?
    Answer: $-\frac{\sqrt{2}}{2}$ 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, $\sin\theta = \frac{y}{r} = \frac{-1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$.
20. Let $P$ be the point where the terminal ray of an angle $\theta$ in standard position intersects the unit circle. What are the coordinates of point $P$?
    Answer: $(\cos\theta, \sin\theta)$ 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 $(\cos\theta, \sin\theta)$.

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