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Module 2 · L14 of 15 ~40 min ⚡ +100 XP available

Modelling with Trigonometric Functions

Tides rise and fall. Temperatures peak in summer and dip in winter. Sound waves travel through air. All of these phenomena can be modelled by sinusoidal functions. In this lesson, you will learn how to extract real-world data, build trigonometric models, and use them to make predictions.

Today's hook — The height of the tide at a particular beach varies between a low of 1.2 metres and a high of 3.6 metres. If the tide follows a roughly sinusoidal pattern, what do you think the average tide height is? And what is the maximum distance the tide deviates from this average?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

Quick warm-up — the height of the tide at a particular beach varies between a low of 1.2 m and a high of 3.6 m. What is the average tide height? What is the maximum deviation from this average?

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02
The two moves
+5 XP to read

There are only two moves to building any trigonometric model from real-world data: extract the amplitude and midline from max and min, then find $b$ from the period. Everything else is choosing the right starting function.

Every periodic model follows the same general form: $y = a\sin(b(x-c)) + d$. The parameter $a = \dfrac{\text{max}-\text{min}}{2}$ is the amplitude, and $d = \dfrac{\text{max}+\text{min}}{2}$ is the midline. The coefficient $b = \dfrac{2\pi}{P}$ comes from the period $P$.

midline d t
$y = a\sin(b(x-c)) + d$
general form
Amplitude from data
$a = \dfrac{\text{max} - \text{min}}{2}$. This is half the total range, not the full range. The amplitude measures maximum deviation from the centre line.
Midline = average
$d = \dfrac{\text{max} + \text{min}}{2}$. The midline is exactly the average value. It represents the centre around which the phenomenon oscillates.
Period to $b$
$b = \dfrac{2\pi}{P}$ where $P$ is the period. A longer period means a smaller $b$, which stretches the graph horizontally.
03
What you will master
Know

Key facts

  • How to calculate amplitude and midline from maximum and minimum values
  • How to determine the period from real-world cycles
  • How to write a trigonometric model from given data
Understand

Concepts

  • Why periodic phenomena are naturally modelled by trig functions
  • How the parameters $a$, $b$, $c$, $d$ correspond to physical quantities
  • The limitations of simple sinusoidal models
Can do

Skills

  • Build a trig model from maximum, minimum, and period data
  • Use the model to predict future values
  • Interpret the meaning of each parameter in context
04
Key terms
Trigonometric RatioThe ratio of sides in a right-angled triangle (sin, cos, tan).
RadianA unit of angle measure where one radian subtends an arc equal to the radius.
Sine RuleA formula relating sides and angles in any triangle: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$.
Cosine RuleA formula for finding sides or angles: $c^2 = a^2 + b^2 - 2ab\cos C$.
PeriodThe length of one complete cycle of a periodic function.
AmplitudeThe maximum displacement from the centre line of a periodic function.
05
Building trigonometric models
core concept

Many real-world phenomena repeat in regular cycles. If you can identify the maximum value, minimum value, and period of the cycle, you can build a sinusoidal model.

Step 1: Find the amplitude and midline

Given the maximum and minimum values of the phenomenon:

$$a = \frac{\text{max} - \text{min}}{2}, \quad d = \frac{\text{max} + \text{min}}{2}$$

Step 2: Find $b$ from the period

If the phenomenon repeats every $P$ units of time (or angle), then:

$$b = \frac{2\pi}{P}$$

Step 3: Determine the phase shift

The phase shift $c$ depends on when the cycle starts. If the model uses sine, the cycle normally starts at the midline going up. If it starts at a maximum, cosine might be more natural. You can always convert between sine and cosine using phase shifts.

