AP Precalculus Trigonometric Functions — Questions and Answers
Question 1: What is the period of f(x) = 4 sin(πx/3) + 2?
Correct answer: 6
For f(x) = a sin(bx) + d, the period is 2π/|b|. Here b = π/3, so the period is 2π ÷ (π/3) = 6.
A sinusoid of the form a sin(bx + c) + d completes one full cycle when bx increases by 2π. This means the period is 2π/|b|. The coefficient of x here is π/3, so the period is 2π ÷ (π/3) = 2π · 3/π = 6. The π cancels, which is why the answer is a plain number and not a multiple of π. The amplitude (4) and vertical shift (2) do not affect the period.
Question 2: For g(x) = −2 cos(x) + 5, which pair gives the midline and the amplitude?
- Midline y = 5, amplitude 2 (Correct answer)
- Midline y = 5, amplitude −2
- Midline y = −2, amplitude 5
- Midline y = 3, amplitude 2
Correct answer: Midline y = 5, amplitude 2
The midline is y = d = 5. The amplitude is |a| = |−2| = 2, so the graph oscillates between 3 and 7.
In y = a cos(bx + c) + d, the vertical shift d gives the midline y = d. The amplitude is |a|, half the distance between the maximum and minimum values. The negative sign on a reflects the graph over its midline but does not change the amplitude. Here the maximum is 5 + 2 = 7 and the minimum is 5 − 2 = 3, and the midline y = 5 is halfway between them.
Question 3: Which set contains all solutions of 2 sin x = √3 on the interval [0, 2π)?
- x = π/6 and x = 5π/6
- x = π/3 and x = 2π/3 (Correct answer)
- x = π/3 and x = 4π/3
- x = 2π/3 and x = 5π/3
Correct answer: x = π/3 and x = 2π/3
Dividing by 2 gives sin x = √3/2. Sine equals √3/2 at the reference angle π/3 in Quadrant I and at π − π/3 = 2π/3 in Quadrant II.
First isolate the trig function: sin x = √3/2. The reference angle with sine value √3/2 is π/3. Sine is positive in Quadrants I and II, so the solutions are x = π/3 and x = π − π/3 = 2π/3. The values π/6 and 5π/6 correspond to sin x = 1/2. The values 4π/3 and 5π/3 lie in quadrants where sine is negative.
Question 4: What is the value of arcsin(sin(5π/6))?
- 5π/6
- −π/6
- 7π/6
- π/6 (Correct answer)
Correct answer: π/6
sin(5π/6) = 1/2, and arcsin returns the angle in [−π/2, π/2] whose sine is 1/2, which is π/6.
The inverse sine function only outputs angles in [−π/2, π/2]. First evaluate the inside: sin(5π/6) = 1/2. Then ask which angle in [−π/2, π/2] has sine 1/2. That angle is π/6. The expression does not simplify back to 5π/6, because 5π/6 lies outside the range of arcsin. The identity arcsin(sin x) = x holds only when x is in [−π/2, π/2].
Question 5: What is the range of y = sec x?
- (−∞, −1] ∪ [1, ∞) (Correct answer)
- (−1, 1)
- [−1, 1]
- (−∞, ∞)
Correct answer: (−∞, −1] ∪ [1, ∞)
Since sec x = 1/cos x and |cos x| ≤ 1 (with cos x ≠ 0), |sec x| ≥ 1. The range is therefore (−∞, −1] ∪ [1, ∞).
The secant function is the reciprocal of cosine. Cosine takes values in [−1, 1], and sec x is undefined where cos x = 0. The reciprocal of any nonzero number with absolute value at most 1 has absolute value at least 1. So sec x is never strictly between −1 and 1. It reaches exactly 1 and −1 where cos x is 1 and −1. It grows without bound near the vertical asymptotes where cos x approaches 0.
Question 6: Which describes the vertical asymptotes of y = tan(2x), where k is any integer?
- x = π/2 + kπ
- x = π/4 + kπ/2 (Correct answer)
- x = π/2 + 2kπ
- x = kπ/2
Correct answer: x = π/4 + kπ/2
Tangent is undefined when its input equals π/2 + kπ. Setting 2x = π/2 + kπ gives x = π/4 + kπ/2.
The parent function tan θ has vertical asymptotes at θ = π/2 + kπ. For y = tan(2x), the input is θ = 2x, so 2x = π/2 + kπ. Dividing by 2 gives x = π/4 + kπ/2. The period of tan(2x) is π/2, so its asymptotes are spaced π/2 apart. That is half the spacing of the parent function, which has asymptotes π apart.
