IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~7 min read

Modelling with Trigonometric Functions

Anything that oscillates — tides, daylight hours, Ferris wheels, AC voltage, body temperature — fits a trig model. The form is the same as before: f(t) = a sin(b(t − c)) + d or with cos. The x-axis is now time, not angle, and each parameter has a real-world meaning.

📘 What you need to know

Standard form: f(t) = a sin(b(t − c)) + d (or with cos).
a (amplitude) = (max − min)/2. Half the total swing.
d (principal axis / vertical shift) = (max + min)/2. The “midline” of the oscillation.
b (frequency parameter): period = 2π/|b| (radians) or 360°/|b| (degrees). Period = time for one full cycle.
c (phase shift): shifts the timing. For sin, c is when the graph crosses the axis going up; for cos, c is when it hits its maximum.
Common contexts: water depth, temperature, height on a wheel, sound waves, AC voltage.
Limitations: real systems aren’t perfectly periodic — amplitudes, periods, and phases drift over time.

The standard model

Trigonometric model f(t) = a sin(b(t − c)) + d or f(t) = a cos(b(t − c)) + d

From context to parameters

a = (max − min)/2
d = (max + min)/2

range of values gives a and d directly

Period to b

b = 2π/period
(or 360°/period)

long period → small b; short period → large b

Limitations of trig models

Real-world periodic systems are never perfectly sinusoidal. Examination questions often ask for limitations — common ones include:

Assumption	Why it can fail in real life
Constant amplitude	tides may shrink during neap tides; daily temperature ranges vary by season
Constant period	day length isn’t exactly 24 hours; tidal periods drift slightly
Smooth sinusoidal shape	real signals have noise, spikes, and transient effects
Indefinite repetition	weather, biology, and machinery all have long-term drift or wear

🧭 Recipe — set up a trig model from a context

Identify max and min of the quantity. Then a = (max − min)/2 and d = (max + min)/2.
Identify the period (e.g., 24 h for daily, 365 d for yearly). Then b = 2π/period.
Choose sin or cos: cos hits its max at t = c; sin crosses the axis going up at t = c.
Find c from a known max/min/zero in the data.
Verify by substituting a known data point.

Worked examples

WE 1

Ferris wheel — read off basic features

The height (in metres) of a passenger above the ground, t minutes after boarding a Ferris wheel, is given by H(t) = −30 cos(πt/4) + 32. Find (a) the height at boarding, (b) the maximum height, and (c) the time for one complete revolution.

(a) Boarding is t = 0 H(0) = −30 cos(0) + 32 = −30 + 32 = 2 m (a) 2 m (b) Max occurs when cos(…) = −1 (because of the negative coefficient) max = d + |a| = 32 + 30 = 62 m (b) 62 m (c) Period = 2π/|b| period = 2π/(π/4) = 8 minutes (c) 8 minutes “−cos” starts at minimum, which fits boarding at the bottom of the wheel

WE 2

Temperature model

The mean daily temperature (°C) in a city d days after January 1st is given by T(d) = 8 sin(2πd/365) + 18. Find (a) the temperature on January 1st, (b) the maximum and minimum temperatures, and (c) the day of the year on which the maximum occurs.

(a) d = 0 T(0) = 8 sin(0) + 18 = 18°C (a) 18°C (b) Max and min using d ± |a| max = 18 + 8 = 26°C; min = 18 − 8 = 10°C (b) max 26°C, min 10°C (c) Sin reaches max when its argument = π/2 2πd/365 = π/2 → d = 365/4 = 91.25 (c) day 91 (≈ April 1)

WE 3

Find the times when a depth is reached

The depth of water (in metres) at a harbour is modelled by D(t) = 4 cos(πt/6) + 10, where t is the number of hours after high tide. Find the times during the first 12 hours when the depth is exactly 11 metres.

