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D3.3 · Model periodic phenomena that do not involve angles

Learn to model periodic phenomena that do not involve angles through clear examples and targeted practice.

Ontario Grade 11 Mathematics

Trigonometric Functions

Using Sinusoidal Functions to Represent Real-World Cycles

Many things in the natural world repeat on a regular schedule — the rise and fall of ocean tides, the hours of daylight through the year, a person's body temperature across a 24-hour day, and the up-and-down motion of a point on a spinning wheel. These repeating patterns are called periodic phenomena. In this lesson you will learn how to take data from one of these real-world situations and write a sinusoidal equation that models it — all without ever measuring an angle directly. The equation you build becomes a tool: once you have it, you can predict values at any point in the cycle.

What you will learn

Prerequisite Bridge: Key Features of Sinusoidal Functions

Before building a model, you need to be comfortable reading the key features of a sinusoidal graph. Recall from earlier in this course that the sine and cosine functions produce smooth, wave-shaped graphs that oscillate above and below a central horizontal line.
The midline (also called the vertical shift, dd) is the horizontal line that runs exactly halfway between the highest and lowest points of the wave. Its value is found using d=max+min2d = \frac{\text{max} + \text{min}}{2}.
The amplitude (aa) measures how far the wave reaches above (or below) the midline. It is always a positive number, found using a=max−min2a = \frac{\text{max} - \text{min}}{2}.
The period (TT) is the length of one complete cycle — the horizontal distance before the pattern repeats exactly. In an equation of the form y=asin⁡(bx)+dy = a\sin(bx) + d, the period and the coefficient bb are related by T=360∘bT = \frac{360^\circ}{b}, so b=360∘Tb = \frac{360^\circ}{T}.
The phase shift (cc) moves the entire wave left or right. In y=asin⁡(b(x−c))+dy = a\sin(b(x - c)) + d, a positive cc shifts the graph to the right by cc units. Keeping these four features in mind — amplitude, period, midline, and phase shift — is the complete toolkit for building a sinusoidal model.
d=max+min2,a=max−min2,b=360∘Td = \frac{\text{max}+\text{min}}{2}, \quad a = \frac{\text{max}-\text{min}}{2}, \quad b = \frac{360^\circ}{T}

What Makes a Phenomenon Periodic (and Sinusoidal)?

A phenomenon is periodic if it repeats the same pattern over equal intervals of time (or distance, or any other input). Not every repeating pattern is sinusoidal — a bouncing ball loses energy and its bounces get shorter, so it is not a true periodic function. A sinusoidal model is suitable when the data rises and falls smoothly and symmetrically, like a wave.
To decide whether a sinusoidal model is reasonable, ask three questions: Does the quantity oscillate between a clear maximum and a clear minimum? Is the time between peaks roughly equal every cycle? Is the rise and fall roughly smooth rather than sudden? If the answer to all three is yes, a sine or cosine function is likely a good fit.
Typical examples in this course include: hours of daylight in a city over a calendar year, the height of a rider on a Ferris wheel over time, water depth at a harbour over a tidal cycle, and average monthly temperature in a Canadian city. In every case, the horizontal axis represents time (or another non-angle quantity) and the vertical axis represents the measured value.

Building the Model Equation Step by Step

Once you have confirmed that data is sinusoidal, follow a consistent four-step process to write the equation. The standard form is y=acos⁡(b(x−c))+dy = a\cos(b(x - c)) + d or y=asin⁡(b(x−c))+dy = a\sin(b(x - c)) + d. You choose sine or cosine based on which one matches the starting behaviour of your data more easily.
Step 1 — Find the midline dd: Locate the maximum and minimum values in the data and apply d=max+min2d = \frac{\text{max} + \text{min}}{2}. This gives the vertical centre of the wave.
Step 2 — Find the amplitude aa: Apply a=max−min2a = \frac{\text{max} - \text{min}}{2}. The amplitude is always positive. If the data starts by going downward from the midline, write −a-a in the equation.
Step 3 — Find bb from the period: Estimate the period TT from the data (the time from one peak to the next, or from one complete cycle). Then calculate b=360∘Tb = \frac{360^\circ}{T}. The unit of bb will be degrees per unit of xx.
Step 4 — Find the phase shift cc: Choose where a standard sine (or cosine) curve would start its cycle and compare that to where your data starts its cycle. The horizontal distance you need to shift is cc. Substitute your values of aa, bb, cc, and dd into the standard form, then verify by checking that the equation gives the correct maximum, minimum, and a midline-crossing point.
y=asin⁡(b(x−c))+dy = a\sin(b(x - c)) + d

