Unit 3 · Topic 3.5

Sinusoidal Functions

A sinusoidal function is a sine or cosine graph after dilations and translations.

Practice 2: Multiple RepresentationsPractice 3: Communication and Reasoning

Part 1 · Hook

Why this matters

The depth of water at a harbor, the temperature over a day and the hum of a guitar string all rise and fall in smooth, even waves. Before you can write a formula for one, you need four numbers that describe any such wave: how far it swings, where its center line is, how long one cycle takes, and how many cycles fit in one unit of time.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is the period of y=cos⁡θy = \cos\theta?

  1. 2π2\pi
  2. π\pi
  3. 11
  4. π2\frac{\pi}{2}
Show the answer

The point on the unit circle returns to the same place after one full turn, 2π2\pi.

  • Correct: 2π2\pi:
  • π\pi:
  • 11:
  • π2\frac{\pi}{2}:

2. The graph of g(x)=f(x)+3g(x) = f(x) + 3 is the graph of ff

  1. moved up 3 units
  2. moved right 3 units
  3. stretched vertically by a factor of 3
  4. moved left 3 units
Show the answer

Adding 3 to every output moves each point up 3 units.

  • Correct: moved up 3 units:
  • moved right 3 units:
  • stretched vertically by a factor of 3:
  • moved left 3 units:

Part 4 · See it

See it first

The graph of y = 3 sin(pi t / 4) + 5 for t from 0 to 10. The dashed midline y = 5 runs halfway between the maximum 8 and the minimum 2. A vertical arrow from the midline up to the maximum shows the amplitude, 3. A horizontal arrow from t = 0 to t = 8 shows one period, 8; the frequency is 1/8 of a cycle per unit of t.
For y=3sin⁡(πt4)+5y = 3\sin\left(\frac{\pi t}{4}\right) + 5: midline y=5y = 5, amplitude 3 (midline to maximum), period 8, frequency 18\frac{1}{8}. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Stretch or shrink the graph of sine or cosine and slide it aroundyou get a sinusoidal function: the same smooth wave shape, with new heights and a new cycle length
  2. Average the greatest and least valuesthe midline y=max⁡+min⁡2y = \frac{\max + \min}{2}, the horizontal line the wave oscillates around
  3. Take half the distance from the least value to the greatestthe amplitude max⁡−min⁡2\frac{\max - \min}{2}, how far the wave reaches above and below its midline
  4. Measure one full cycle along the input axisthe period PP; the frequency 1P\frac{1}{P} counts the cycles per unit of input

Part 6 · Key ideas

Key ideas

  • A sinusoidal function has the shape of a sine graph after dilations and translations. y=sin⁡θy = \sin\theta and y=cos⁡θy = \cos\theta are both sinusoidal, each with amplitude 1, midline y=0y = 0 and period 2π2\pi.
  • Amplitude is a distance, so it is never negative. Midline is a line, so write it as an equation: y=5y = 5.
  • Frequency =1period= \frac{1}{\text{period}}. A period of 8 seconds means 18\frac{1}{8} of a cycle per second; 220 cycles per second means a period of 1220\frac{1}{220} second.
  • Sine is odd, sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin\theta; cosine is even, cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta.

Part 7 · Misconception

A common mistake

The wrong idea: The amplitude is the distance from the lowest point to the highest point.

What actually happens: That distance is twice the amplitude. The amplitude is measured from the midline to the highest point (or to the lowest), so it is half of max⁡−min⁡\max - \min.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

A sinusoidal function ff

123456789101112−3−2−11tyf

1. What is the amplitude of ff?

  1. 22
  2. 44
  3. 11
  4. 33
Show the answer

The greatest value is 11 and the least is −3-3: amplitude =1−(−3)2=2= \frac{1 - (-3)}{2} = 2.