Step 1: Set D(t) = 11 and solve for the cos value 4 cos(πt/6) + 10 = 11 cos(πt/6) = 1/4 = 0.25 Step 2: Principal value (radian mode) πt/6 = arccos(0.25) = 1.318… t₁ = 6 × 1.318/π = 2.517… Step 3: Second solution (cos: 2π − x) πt/6 = 2π − 1.318 = 4.965 t₂ = 6 × 4.965/π = 9.483… t ≈ 2.52 hours and t ≈ 9.48 hours period 2π/(π/6) = 12 hours, so two solutions in one cycle ✓

WE 4

Total time above a threshold

Using the same tide model D(t) = 4 cos(πt/6) + 10, find the total length of time during the first 12 hours that the depth is at least 12 metres.

Step 1: Set up the inequality 4 cos(πt/6) + 10 ≥ 12 → cos(πt/6) ≥ 0.5 Step 2: Solve cos(u) ≥ 0.5 for u in [0, 2π] u ∈ [0, π/3] ∪ [5π/3, 2π] Step 3: Convert back to t (since u = πt/6) [0, π/3] → t ∈ [0, 2] [5π/3, 2π] → t ∈ [10, 12] Step 4: Total length 2 + 2 = 4 hours depth ≥ 12 m for a total of 4 hours two windows around high tide (which occurs at t = 0 and t = 12)

WE 5

Build the equation from a context

The temperature in a town varies sinusoidally between 5°C and 25°C, with the maximum occurring at 3 pm. Assuming a 24-hour cycle, find a model of the form T(t) = a cos(b(t − c)) + d giving the temperature t hours after midnight.

Step 1: Find a and d from max and min a = (25 − 5)/2 = 10 d = (25 + 5)/2 = 15 Step 2: Find b from period period = 24 hours b = 2π/24 = π/12 Step 3: Find c — cos hits max at t = c 3 pm = t = 15 → c = 15 T(t) = 10 cos((π/12)(t − 15)) + 15 Step 4: Verify at midnight (t = 0) T(0) = 10 cos(−5π/4) + 15 = 10(−√2/2) + 15 ≈ 7.93°C ✓

WE 6

AC voltage — peak, frequency, and a limitation

The voltage in a domestic mains supply is modelled by V(t) = 240 sin(100πt), where V is in volts and t is in seconds. (a) State the peak voltage. (b) Find the period and frequency. (c) Calculate V when t = 0.005 s. (d) State one limitation of this model.

(a) Peak voltage = amplitude peak = 240 V (a) 240 V (b) Period and frequency period = 2π/(100π) = 1/50 = 0.02 s frequency = 1/period = 50 Hz (b) period 0.02 s, frequency 50 Hz (c) Substitute t = 0.005 V = 240 sin(100π × 0.005) = 240 sin(π/2) = 240 V (c) V = 240 V (peak) (d) Limitation (d) real mains voltage has noise, harmonics, and small fluctuations — not a pure sine wave 50 Hz is the standard mains frequency in most of the world (60 Hz in North America)

💡 Top tips

Always identify a, b, c, d first. Once you have them, every other quantity follows.
Pick sin or cos based on the starting condition: if t = 0 corresponds to a max or min, cos is natural; if it crosses the axis going up, sin is natural.
Check using a known data point. Substitute t = 0 (or another given time) and compare to the value in the question.
Match the unit of time to the period: hours/24, days/365, minutes for short cycles, seconds for AC.
For “how long” questions, set up an inequality, solve for the boundary times, and subtract.

⚠ Common mistakes

Calculator in the wrong mode. Most modelling questions use radians — set RAD before any computation.
Forgetting to factor b when reading the phase shift c.
Mixing units. If t is in hours, the period must be in hours too.
Stopping after the first solution when finding times. Trig equations have multiple solutions per period.
Vague limitations. “It’s not perfect” doesn’t earn marks — say what specifically: “the amplitude isn’t constant”, “the period drifts seasonally”, etc.

That closes the Trigonometric Functions & Graphs section. You’ve now covered the full toolkit: the three base graphs, solving equations, transforming, and modelling. The next section dives into Trigonometric Equations & Identities — algebraic relationships between trig functions, like sin²θ + cos²θ = 1, that let you simplify expressions and solve harder equations.

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