Trigonometric Models (AI HL)
Ferris wheels, tides, sound waves and daylight hours all rise and fall in the same repeating pattern, so IB describes them with a single sine or cosine model. This page focuses on building and reading that model - working out the amplitude, period and principal axis from a real-world description before you ever have to solve anything with it. It's part of the broader Trigonometric Models topic.
29 questions on this sub-topic.
The three parameters
Covered under IB syllabus reference SL2.5: sinusoidal models \(f(x)=a\sin(bx)+d\), \(f(x)=a\cos(bx)+d\). Students find amplitude (\(a\)), period (\(\tfrac{360^\circ}{b}\) in degrees, or \(\tfrac{2\pi}{b}\) in radians), or the equation of the principal axis (\(y=d\)). Contexts include tides, weather patterns, and the motion of Ferris and bicycle wheels.
Amplitude and principal axis
\(a = \dfrac{\text{max} - \text{min}}{2}\), \(\quad d = \dfrac{\text{max} + \text{min}}{2}\)
The amplitude is how far the graph swings above and below the middle line; the principal axis \(y=d\) is that middle line itself.
Period
Period \(= \dfrac{360^\circ}{b}\) (degrees) or \(\dfrac{2\pi}{b}\) (radians)
A larger \(b\) squashes the graph horizontally, giving a shorter time between repeats - useful for checking whether a modelled cycle (e.g. a 12-hour tide) matches the value of \(b\) you've written down.
Once the model is set up, need to solve it for a specific time or value? See Solving Trig Equations.
Worked examples
The height of a passenger on a Ferris wheel is modelled by \(h(t) = 15 - 14\cos\!\left(\dfrac{\pi}{6}t\right)\) metres, where \(t\) is the time in minutes after the ride starts.
(a) State the maximum height reached by the passenger.
(b) Find the height of the passenger after 3 minutes.
Worked solution
(a) \(h_{max} = 15 + 14 = 29\) m. A1
(b) \(h(3) = 15 - 14\cos\!\left(\dfrac{\pi}{6}(3)\right) = 15 - 14\cos\!\left(\dfrac{\pi}{2}\right)\) M1
\(= 15 - 14(0) = 15\) m. A1
Explain why the average value of \(h(t) = 3\sin\!\left(\tfrac{\pi}{6}t\right) + 5\) over one full period is 5.
Worked solution
Over a full period the sine term averages to 0 (equal positive and negative areas). M1 A1
So the mean is the midline, M1
\(d = 5.\) A1
Common mistakes
- Averaging max and min incorrectly for \(d\). The principal axis is \(\dfrac{\text{max}+\text{min}}{2}\), not the maximum itself and not just the minimum - a very common slip when a question gives both values in the same sentence.
- Using degrees for \(b\) when the argument is written with \(\pi\). If the model is \(a\sin(bx)+d\) with \(x\) multiplied by a fraction of \(\pi\), the period formula needs \(\tfrac{2\pi}{b}\), not \(\tfrac{360}{b}\) - mixing the two gives a period that's out by a factor of roughly 57.
- Assuming a cosine model always starts at a maximum. \(a\cos(bx)+d\) starts at a maximum only when \(a\) is positive; a negative \(a\) flips it to start at a minimum, which changes how you read off features like the first zero or first maximum.
Ready to practise properly?
11 trigonometric-model questions, marked instantly like the real exam.
Quick answers
What do a, b and d control in a sinusoidal model?
In \(f(x) = a\sin(bx) + d\) (or \(a\cos(bx)+d\)), \(a\) is the amplitude (how far the graph swings above and below the middle), \(b\) sets the period, and \(d\) shifts the whole graph up or down to the principal axis \(y = d\).
How do you find the amplitude and principal axis from a maximum and minimum?
The principal axis is the average of the maximum and minimum values, \(d = \dfrac{\text{max} + \text{min}}{2}\), and the amplitude is half the distance between them, \(a = \dfrac{\text{max} - \text{min}}{2}\).