Question 7: A cell-phone signal strength reading s(t) oscillates sinusoidally with a maximum of 9 and a minimum of 3. It reaches a maximum at t = 0 and completes one cycle every 8 seconds. Which function models s(t)?
- s(t) = 3 cos(πt/4) + 6 (Correct answer)
- s(t) = 3 sin(πt/4) + 6
- s(t) = 6 cos(πt/4) + 3
- s(t) = 3 cos(πt/8) + 6
Correct answer: s(t) = 3 cos(πt/4) + 6
The midline is (9+3)/2 = 6 and the amplitude is (9−3)/2 = 3. Starting at a maximum calls for cosine, and b = 2π/8 = π/4.
The midline is the average of the maximum and minimum: (9 + 3)/2 = 6. The amplitude is half their difference: (9 − 3)/2 = 3. A positive cosine function starts at its maximum when t = 0, so cosine fits without a phase shift. A period of 8 gives b = 2π/8 = π/4. The model is s(t) = 3 cos(πt/4) + 6.
Question 8: A Ferris wheel has a diameter of 40 m, its center is 25 m above the ground, and it makes one full revolution every 10 minutes. A rider is at the lowest point at t = 0. Which function gives the rider's height h(t) in meters?
- h(t) = 20 cos(πt/5) + 25
- h(t) = −20 sin(πt/5) + 25
- h(t) = −20 cos(2πt/5) + 25
- h(t) = −20 cos(πt/5) + 25 (Correct answer)
Correct answer: h(t) = −20 cos(πt/5) + 25
The radius is 20 m and the center is at 25 m, so amplitude is 20 and midline is 25. Starting at the minimum requires −cos, and b = 2π/10 = π/5.
The amplitude is the radius, 20 m, and the midline is the height of the center, 25 m. The rider starts at the lowest point (25 − 20 = 5 m), so the model starts at a minimum, which is −cos at t = 0. The period is 10 minutes, so b = 2π/10 = π/5. This gives h(t) = −20 cos(πt/5) + 25. Using 2π/5 would give a period of 5 minutes, and a positive cosine would start at the top.
Question 9: For which values of x in [0, 2π) is cos x > 1/2?
- 0 ≤ x < π/3 or 5π/3 < x < 2π (Correct answer)
- π/3 < x < 5π/3
- 0 ≤ x < π/6 or 11π/6 < x < 2π
- π/3 < x < 2π/3
Correct answer: 0 ≤ x < π/3 or 5π/3 < x < 2π
cos x = 1/2 at x = π/3 and x = 5π/3. Cosine is above 1/2 near x = 0, so the solution is [0, π/3) together with (5π/3, 2π).
Solve the equation cos x = 1/2 first, which gives x = π/3 and x = 5π/3 on [0, 2π). These two points split the interval into three parts. At x = 0, cos x = 1, which is greater than 1/2, so the intervals touching 0 and 2π satisfy the inequality. In the middle interval (π/3, 5π/3), cosine is below 1/2. The solution set is therefore [0, π/3) ∪ (5π/3, 2π).
Question 10: If cos θ = −3/5 and θ is in Quadrant II, what is cot θ?
- 3/4
- −4/3
- −3/4 (Correct answer)
- −5/3
Correct answer: −3/4
With cos θ = −3/5, the Pythagorean identity gives sin θ = 4/5 (positive in Quadrant II). Then cot θ = cos θ / sin θ = (−3/5)/(4/5) = −3/4.
From sin²θ + cos²θ = 1, sin²θ = 1 − 9/25 = 16/25, so sin θ = ±4/5. In Quadrant II sine is positive, so sin θ = 4/5. Cotangent is cos θ / sin θ, which equals (−3/5) ÷ (4/5) = −3/4. The result is negative because cosine is negative and sine is positive in this quadrant. The value −4/3 would be tan θ, and −5/3 would be sec θ.
Question 11: On the interval (π/2, π), the graph of y = sin x is which of the following?
- Increasing and concave up
- Increasing and concave down
- Decreasing and concave down (Correct answer)
- Decreasing and concave up
Correct answer: Decreasing and concave down
Sine falls from 1 at π/2 to 0 at π, so it is decreasing. The graph bends downward there, so it is concave down.
The sine curve reaches its maximum of 1 at x = π/2 and then falls to 0 at x = π, so it is decreasing on (π/2, π). Sine is concave down wherever its values are positive, since the curve bends below its tangent lines like the top of a hill. On (0, π) the sine graph is positive and concave down throughout. It changes to concave up on (π, 2π), where the values are negative.