1m 2m 3m 4m 0h 4h 8h 12h 16h 20h 24h Time (hours) Height (m) midline d=2.4 h = 2.4 − 1.2 sin(πt/6)

Tide height modelled by a sinusoidal function with period 12 hours, amplitude 1.2 m, and midline 2.4 m

Tides in action. In many coastal locations, tides follow a roughly sinusoidal pattern with a period of about 12.4 hours (semidiurnal tide). The amplitude varies with the lunar cycle — during spring tides, the amplitude is largest; during neap tides, it is smallest. Trigonometric models help harbour masters schedule ship arrivals and fishermen plan their trips.

General form: $y = a\sin(b(x-c)) + d$; Amplitude: $a = \dfrac{\text{max} - \text{min}}{2}$ — half the total range

Pause — copy the four-step modelling process ($a$, $b$, $c$, $d$ from max/min data) and the amplitude formula $a = \frac{\text{max} - \text{min}}{2}$ into your book.

Did you get this? True or false: the amplitude of a sinusoidal model is calculated as $\text{max} - \text{min}$ (the full range).

PROBLEM 1 · TIDE HEIGHT MODEL

The tide height at a harbour varies between a low of 1.2 m and a high of 3.6 m, with a period of 12 hours. Write a model for the tide height $h$ (in metres) $t$ hours after low tide, using a sine function.

1
Find amplitude and midline
$a = \dfrac{3.6 - 1.2}{2} = 1.2$, $\quad d = \dfrac{3.6 + 1.2}{2} = 2.4$
PROBLEM 2 · TEMPERATURE MODEL

The average monthly temperature in a city varies from a minimum of $8^\circ$C in July to a maximum of $24^\circ$C in January. The cycle repeats annually. Model the temperature $T$ as a function of time $t$ in months, with $t = 0$ representing January, using a cosine function.

1
Find amplitude and midline
$a = \dfrac{24 - 8}{2} = 8$, $\quad d = \dfrac{24 + 8}{2} = 16$
PROBLEM 3 · USING THE MODEL TO PREDICT

Using the temperature model $T = 8\cos\!\left(\dfrac{\pi}{6}t\right) + 16$, find the temperature in April ($t = 3$).

1
Substitute $t = 3$
$T = 8\cos\!\left(\dfrac{\pi}{6} \times 3\right) + 16 = 8\cos\!\left(\dfrac{\pi}{2}\right) + 16$

Quick check: A ferris wheel has a minimum height of 2 m and a maximum height of 18 m. What is the amplitude of its height model?

Trap 01
Using $a = \text{max} - \text{min}$ instead of half
The amplitude is half the distance between the maximum and minimum, not the full distance. Fix: Amplitude = $\dfrac{\text{max} - \text{min}}{2}$.
Trap 02
Forgetting to convert the period to $b$ correctly
Students sometimes use $b = P$ or $b = \dfrac{P}{2\pi}$ instead of $b = \dfrac{2\pi}{P}$. Fix: $b = \dfrac{2\pi}{P}$. The larger the period, the smaller $b$ is.
Trap 03
Not choosing the right starting function (sine vs cosine)
If the data starts at a maximum, cosine is usually easier. If it starts at the midline going up, sine is more natural. Using the wrong one leads to unnecessary phase shift complications. Fix: Match the function to the starting point of your data.

Fill in the blank: If a sinusoidal phenomenon has a period of 24 hours, then $b = \dfrac{2\pi}{\square}$ = (leave in terms of $\pi$).

Work mode · how are you completing this lesson?
1

A ferris wheel has a minimum height of 5 m and a maximum height of 25 m. It completes one revolution every 4 minutes. Let $h$ be the height $t$ minutes after reaching the minimum height. Write a model.

2

The number of daylight hours in a town varies from a minimum of 9 hours in June to a maximum of 15 hours in December, repeating annually. Let $D$ be the number of daylight hours $t$ months after December. Write a model.

3

A pendulum swings so that its angle from the vertical varies from $-15^\circ$ to $+15^\circ$, completing one full swing every 2 seconds. Let $\theta$ be the angle $t$ seconds after passing through the vertical in the positive direction. Write a model.