Using the Model to Make Predictions

Once you have a sinusoidal equation, you can substitute any input value of xx to predict the output. This is the main reason for building the model — it lets you estimate values that were not measured directly.
For example, if your model gives the average daily temperature TT (in °C) as a function of the day number nn in the year, you can substitute n=200n = 200 to estimate the temperature on the 200th day of the year. You evaluate the equation just like any other function: substitute, simplify, and report the answer with appropriate units.
Remember that all sinusoidal models are approximations. Real data rarely fits a perfect sine curve, so treat predictions as estimates. You should also check whether your predicted value is reasonable — it should lie between the maximum and minimum of the phenomenon.

Choosing Between Sine and Cosine — and Checking Your Work

Both y=asin⁡(b(x−c))+dy = a\sin(b(x-c)) + d and y=acos⁡(b(x−c))+dy = a\cos(b(x-c)) + d can describe the same wave; they differ only in the phase shift used. Use cosine when the data starts at a maximum (or minimum) value at x=0x = 0, because the cosine curve starts at its peak. Use sine when the data crosses the midline going upward at x=0x = 0.
If neither case fits neatly, choose either function and adjust the phase shift cc accordingly. It is often easiest to identify the first maximum value in the data, then use a cosine model with the phase shift equal to the xx-value of that maximum.
After writing the equation, always run two checks. First, substitute the xx-value of the known maximum and confirm the equation gives the maximum yy-value. Second, substitute the xx-value of the known minimum and confirm the equation gives the minimum yy-value. If both checks pass, your model is correct.
y=acos⁡(b(x−c))+dy = a\cos(b(x - c)) + d

Summary of the Four Model Parameters

ParameterSymbolHow to Calculate ItWhat It Does to the Graph
Amplitudeaamax−min2\frac{\text{max} - \text{min}}{2}Sets the height of the wave above and below the midline
Vertical shift (midline)ddmax+min2\frac{\text{max} + \text{min}}{2}Moves the entire wave up or down
Period coefficientbb360∘T\frac{360^\circ}{T}Controls how quickly the wave completes one cycle
Phase shiftccxx-value of first peak (cosine) or first upward midline crossing (sine)Slides the wave left or right

