← Back to the Graphs Of Sine And Cosine worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: F-TF.B.5, F-IF.C.7e · Level: Grade 11 Algebra 2
Problem 1 Amplitude and period of sine
In y = A sin(Bx), A = 4 and B = 1. Amplitude 4, period 2π.
Problem 2 Period of a compressed cosine
Period = 2π/3 since B = 3, and the amplitude is 1. Period 2π/3, amplitude 1.
Problem 3 Reflected sine
The amplitude is |−2| = 2 and the period is 2π. The negative sign reflects the graph across the x-axis, so it dips first. Amplitude 2, period 2π, reflected.
Problem 4 Vertical shift
The midline is y = 5. The maximum is 5 + 1 = 6 and the minimum is 5 − 1 = 4. Midline y = 5; max 6; min 4.
Problem 5 Range of a cosine function
Midline −1 and amplitude 3: max = −1 + 3 = 2 and min = −1 − 3 = −4. Max 2, min −4, range −4 ≤ y ≤ 2.
Problem 6 Period with π in the coefficient
Here B = π/4, so the period is 2π ÷ (π/4) = 8. Period 8.
Problem 7 Write an equation from a graph
The maximum is 3, so A = 3. One full wave takes π (zeros at 0, π/2, π), so B = 2π/π = 2. y = 3 sin 2x.
Problem 8 Phase shift to the right
The graph shifts right π/3. The upward midline crossing moves from 0 to π/3. The maximum moves from π/2 to π/2 + π/3 = 5π/6. Shift right π/3; crossing at π/3; max at 5π/6.
Problem 9 Shift left and down
Write x + π/2 = x − (−π/2), so the shift is π/2 to the left. Amplitude 2, midline y = −3. Amplitude 2, shift left π/2, midline y = −3.
Problem 10 Read features from a graph
Maximum 6 at x = 0 and 6; minimum 2 at x = 3 and 9. Midline = (6 + 2)/2 = 4, amplitude = (6 − 2)/2 = 2, and the period is 6 (peaks at 0 and 6). Max 6, min 2, midline 4, amplitude 2, period 6.
Problem 11 Cosine from a description
B = 2π/12 = π/6. y = 4 cos(πx/6) + 6. Check: at x = 0, y = 4 + 6 = 10, a maximum.
Problem 12 Sine from a description
B = 2π/10 = π/5. y = 2.5 sin(πx/5). Check: at x = 2.5 it equals 2.5 sin(π/2) = 2.5.
Problem 13 All features at once
Factor the inside: 2x − π/2 = 2(x − π/4). Amplitude 3, period 2π/2 = π, shift π/4 to the right. Amplitude 3, period π, phase shift π/4.
Problem 14 Locate maximum points
A maximum occurs when 4x = π/2 + 2πk, so x = π/8 + πk/2. For k = 0 and 1 this gives π/8 and 5π/8; k = 2 gives 9π/8, which is beyond π. x = π/8 and 5π/8, each with y = 5.
Problem 15 Ferris wheel model
h(15) = −20 cos(3π/4) + 24 = −20(−0.70711) + 24 = 38.1421. ≈ 38.14 feet.
Problem 16 Tide model
Midline = (7.5 + 1.5)/2 = 4.5, amplitude 3, period 12, so B = π/6. d(t) = 3 cos(πt/6) + 4.5. d(2) = 3 cos(π/3) + 4.5 = 6. The depth is 4.5 when cos(πt/6) = 0, so πt/6 = π/2 and t = 3. d(t) = 3 cos(πt/6) + 4.5; 6 feet; at t = 3 hours.
Problem 17 Daylight hours
Midline 12.1 and amplitude 2.8: longest 14.9 hours, shortest 9.3 hours. The period is 2π ÷ (2π/365) = 365 days. D(120) = 2.8 sin(2π · 40/365) + 12.1 = 2.8(0.6362) + 12.1 = 13.879. Longest 14.9 h, shortest 9.3 h, period 365 days, D(120) ≈ 13.9 h.
Problem 18 Daily temperature
Max = 68 + 14 = 82°F and min = 68 − 14 = 54°F. The sine peaks when π(t − 9)/12 = π/2, so t − 9 = 6 and t = 15 (3 PM). T(12) = 14 sin(π/4) + 68 = 77.899. 82°F at 3 PM; 54°F; T(12) ≈ 77.9°F.
Problem 19 Show two forms are equal
At x = 0: cos 0 = 1 and sin(π/2) = 1. At π/2: cos = 0 and sin(π) = 0. At π: cos π = −1 and sin(3π/2) = −1. Shifting the sine graph left π/2 moves its maximum from π/2 to 0, matching cosine. The two functions are identical.
Problem 20 Find a model from extreme values
Midline (10 + (−2))/2 = 4, amplitude 6. Half a period is 4 − 1 = 3, so the period is 6 and B = π/3. The maximum is at t = 1, so y = 6 cos(π(t − 1)/3) + 4. Check: at t = 4, 6 cos π + 4 = −2. y = 6 cos(π(t − 1)/3) + 4.