Mathematics Advanced • Year 11 • Module 2 • Lesson 14

Modelling with Trigonometric Functions

Build fluency in extracting $a$, $b$, $c$, $d$ from "max / min / period / starting condition" data and writing the corresponding sinusoidal model.

Build · Skill Drill

1. Quick recall

Three warm-ups. 1 mark each

Q1.1 Complete the modelling formulas:

Amplitude: $a = $ _______________________ (in terms of max and min).

Midline: $d = $ _______________________ (in terms of max and min).

Q1.2 If the cycle repeats every $P$ units of time, then $b = $ _______________________.

(Hint: rearrange Period $= 2\pi/|b|$.)

Q1.3 Choosing sine vs cosine. In one sentence each:

Use cosine when the data starts at ________________________________________.

Use sine when the data starts at ________________________________________.

Stuck? Revisit lesson § The two moves — and § Trap 03 on choosing the right starting function.

2. Worked example — tide-height model

Problem. The tide at a harbour ranges between a low of $1.2\,\text{m}$ and a high of $3.6\,\text{m}$, with period $12\,\text{h}$. Build a sinusoidal model for tide height $h$ (in metres) $t$ hours after low tide.

Step 1 — Amplitude and midline.

$a = (3.6 - 1.2)/2 = 1.2\,\text{m}$.

$d = (3.6 + 1.2)/2 = 2.4\,\text{m}$.

Step 2 — $b$ from period.

$b = 2\pi/12 = \pi/6\,\text{rad/h}$.

Step 3 — Choose function and phase.

At $t = 0$ we want the minimum. A negative-amplitude sine ($-a\sin$) starts at the midline going down, reaching its minimum at $T/4$ — not what we want. A negative cosine ($-a\cos$) starts at its minimum at $t = 0$ exactly — perfect.

Use $h(t) = -a\cos(bt) + d = -1.2\cos(\pi t/6) + 2.4$.

Step 4 — Verify at $t = 0$ and $t = 6$.

$h(0) = -1.2(1) + 2.4 = 1.2$ ✓ (low). $h(6) = -1.2(-1) + 2.4 = 3.6$ ✓ (high).

Conclusion. $h(t) = $ $-1.2\cos(\pi t/6) + 2.4$ (or equivalently $2.4 - 1.2\cos(\pi t/6)$).

3. Faded example — fill in the missing steps

The temperature in a city varies between $8^\circ$C (in July) and $24^\circ$C (in January), with annual period. Build a cosine model for $T$ as a function of months $t$ since January. 5 marks

Step 1 — Amplitude and midline.

$a = (\text{max} - \text{min})/2 = (24 - 8)/2 = $ ________.

$d = (\text{max} + \text{min})/2 = (24 + 8)/2 = $ ________.

Step 2 — $b$ from period.

Period $P = $ ________ months $\Rightarrow b = 2\pi/P = $ ________.

Step 3 — Choose function and phase.

At $t = 0$ (January) the temperature is at its maximum, so cosine with no phase shift fits.

$T(t) = $ ________ $\cdot \cos($ ________ $\cdot t) + $ ________.

Step 4 — Verify.

$T(0) = $ ________ ($24^\circ$C, max). $T(6) = $ ________ ($8^\circ$C, min). ✓

Conclusion. $T(t) = $ ________________________________.

Stuck? Revisit lesson § Worked Example 2 (Temperature Model).

4. Graduated practice

For each, extract $a$, $b$, $c$ (if any), $d$ and write the model.

Foundation — extract parameters only (4 questions)

Q	Data	$a$	$d$
4.1 1	Max $= 10$, min $= 4$
4.2 1	Max $= 25$ m, min $= 5$ m
4.3 1	Period $4$ minutes $\Rightarrow b = $ ?
4.4 1	Period $24$ hours $\Rightarrow b = $ ?

Standard — build full models (6 questions)

4.5 A ferris wheel has min height $5\,\text{m}$, max height $25\,\text{m}$, period $4\,\text{min}$. The rider starts at the minimum at $t = 0$. Write $h(t)$. 3 marks

4.6 Daylight hours in a town vary from min $9\,\text{h}$ (June) to max $15\,\text{h}$ (December), annual period. Write $D(t)$ where $t$ is months after December. 3 marks

4.7 A piston in an engine moves between $-2\,\text{cm}$ and $+2\,\text{cm}$ from its central position, with period $1/50\,\text{s}$. At $t = 0$ the piston is at $+2\,\text{cm}$. Write $x(t)$. 3 marks

4.8 A sound wave has amplitude $0.5$ (pressure units), midline $0$, and period $1/440\,\text{s}$. At $t = 0$ the wave is on its midline rising. Write $p(t)$. 3 marks

4.9 Using the tide model $h(t) = -1.2\cos(\pi t/6) + 2.4$, find $h$ at $t = 3\,\text{h}$ (mid-period). 2 marks

4.10 Using the city temperature model $T(t) = 8\cos(\pi t/6) + 16$, find $T$ at $t = 3$ months (April). 2 marks

Extension — interpret limitations (2 questions)

4.11 A water wheel of radius $1.5\,\text{m}$ has its centre $0.5\,\text{m}$ below the water surface. A bucket's height above water (in metres) is given by $h(\theta) = 1.5\sin\theta - 0.5$ as the wheel rotates through angle $\theta$ (rad). Find the values of $\theta \in [0, 2\pi]$ where the bucket is submerged ($h \leq 0$). 3 marks

4.12 A simple sinusoidal model is used for monthly rainfall: $R(t) = 50 + 30\sin(\pi t/6)$ mm, $t$ in months from October. List two limitations of using a pure sine model for rainfall data. 3 marks

Stuck on 4.11? Set $1.5\sin\theta - 0.5 \leq 0$ and solve for $\theta$.

