Sinusoidal Graphs Worksheet Answer Key – Precalculus

Problem 1 Amplitude and Period

(a) Amplitude 3, period 2π.

(b) Amplitude 2 (the sign only reflects), period 2π/4 = π/2.

(c) Amplitude 1/2, period 2π ÷ (1/3) = 6π.

(d) Amplitude 5, period 2π/π = 2.

Problem 2 Midline and Range

(a) Midline y = 3; maximum 3 + 2 = 5; minimum 3 − 2 = 1.

(b) Midline y = −2; amplitude 1, so the maximum is −1 and the minimum is −3.

Problem 3 Key Points of a Sine Wave

Sine is 0, 1, 0, −1, 0 at these inputs; multiplying by 3 gives 0, 3, 0, −3, 0.

Problem 4 Read the Phase Shift

(a) π/3 to the right.

(b) π/2 to the left.

(c) Factor: sin(2(x − π/2)), so π/2 to the right (not π).

Problem 5 Write an Equation from a Graph

The peak is 3, so A = 3. The curve starts at 0 and reaches its first peak at π/4, a quarter of a period, so the period is 4 · π/4 = π. Then B = 2π/π = 2.

y = 3 sin(2x). Check: at x = π/4, 3 sin(π/2) = 3 ✓.

Problem 6 Tangent Properties

Asymptotes where cosine is 0: x = −π/2 and x = π/2.

Zeros where sine is 0: x = −π, 0, π.

tan(π/4) = 1.

Problem 7 Write a Cosine from a Description

B = 2π/6 = π/3. With a maximum at x = 0 there is no shift.

y = 2.5 cos(πx/3) − 1. Check: maximum 2.5 − 1 = 1.5 at x = 0.

Problem 8 Equation from a Maximum and a Minimum

A = (5 − (−1))/2 = 3; D = (5 + (−1))/2 = 2.

Max to min is half a period: 7π/6 − π/6 = π, so the period is 2π and B = 1. The shift is the maximum’s location, π/6.

y = 3 cos(x − π/6) + 2.

Problem 9 Reflect, Shift and Compress

Factor: 3x − π = 3(x − π/3), so y = −2 sin(3(x − π/3)) + 4.

Amplitude 2 (reflected over the midline), period 2π/3 = 2π/3, shift π/3 to the right, midline y = 4, maximum 6, minimum 2.

Problem 10 Tangent Transformations

(a) Period π ÷ (1/6) = 6π. Asymptotes where x/6 = ±π/2: x = ±3π.

(b) Period π. Asymptotes where x + π/6 = ±π/2: x = π/3 and x = −2π/3. The curve passes through (−π/6, −1).

Problem 11 Pier Water Depth

Period 2π ÷ (π/6) = 12 hours. Greatest depth 9 + 4 = 13 ft; least depth 9 − 4 = 5 ft.

d(2) = 4 sin(π/3) + 9 = 4(0.866) + 9 ≈ 12.5 ft.

Problem 12 Ferris Wheel Height

H(0) = 30 − 25 = 5 ft; H(10) = 30 − 25 cos(π/2) = 30 ft; H(20) = 30 + 25 = 55 ft.

Solve 30 − 25cos(πt/20) = 42.5: cos(πt/20) = −1/2, so πt/20 = 2π/3 and t = 40/3 ≈ 13.3 seconds.

Problem 13 Model from a Table

Maximum 9, minimum 1: amplitude 4, midline y = 5. The period is 8 (minimum at 0 and again at 8), so B = π/4.

The curve starts at its minimum, so use a reflected cosine: y = −4cos(πx/4) + 5. Check: x = 4 gives −4(−1) + 5 = 9 ✓.

Problem 14 Graph of a Compressed Tangent

Period π/5. The asymptotes shown are x = ±π/10, where 5x = ±π/2.

Zeros where 5x = 0 or ±π: x = 0 and, at the ends of the interval, x = ±π/5.

At x = π/20: tan(5π/20) = tan(π/4) = 1.

Problem 15 Describe the Transformations

(1) Stretch vertically by a factor of 2 and reflect over the x-axis. (2) Stretch horizontally by a factor of 2, making the period 2π ÷ (1/2) = 4π. (3) Shift up 3.

Amplitude 2; range [1, 5].

Problem 16 Pedal Height

Lowest 14 − 8 = 6 in; highest 14 + 8 = 22 in.

h(0.3) = 14 − 8cos(π/2) = 14 in.

Solve 14 − 8cos(5πt/3) = 18: cos(5πt/3) = −1/2, so 5πt/3 = 2π/3 and t = 0.4 s.

Problem 17 Hours of Daylight

A = (15.4 − 8.6)/2 = 3.4, midline (15.4 + 8.6)/2 = 12.

D(80) = 3.4cos(2π(−92)/365) + 12 ≈ 11.96 hours.

Solve 3.4cos(·) + 12 ≥ 14: cos(·) ≥ 0.588, so |2π(d − 172)/365| ≤ 0.942, giving |d − 172| ≤ 54.7. That is days 118 to 226, which is 109 days.

Problem 18 Mass on a Spring

Amplitude 6 cm; period 2π/4π = 0.5 s; frequency 1/0.5 = 2 cycles per second.

Solve 6 sin(4πt) = 3: sin(4πt) = 1/2, so 4πt = π/6 or 5π/6, giving t = 1/24 s and t = 5/24 s.

Problem 19 Household Voltage

Peak 170 V; period 2π/120π = 1/60 s.

V(1/240) = 170 sin(π/2) = 170 V.

170 sin(120πt) = 85 gives sin(120πt) = 1/2, so 120πt = π/6 or 5π/6: t = 1/720 s and 1/144 s.

Problem 20 Model the Tide from Data

Half a period is 6.2 hours, so the period is 12.4 and B = 2π/12.4 ≈ 0.5067. A = (7.8 − 0.6)/2 = 3.6 and D = (7.8 + 0.6)/2 = 4.2.

h(t) = 3.6cos(0.5067(t − 3)) + 4.2.

h(6) ≈ 4.4 ft; h(12) ≈ 3.7 ft.