  • Correct: 22: Half of max⁡−min⁡=4\max - \min = 4.
  • 44: 44 is the full distance from the lowest to the highest point, twice the amplitude.
  • 11: 11 is the greatest value, measured from the tt-axis instead of from the midline.
  • 33: 33 is the size of the least value, measured from the tt-axis instead of from the midline.

2. What is the midline of the graph of ff?

  1. y=−1y = -1
  2. y=0y = 0
  3. y=−3y = -3
  4. y=2y = 2
Show the answer

The midline is halfway between 11 and −3-3: y=1+(−3)2=−1y = \frac{1 + (-3)}{2} = -1.

  • Correct: y=−1y = -1: Average of the greatest and least values.
  • y=0y = 0: The tt-axis is not the midline here: the wave reaches 1 above it but 3 below it.
  • y=−3y = -3: −3-3 is the least value, the bottom of the wave, not its center line.
  • y=2y = 2: 22 is the amplitude, not the height of the midline.

3. What is the period of ff?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Highest points at t=0t = 0 and t=6t = 6: one full cycle takes 6 units.

  • Answer: 6

4. What is the frequency of ff, in cycles per unit of tt?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Frequency =1period=16= \frac{1}{\text{period}} = \frac{1}{6} cycle per unit of tt.

  • Answer: 16\frac{1}{6} ≈ 0.167

5. What is the average rate of change of ff on [0,3][0, 3]?

  1. −43-\frac{4}{3}
  2. −4-4
  3. −34-\frac{3}{4}
  4. −23-\frac{2}{3}
Show the answer

f(3)−f(0)3−0\frac{f(3) - f(0)}{3 - 0}=−3−13= \frac{-3 - 1}{3}=−43= -\frac{4}{3}, the slope of the secant line through (0,1)(0, 1) and (3,−3)(3, -3).

  • Correct: −43-\frac{4}{3}: Change in output −4-4 over change in input 33.
  • −4-4: −4-4 is the change in output; divide by the 3 units of input.
  • −34-\frac{3}{4}: This is the change in input over the change in output, upside down.
  • −23-\frac{2}{3}: This uses the amplitude 2 as the change in output; the graph drops 4, from the top to the bottom.

6. A guitar string vibrates with a frequency of 220 cycles per second. What is the period of its vibration, in seconds?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in seconds.

Show the answer

Period =1frequency=1220= \frac{1}{\text{frequency}} = \frac{1}{220} second.

  • Answer: 1220\frac{1}{220} ≈ 0.005 seconds

7. A pendulum's horizontal position, measured from a mark on the floor, swings smoothly between −12-12 cm and 1818 cm. Its position is a sinusoidal function of time. What are its amplitude and midline?

  1. Amplitude 15 cm, midline x=3x = 3
  2. Amplitude 30 cm, midline x=3x = 3
  3. Amplitude 15 cm, midline x=0x = 0
  4. Amplitude 18 cm, midline x=6x = 6
Show the answer

Amplitude 18−(−12)2=15\frac{18 - (-12)}{2} = 15 cm; midline x=18+(−12)2=3x = \frac{18 + (-12)}{2} = 3. Check: 3±153 \pm 15 gives 1818 and −12-12.

  • Correct: Amplitude 15 cm, midline x=3x = 3: Half the swing, and the average of the two extremes.
  • Amplitude 30 cm, midline x=3x = 3: 30 cm is the whole swing, twice the amplitude.
  • Amplitude 15 cm, midline x=0x = 0: The mark on the floor is not the center of the swing: the pendulum goes further right than left.
  • Amplitude 18 cm, midline x=6x = 6: 18 is the greatest position, not the amplitude, and 18−122=3\frac{18 - 12}{2} = 3, not 6.

Part 9 · Summary

Summary

A sinusoidal function is a sine or cosine graph after dilations and translations. Four numbers describe it: the midline y=max⁡+min⁡2y = \frac{\max + \min}{2}, the amplitude max⁡−min⁡2\frac{\max - \min}{2}, the period (the length of one cycle) and the frequency, one over the period. Sine is odd and cosine is even.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic