IB Maths AA SL Topic 3 — Geometry & Trig Paper 2 ~10 min read

Modelling with Trigonometric Functions

Anything that goes up and down in a regular pattern — tides, temperatures, your height on a Ferris wheel — can be modelled with a trig function. The maths is exactly the same as transforming a sine curve. The trick is reading the equation in real-world units and answering questions about what happens at specific times.

📘 What you need to know

Trig models take the form f(x) = a sin(b(x − c)) + d (or with cos).
The x-axis is usually time, not an angle. The y-axis represents whatever’s oscillating (depth, height, temperature).
Maximum value = d + |a|, Minimum value = d − |a|.
Time for one cycle = period = 360°/|b| (or 2π/|b|).
To find a value at a specific time, substitute. To find times when a value is reached, solve the equation.
Models have limitations — real life doesn’t repeat perfectly forever.

What can trig functions model?

Any quantity that oscillates — meaning it goes up and down between fixed values in a regular cycle — can be modelled with a transformed sine or cosine wave. The classic IB examples:

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Tide depth

D(t) = depth in metres at time t hours after midnight. Cycle ≈ 12 hours.

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Temperature

T(d) = temperature in °C on day d of the year. Cycle = 365 days.

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Ferris wheel height

H(t) = height in metres at time t minutes after boarding.

Spot the pattern: the input variable is usually time, and the output is whatever is fluctuating. The x-axis is no longer an angle in degrees or radians — it’s hours, days, or minutes.

The general form of a trig model

Trig model — general form f(x) = a sin(b(x − c)) + d

Each parameter has a clear real-world meaning:

Amplitude

Half the gap between max and min. Bigger a = bigger swings (more extreme tide range, hotter summers).

Period parameter

Period = 360°|b| or 2π|b|. Smaller b = slower cycles (longer to repeat).

Horizontal shift

When the cycle “starts”. Positive c shifts the graph right.

Principal axis (vertical shift)

The middle value the function oscillates around. Average tide depth, average temperature, axle height of the Ferris wheel.

For Ferris wheels modelled with cos, you’ll often see a negative a. That’s because at t = 0 (boarding), you’re at the bottom — but cos starts at the top. The negative sign flips the curve so it starts at the minimum instead.

Example — reading a Ferris wheel model

Suppose H(t) = −20 cos(2t) + 21 gives the height (metres) of a passenger above the ground, t minutes after boarding.

Reading the parameters a = −20 → amplitude = 20 (negative flips the cos so it starts low)
b = 2 → period = 2π/2 = π ≈ 3.14 minutes per cycle
c = 0 → no horizontal shift
d = 21 → axle height = 21 m

Quick reads:
• Highest point above ground = 21 + 20 = 42 m
• Lowest point above ground = 21 − 20 = 1 m
• Time for one full revolution ≈ 3.14 minutes

Visualising a model — the tide example

D(t) = 3 sin(15°(t − 2)) + 12 over 24 hours

A 24-hour tide cycle: midnight depth ≈ 10.5 m, peak at 8 a.m. (15 m), trough at 8 p.m. (9 m).

Answering modelling questions — the method

📋 Method — modelling problems

Read the parameters — identify a, b, c, d from the equation.
Translate to real-world meaning — what’s max, min, principal axis, period?
“Find value at time T” → substitute t = T into the equation.
“Find time when value is V” → set the equation equal to V and solve. Use the graph’s symmetry to find all answers.
“How long is value at least V” → solve for the two times when value = V, then find the difference.
For min/max questions, use max = d + |a|, min = d − |a|.

🤔 Why use sin/cos for periodic things?

Because they’re the simplest functions that go up and down forever in a smooth, regular way. Real periodic phenomena (waves, rotations, seasons) match this shape closely. The four parameters a, b, c, d let us “tune” the basic curve to fit any specific situation — bigger swings, faster cycles, different starting positions, different averages.

Limitations of trig models

No model is perfect. Trig models assume the world repeats exactly the same way forever — but reality drifts. Be ready to mention these in extended-response questions:

⚠ Constant amplitude assumption

The model predicts the same maximum and minimum every cycle. In reality, tide ranges vary with the moon, summer temperatures vary year to year, Ferris wheels slow down. The amplitude can drift over time.

⚠ Constant period assumption

The model predicts each cycle takes exactly the same time. In reality, periods can change too — climate shifts, tidal anomalies, mechanical wear. The model is most accurate over short time spans.

Worked examples

WE 1

Find the value at a given time

The temperature in a city is modelled by T(d) = 10 sin(2π365(d − 80)) + 18, where T is in °C and d is the day of the year. Find the temperature on day 80 (about 21 March).

