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.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
Wrong: (a + b)² = a² + b².
Right: (a + b)² = a² + 2ab + b²; the middle term 2ab is essential and commonly forgotten.
📚 Core Content
Many real-world phenomena repeat in regular cycles. If we can identify the maximum value, minimum value, and period of the cycle, we can build a sinusoidal model.
Given the maximum and minimum values of the phenomenon:
$$a = \frac{\text{max} - \text{min}}{2}, \quad d = \frac{\text{max} + \text{min}}{2}$$
If the phenomenon repeats every $P$ units of time (or angle), then:
$$b = \frac{2\pi}{P}$$
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.
🧮 Worked Examples
🧪 Activities
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.
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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.
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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.
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1 For the ferris wheel model, what is the height after 2 minutes?
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2 For the daylight model, how many daylight hours are there in June ($t = 6$)?
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Earlier you were asked about the tide heights of 1.2 m and 3.6 m.
The average tide height (midline) is $\frac{1.2 + 3.6}{2} = 2.4$ metres. The maximum deviation from this average (amplitude) is $\frac{3.6 - 1.2}{2} = 1.2$ metres. These two numbers, $a = 1.2$ and $d = 2.4$, are the keys to building the tide model.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
8. 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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9. The depth of water in a harbour is modelled by $D = 3\sin\left(\frac{\pi}{6}t\right) + 5$, 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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10. 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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Answer in your workbook.
1. $a = 10$, $d = 15$, $b = \frac{\pi}{2}$. Minimum at $t = 0$, so $h = 15 - 10\sin\left(\frac{\pi}{2}t\right)$ or $h = 15 + 10\cos\left(\frac{\pi}{2}(t - 1)\right)$.
2. $a = 3$, $d = 12$, $b = \frac{\pi}{6}$. December ($t = 0$) is maximum, so $D = 3\cos\left(\frac{\pi}{6}t\right) + 12$.
3. $a = 15$, $d = 0$, $b = \pi$. Passes through vertical at $t = 0$ going positive, so $\theta = 15\sin(\pi t)$.
1. $h = 15 - 10\sin\left(\frac{\pi}{2} \times 2\right) = 15 - 10\sin(\pi) = 15$ m.
2. $D = 3\cos\left(\frac{\pi}{6} \times 6\right) + 12 = 3\cos(\pi) + 12 = 9$ hours.
1. A — $a = \frac{20 - 8}{2} = 6$.
2. A — $d = \frac{20 + 8}{2} = 14$.
3. A — Period 12 months gives $b = \frac{\pi}{6}$.
4. A — $h = 10 + 4\sin\left(\frac{\pi}{6}t\right)$ has max 14 m, min 6 m.
5. A — $T = 5\cos\left(\frac{\pi}{6}t\right) + 15$; at $t = 3$, $T = 15^\circ$C.
Q8 (4 marks): $a = \frac{2400 - 800}{2} = 800$ [0.5], $d = \frac{2400 + 800}{2} = 1600$ [0.5]. Period = 12 months, so $b = \frac{\pi}{6}$ [1]. March ($t = 2$) is minimum; using cosine shifted: $P = 1600 - 800\cos\left(\frac{\pi}{6}(t - 2)\right)$ or equivalent [2].
Q9 (4 marks): (a) Max = $3 + 5 = 8$ m, Min = $-3 + 5 = 2$ m [2]. (b) At $t = 9$: $D = 3\sin\left(\frac{3\pi}{2}\right) + 5 = -3 + 5 = 2$ m [2].
Q10 (3 marks): Period = $\frac{2\pi}{30} = \frac{\pi}{15}$ hours $\approx 12.57$ min [1]. The student is incorrect; the period is $\frac{\pi}{15}$ hours, not 12 hours [1]. Max = $2 + 1 = 3$ m, Min = $2 - 1 = 1$ m [1].
Modelling with Trigonometric Functions
Tick when you've finished all activities and checked your answers.