Worked example

Example 1 — Hours of Daylight in Toronto

The approximate number of hours of daylight in Toronto reaches a maximum of 15.4 hours in June (month 6) and a minimum of 9.0 hours in December (month 12). The pattern repeats every 12 months. Write a sinusoidal equation to model the number of daylight hours hh as a function of the month number mm, then use it to predict the hours of daylight in month 9 (September).
  1. Identify the maximum, minimum, and period
    The problem states the maximum is 15.415.4 hours and the minimum is 9.09.0 hours. The cycle repeats every 12 months, so the period is T=12T = 12 months.
    max=15.4,min=9.0,T=12\text{max} = 15.4, \quad \text{min} = 9.0, \quad T = 12
  2. Calculate the midline d
    Add the maximum and minimum, then divide by 2. This gives the horizontal centre of the wave.
    d=15.4+9.02=24.42=12.2d = \frac{15.4 + 9.0}{2} = \frac{24.4}{2} = 12.2
  3. Calculate the amplitude a
    Subtract the minimum from the maximum, then divide by 2. The amplitude tells you how far the wave reaches above or below the midline.
    a=15.4−9.02=6.42=3.2a = \frac{15.4 - 9.0}{2} = \frac{6.4}{2} = 3.2
  4. Calculate b from the period
    Use the relationship b=360∘Tb = \frac{360^\circ}{T} with T=12T = 12. Dividing 360∘360^\circ by 12 gives the number of degrees the input variable advances per month.
    b=360∘12=30∘b = \frac{360^\circ}{12} = 30^\circ
  5. Determine the phase shift c
    The maximum occurs at month 6. For a cosine model, the standard cosine curve peaks at x=0x = 0. To shift the peak to m=6m = 6, set c=6c = 6. This moves the graph 6 units to the right.
    c=6c = 6
  6. Write the full equation
    Substitute a=3.2a = 3.2, b=30∘b = 30^\circ, c=6c = 6, and d=12.2d = 12.2 into the cosine model h=acos⁡(b(m−c))+dh = a\cos(b(m - c)) + d.
    h=3.2cos⁡(30∘(m−6))+12.2h = 3.2\cos(30^\circ(m - 6)) + 12.2
  7. Verify with the maximum and minimum
    At m=6m = 6: the angle inside the cosine is 30∘(6−6)=0∘30^\circ(6 - 6) = 0^\circ, so h=3.2cos⁡(0∘)+12.2=3.2(1)+12.2=15.4h = 3.2\cos(0^\circ) + 12.2 = 3.2(1) + 12.2 = 15.4 ✓. At m=12m = 12: the angle is 30∘(12−6)=180∘30^\circ(12 - 6) = 180^\circ, so h=3.2cos⁡(180∘)+12.2=3.2(−1)+12.2=9.0h = 3.2\cos(180^\circ) + 12.2 = 3.2(-1) + 12.2 = 9.0 ✓. Both known values are reproduced correctly.
    h(6)=15.4,h(12)=9.0h(6) = 15.4, \quad h(12) = 9.0
  8. Predict daylight hours for month 9
    Substitute m=9m = 9 into the equation. First compute the angle: 30∘(9−6)=30∘×3=90∘30^\circ(9 - 6) = 30^\circ \times 3 = 90^\circ. Since cos⁡(90∘)=0\cos(90^\circ) = 0, the calculation gives h=3.2(0)+12.2=12.2h = 3.2(0) + 12.2 = 12.2 hours.
    h=3.2cos⁡(90∘)+12.2=0+12.2=12.2h = 3.2\cos(90^\circ) + 12.2 = 0 + 12.2 = 12.2
Answer: The equation is h=3.2cos⁡(30∘(m−6))+12.2h = 3.2\cos(30^\circ(m - 6)) + 12.2. In September (month 9) there are approximately 12.2 hours of daylight.
Check: The predicted value of 12.2 hours lies exactly on the midline, which makes sense because month 9 is three months past the peak (month 6) — exactly one quarter of the 12-month cycle. At that point the cosine function equals zero, so the output equals the midline value. This is consistent with the known fact that around the September equinox, day and night are nearly equal in length.

