Trigonometric Functions Flashcards

What is the period of f(x) = 4 sin(πx/3) + 2?

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.

For g(x) = −2 cos(x) + 5, which pair gives the midline and the amplitude?

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.

Which set contains all solutions of 2 sin x = √3 on the interval [0, 2π)?

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.

What is the value of arcsin(sin(5π/6))?

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.

What is the range of y = sec x?

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, ∞).

Which describes the vertical asymptotes of y = tan(2x), where k is any integer?

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.

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)?

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.

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?

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.

For which values of x in [0, 2π) is cos x > 1/2?

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).

If cos θ = −3/5 and θ is in Quadrant II, what is cot θ?

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.

On the interval (π/2, π), the graph of y = sin x is which of the following?

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.