4

Using your ferris wheel model, find the height at $t = 2$ minutes.

5

Using your daylight model, find the number of hours at $t = 6$ months (June).

Odd one out: Three of these are correct steps in building a trig model. Which one is incorrect?

Match each parameter to its meaning in $y = a\sin(b(x-c)) + d$:

$a$
$d$
$b$
$c$
Amplitude (max deviation from midline)
Phase shift (horizontal translation)
Controls the period: $P = 2\pi/b$
Midline (vertical shift)
12
Revisit your thinking

Earlier you were asked about tide heights of 1.2 m and 3.6 m.

The average tide height (midline) is $\dfrac{1.2 + 3.6}{2} = 2.4$ metres. The maximum deviation from this average (amplitude) is $\dfrac{3.6 - 1.2}{2} = 1.2$ metres. These two numbers, $a = 1.2$ and $d = 2.4$, are the keys to building the tide model.

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 44 marks

Q1. A population of rabbits varies sinusoidally between a minimum of 800 in March and a maximum of 2400 in September, repeating annually. Let $P$ be the population $t$ months after January. Write a model for $P$ in terms of $t$. (4 marks)

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ApplyBand 44 marks

Q2. The depth of water in a harbour is modelled by $D = 3\sin\!\left(\dfrac{\pi}{6}t\right) + 5$, where $D$ is in metres and $t$ is hours after midnight. (a) Find the maximum and minimum depths. (b) Find the depth at 9:00 am. (4 marks)

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AnalyseBand 53 marks

Q3. A student claims that the tide model $h = 2 + \sin(30t)$ (where $h$ is in metres and $t$ is in hours) has a period of 12 hours. Verify this claim and find the maximum and minimum tide heights predicted by the model. (3 marks)

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📖 Comprehensive answers (click to reveal)

Drill 1: $a = 10$, $d = 15$, $b = \dfrac{\pi}{2}$. Minimum at $t = 0$, so $h = 15 - 10\sin\!\left(\dfrac{\pi}{2}t\right)$.

Drill 2: $a = 3$, $d = 12$, $b = \dfrac{\pi}{6}$. December ($t = 0$) is maximum, so $D = 3\cos\!\left(\dfrac{\pi}{6}t\right) + 12$.

Drill 3: $a = 15$, $d = 0$, $b = \pi$. Passes through vertical at $t = 0$ going positive, so $\theta = 15\sin(\pi t)$.

Drill 4: $h = 15 - 10\sin\!\left(\dfrac{\pi}{2} \times 2\right) = 15 - 10\sin(\pi) = 15$ m.

Drill 5: $D = 3\cos\!\left(\dfrac{\pi}{6} \times 6\right) + 12 = 3\cos(\pi) + 12 = 9$ hours.

Q1 (4 marks): $a = \dfrac{2400 - 800}{2} = 800$ [0.5], $d = \dfrac{2400 + 800}{2} = 1600$ [0.5]. Period = 12 months, so $b = \dfrac{\pi}{6}$ [1]. March ($t = 2$) is minimum; $P = 1600 - 800\cos\!\left(\dfrac{\pi}{6}(t - 2)\right)$ or equivalent [2].

Q2 (4 marks): (a) Max = $3 + 5 = 8$ m, Min = $-3 + 5 = 2$ m [2]. (b) At $t = 9$: $D = 3\sin\!\left(\dfrac{3\pi}{2}\right) + 5 = -3 + 5 = 2$ m [2].

Q3 (3 marks): Period = $\dfrac{2\pi}{30} = \dfrac{\pi}{15}$ hours $\approx 12.57$ min [1]. The student is incorrect — the period is $\dfrac{\pi}{15}$ hours, not 12 hours [1]. Max = 3 m, Min = 1 m [1].

01
Boss battle · Modelling with Trigonometric Functions
earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering trigonometric modelling questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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