Worked example

Example 2 — Water Depth at a Harbour

At a small harbour, the water depth (in metres) follows a tidal cycle. On a particular day, the depth reaches a maximum of 8.6 m at 2:00 a.m. and a minimum of 2.0 m at 8:00 a.m. The tidal cycle repeats every 12 hours. Let tt represent the number of hours after midnight. Write a sinusoidal equation for the water depth WW as a function of tt, then predict the water depth at 5:00 a.m. (t=5t = 5).
  1. Identify the maximum, minimum, and period
    The maximum depth is 8.6 m at t=2t = 2 (2:00 a.m.) and the minimum depth is 2.0 m at t=8t = 8 (8:00 a.m.). The tidal cycle repeats every 12 hours, so T=12T = 12.
    max=8.6 m,min=2.0 m,T=12\text{max} = 8.6\text{ m}, \quad \text{min} = 2.0\text{ m}, \quad T = 12
  2. Calculate the midline d
    The midline is the average of the maximum and minimum depths.
    d=8.6+2.02=10.62=5.3d = \frac{8.6 + 2.0}{2} = \frac{10.6}{2} = 5.3
  3. Calculate the amplitude a
    The amplitude is half the total range of the wave, telling you how far the depth rises above or drops below the midline.
    a=8.6−2.02=6.62=3.3a = \frac{8.6 - 2.0}{2} = \frac{6.6}{2} = 3.3
  4. Calculate b from the period
    Use the relationship b=360∘Tb = \frac{360^\circ}{T} with T=12T = 12 hours. This gives the number of degrees the input advances per hour.
    b=360∘12=30∘b = \frac{360^\circ}{12} = 30^\circ
  5. Determine the phase shift c
    The first maximum occurs at t=2t = 2. Using a cosine model, the standard cosine peaks at x=0x = 0. Shifting the peak to t=2t = 2 means c=2c = 2.
    c=2c = 2
  6. Write the full equation
    Substitute a=3.3a = 3.3, b=30∘b = 30^\circ, c=2c = 2, and d=5.3d = 5.3 into W=acos⁡(b(t−c))+dW = a\cos(b(t - c)) + d.
    W=3.3cos⁡(30∘(t−2))+5.3W = 3.3\cos(30^\circ(t - 2)) + 5.3
  7. Verify with the maximum and minimum
    At t=2t = 2: the angle is 30∘(2−2)=0∘30^\circ(2 - 2) = 0^\circ, so W=3.3cos⁡(0∘)+5.3=3.3(1)+5.3=8.6W = 3.3\cos(0^\circ) + 5.3 = 3.3(1) + 5.3 = 8.6 ✓. At t=8t = 8: the angle is 30∘(8−2)=180∘30^\circ(8 - 2) = 180^\circ, so W=3.3cos⁡(180∘)+5.3=3.3(−1)+5.3=2.0W = 3.3\cos(180^\circ) + 5.3 = 3.3(-1) + 5.3 = 2.0 ✓. Both known values are reproduced correctly.
    W(2)=8.6,W(8)=2.0W(2) = 8.6, \quad W(8) = 2.0
  8. Predict depth at t = 5
    Substitute t=5t = 5. The angle inside cosine is 30∘(5−2)=30∘×3=90∘30^\circ(5 - 2) = 30^\circ \times 3 = 90^\circ. Since cos⁡(90∘)=0\cos(90^\circ) = 0, the depth is W=3.3(0)+5.3=5.3W = 3.3(0) + 5.3 = 5.3 m.
    W=3.3cos⁡(90∘)+5.3=0+5.3=5.3W = 3.3\cos(90^\circ) + 5.3 = 0 + 5.3 = 5.3
Answer: The equation is W=3.3cos⁡(30∘(t−2))+5.3W = 3.3\cos(30^\circ(t - 2)) + 5.3. At 5:00 a.m. the water depth is approximately 5.3 m.
Check: At t=5t = 5 the tide is exactly halfway through its fall from peak (at t=2t = 2) to trough (at t=8t = 8). That midpoint in time corresponds to the midline of the wave, so the output should equal d=5.3d = 5.3 m. The calculation confirms this, which means the equation and the arithmetic are both correct.

Common mistakes and how to avoid them

Using the period TT directly as bb in the equation instead of computing b=360∘Tb = \frac{360^\circ}{T}.
Correction: Always convert the period to the coefficient bb using b=360∘Tb = \frac{360^\circ}{T} before writing the equation.
Writing cos⁡(bx−c)\cos(bx - c) instead of cos⁡(b(x−c))\cos(b(x - c)), which changes the effective phase shift from cc to cb\frac{c}{b}.
Correction: Keep the phase shift inside brackets with bb factored out: cos⁡(b(x−c))\cos(b(x - c)). If you expand it, the constant subtracted inside must be bcbc, not cc.
Computing a negative amplitude by writing a=min−maxa = \text{min} - \text{max} instead of a=max−mina = \text{max} - \text{min}.
Correction: Amplitude is always positive: a=max−min2a = \frac{\text{max} - \text{min}}{2}. If the wave starts by going downward, place the negative sign in front of aa in the equation, not inside the value of aa itself.
Skipping the verification step and not checking whether the equation reproduces the known maximum and minimum values.
Correction: Always substitute the xx-values of the maximum and minimum back into the completed equation and confirm the outputs match the given data.
Reading the phase shift directly from the data without first deciding whether the reference point is the peak (cosine) or the upward midline crossing (sine).
Correction: Decide which function you are using first. For cosine, use the xx-value of the first peak as cc. For sine, use the xx-value of the first upward midline crossing as cc.