5. Self-check the easy 3

Tick the first three once you've verified each one.

For 4.1 I have $a = 3$, $d = 7$ (half of $(10 - 4)$, average of $(10 + 4)$).

For 4.3 my answer is $b = 2\pi/4 = \pi/2$.

For 4.4 my answer is $b = 2\pi/24 = \pi/12$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — Amplitude and midline formulas

$a = \dfrac{\text{max} - \text{min}}{2}$ (Trap 01 in the lesson: it's half, not the full distance). $d = \dfrac{\text{max} + \text{min}}{2}$ (the average).

Q1.2 — $b$ from period

$b = \dfrac{2\pi}{P}$ (Trap 02: not $P/(2\pi)$, not $P$, not $\pi/P$).

Q1.3 — Sine vs cosine

Use cosine when the data starts at a maximum (or minimum, with $-\cos$). Use sine when the data starts at the midline going up (or going down, with $-\sin$). Wrong choice forces a phase shift — Trap 03 in the lesson.

Q3 — Faded example: City temperature

$a = 8$, $d = 16$, $P = 12$ months $\Rightarrow b = 2\pi/12 = \pi/6$. $T(t) = 8\cos(\pi t/6) + 16$. Verify: $T(0) = 8(1) + 16 = 24$ ✓ (max). $T(6) = 8(-1) + 16 = 8$ ✓ (min). $T(t) = 8\cos(\pi t/6) + 16$.

Q4.1 — Max 10, min 4

$a = (10 - 4)/2 = $ $3$. $d = (10 + 4)/2 = $ $7$.

Q4.2 — Max 25 m, min 5 m

$a = (25 - 5)/2 = $ $10$ m. $d = (25 + 5)/2 = $ $15$ m.

Q4.3 — Period 4 min

$b = 2\pi/4 = $ $\pi/2$ rad/min.

Q4.4 — Period 24 h

$b = 2\pi/24 = $ $\pi/12$ rad/h.

Q4.5 — Ferris wheel (start at min)

$a = 10$, $d = 15$, $b = \pi/2$. Starts at min, so use $-\cos$: $h(t) = -10\cos(\pi t/2) + 15$. Verify: $h(0) = -10 + 15 = 5$ ✓, $h(2) = -10(-1) + 15 = 25$ ✓. $h(t) = 15 - 10\cos(\pi t/2)$.

Q4.6 — Daylight (start at max in December)

$a = 3$, $d = 12$, $b = 2\pi/12 = \pi/6$. Starts at max, use $\cos$: $D(t) = 3\cos(\pi t/6) + 12$.

Q4.7 — Piston (start at max)

$a = 2$, $d = 0$, $b = 2\pi/(1/50) = 100\pi$. Starts at max, use $\cos$: $x(t) = 2\cos(100\pi t)$.

Q4.8 — Sound wave (start at midline rising)

$a = 0.5$, $d = 0$, $b = 2\pi \times 440 = 880\pi$. Starts at midline going up, use $\sin$: $p(t) = 0.5\sin(880\pi t)$.

Q4.9 — Tide at $t = 3$ h

$h(3) = -1.2\cos(\pi/2) + 2.4 = -1.2(0) + 2.4 = $ $2.4$ m (midline — quarter period into the cycle).

Q4.10 — Temperature at $t = 3$ months

$T(3) = 8\cos(\pi/2) + 16 = 0 + 16 = $ $16^\circ$C (midline, around April).

Q4.11 — Submerged bucket

$1.5\sin\theta - 0.5 \leq 0 \Rightarrow \sin\theta \leq 1/3$. Reference angle $\sin^{-1}(1/3) \approx 0.3398$ rad. Sine is $\leq 1/3$ outside the interval $[\sin^{-1}(1/3),\ \pi - \sin^{-1}(1/3)]$, so in $[0, 2\pi]$: $\theta \in [0, 0.3398] \cup [\pi - 0.3398, 2\pi] \approx [0, 0.340] \cup [2.802, 6.283]$. The bucket is submerged for $\theta \in [0, 0.340] \cup [2.802, 2\pi]$ (approximately).

Q4.12 — Limitations of sine model for rainfall

Two of (any reasonable): (a) rainfall cannot be negative, but $R(t)$ would dip below $0$ if amplitude exceeded midline; this model is OK (amplitude $30$ < midline $50$). (b) Real rainfall has year-to-year variability not captured by a pure periodic function. (c) Climate change introduces a trend that a stationary sinusoid cannot model. (d) Rainfall events are discrete (storm-driven), not smoothly varying — a sinusoid models the average only.