Step 1: Substitute d = 80 T(80) = 10 sin(2π365(80 − 80)) + 18 = 10 sin(0) + 18 = 0 + 18 T = 18 °C at d = 80, the inside is 0 → sin(0) = 0 → just the principal axis

WE 2

Read max, min, and cycle time from a Ferris wheel model

The height (metres) of a passenger on a Ferris wheel is modelled by H(t) = −20 cos(2t) + 21, where t is the time in minutes since boarding. Find:

(a) the maximum and minimum heights (b) how long one full revolution takes.

Step 1: Identify a, b, d a = −20, b = 2, d = 21 Step 2: Max and min from amplitude Max = 21 + 20 = 42 m Min = 21 − 20 = 1 m Step 3: Period from b (in radians: 2π/b) Period = 2π2 = π ≈ 3.14 minutes Max 42 m, Min 1 m, one revolution ≈ 3.14 min amplitude is |a| = 20, so the swing is 20 above and 20 below 21

WE 3

Find when the function reaches its maximum

The depth of water at a beach is modelled by D(t) = 4 sin(30°t) + 10, where t is hours after low tide. When does high tide first occur?

Step 1: Max occurs when sin(…) = 1 sin(30°t) = 1 Step 2: sin = 1 when angle = 90° 30°t = 90° t = 9030 = 3 High tide occurs 3 hours after low tide max value = 10 + 4 = 14 m

WE 4

Find times for a specific value

Using D(t) = 4 sin(30°t) + 10 from above, find the times in the first 12 hours when the depth is exactly 12 m.

Step 1: Set D(t) = 12 and rearrange 4 sin(30°t) + 10 = 12 sin(30°t) = 0.5 Step 2: Find principal value 30°t = sin⁻¹(0.5) = 30° t = 1 Step 3: Use sin symmetry — second value Second solution: 30°t = 180° − 30° = 150° t = 5 D = 12 m at t = 1 hr and t = 5 hr two solutions per cycle — the depth passes through 12 m on the way up AND on the way down

WE 5

Full 3-part tide problem (SME-style)

The water depth, D (metres), at a port is modelled by

D(t) = 3 sin(15°(t − 2)) + 12, 0 ≤ t < 24,

where t is hours after midnight. Find:

(a) the depth at midnight (b) the minimum depth and when it occurs (c) how long the depth is at least 13.5 m each day.

PART (a) Substitute t = 0 D(0) = 3 sin(15°(0 − 2)) + 12 = 3 sin(−30°) + 12 = 3(−12) + 12 = 10.5 D = 10.5 mPART (b) Min when sin(…) = −1 Principal axis = 12, amplitude = 3, so min = 12 − 3 = 9 m Find when this happens sin(15°(t − 2)) = −1 15°(t − 2) = −90° t − 2 = −6 → t = −4 Cycle repeats every 24 hours, so add 24: t = −4 + 24 = 20 Min depth 9 m, 20 hours after midnightPART (c) Set D(t) = 13.5 and solve 3 sin(15°(t − 2)) + 12 = 13.5 sin(15°(t − 2)) = 0.5 Find both times in [0, 24] First: 15°(t − 2) = 30° → t = 4 Second (sin symmetry): 15°(t − 2) = 180° − 30° = 150° → t = 12 Find the time difference 12 − 4 = 8 Depth ≥ 13.5 m for 8 hours per day always check both solutions are in the interval before subtracting

💡 Top tips

Read the units carefully. Time in hours, days, minutes? Output in metres, °C, height? Match your answer to the units.
Max = d + |a|, Min = d − |a|. Don’t forget to take the absolute value of a.
“Time for one cycle” means period — use 360°/|b| or 2π/|b| depending on whether the equation uses degrees or radians.
For “find time when value is V” questions, set the function equal to V and solve like a regular trig equation — there’s usually 2 solutions per cycle.
For “how long is value at least V” questions: find the two times when value = V, then subtract.
Sketch the graph before solving — it helps you visualise where the function reaches its target value and how many solutions to expect.
Check the unit on the angle. 15°t means the bracket is in degrees; 2t alone usually means radians.

⚠ Common mistakes

Confusing max with amplitude. Amplitude is the swing size (|a|); max value is amplitude + principal axis (d + |a|).
Wrong unit for the period. If the inside uses degrees (e.g. 15°t), divide by 360°. If it uses radians (e.g. 2t), divide by 2π.
Only finding one solution. “When does the depth equal X?” usually has 2 solutions per cycle for sin/cos.
Forgetting to add the period when the principal value gives a negative time. The cycle repeats — add 24 (or whatever the period is).
Solutions outside the given interval. Always check answers fit the stated domain (e.g. 0 ≤ t < 24).
Calculator in the wrong mode. If the equation has 15°t, your calculator must be in DEG mode.
Not stating units. “20 hours after midnight” is much clearer than just “20”.

Modelling questions feel intimidating because they’re wrapped in real-world language, but underneath they’re just transformed sine and cosine graphs. Read the parameters, picture the wave, then answer in plain English with the right units.

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