Lesson summary

Check your understanding

Question 1

A sinusoidal phenomenon has a maximum value of 20 and a minimum value of 4. What are the amplitude and midline?
  1. Amplitude = 20, midline = 4
  2. Amplitude = 8, midline = 12
  3. Amplitude = 16, midline = 12
  4. Amplitude = 8, midline = 4
Show answer and explanation
Amplitude = 8, midline = 12
Midline d=20+42=12d = \frac{20 + 4}{2} = 12 and amplitude a=20−42=8a = \frac{20 - 4}{2} = 8. Option A confuses the maximum with the amplitude. Option C uses the full range (16) instead of half the range. Option D has the right amplitude but the wrong midline.

Question 2

A tidal pattern repeats every 6 hours. What is the value of bb in the model equation?
  1. b=6∘b = 6^\circ
  2. b=60∘b = 60^\circ
  3. b=30∘b = 30^\circ
  4. b=720∘b = 720^\circ
Show answer and explanation
b=60∘b = 60^\circ
b=360∘T=360∘6=60∘b = \frac{360^\circ}{T} = \frac{360^\circ}{6} = 60^\circ per hour. Choosing b=6b = 6 confuses the period with the coefficient. Choosing b=30∘b = 30^\circ would be correct for a 12-hour period. Choosing b=720∘b = 720^\circ inverts the fraction.

Question 3

A cosine model peaks at t=3t = 3. The midline is d=10d = 10. Which equation correctly places the phase shift?
  1. y=acos⁡(bt)+10y = a\cos(bt) + 10
  2. y=acos⁡(b(t+3))+10y = a\cos(b(t + 3)) + 10
  3. y=acos⁡(b(t−3))+10y = a\cos(b(t - 3)) + 10
  4. y=acos⁡(bt−10)+3y = a\cos(bt - 10) + 3
Show answer and explanation
y=acos⁡(b(t−3))+10y = a\cos(b(t - 3)) + 10
For a cosine model, the phase shift cc equals the tt-value of the first peak, which is 3. This is written as (t−3)(t - 3) inside the function, shifting the peak rightward to t=3t = 3. Option B shifts the peak to the left instead. Option A has no phase shift, placing the peak at t=0t = 0. Option D incorrectly swaps the roles of dd and cc.

Question 4

Using the model h=3.2cos⁡(30∘(m−6))+12.2h = 3.2\cos(30^\circ(m - 6)) + 12.2, what is hh when m=12m = 12?
  1. h=15.4h = 15.4
  2. h=12.2h = 12.2
  3. h=9.0h = 9.0
  4. h=3.2h = 3.2
Show answer and explanation
h=9.0h = 9.0
At m=12m = 12 the angle inside cosine is 30∘(12−6)=180∘30^\circ(12 - 6) = 180^\circ. Since cos⁡(180∘)=−1\cos(180^\circ) = -1, we get h=3.2(−1)+12.2=−3.2+12.2=9.0h = 3.2(-1) + 12.2 = -3.2 + 12.2 = 9.0. This is the minimum value, consistent with December having the fewest daylight hours.

Key terms

Periodic phenomenon
A real-world event or measurement that repeats the same pattern over equal intervals of the input variable (such as time).
Sinusoidal function
A function whose graph has the smooth, symmetric wave shape of a sine or cosine curve.
Amplitude
The distance from the midline to the maximum (or minimum) of a sinusoidal wave; always a positive number.
Midline
The horizontal line exactly halfway between the maximum and minimum of a sinusoidal wave; its equation is y=dy = d.
Period
The horizontal length of one complete cycle of a periodic function — the smallest value TT such that the pattern repeats.
Phase shift
A horizontal translation of the sinusoidal graph; in y=acos⁡(b(x−c))+dy = a\cos(b(x - c)) + d, the graph shifts cc units to the right when c>0c > 0.
Period coefficient (b)
The value bb in a sinusoidal equation that controls the period; calculated as b=360∘Tb = \frac{360^\circ}{T}.
Sinusoidal model
A sine or cosine equation fitted to real-world periodic data so that it can be used to describe and predict the behaviour of that phenomenon.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Mathematics (MCR3U), expectation D3.3. It is a study resource, not an official curriculum publication.

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