Applications Of Trigonometric Functions
Learn to model periodic phenomena with a sinusoidal function for Queensland Year 11 Mathematical Methods (QCAA). Real cycles such as tides and a Ferris wheel repeat in a smooth wave a sine model captures.
You will learn to read the amplitude, period and mean from a context, build the equation, evaluate it to predict a value, and solve for when a level is reached — genuine applications of trigonometry in the QCAA course.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), many repeating real-world quantities — tides, daily temperature, the height of a Ferris-wheel seat — are modelled by a sinusoid \(y=a\sin n(t-e)+b\). The amplitude \(|a|\), period \(\dfrac{2\pi}{n}\) and mean \(b\) carry meaning in the context, and solving the equation tells you when a value is reached.
A sinusoidal model \(y=a\sin n(t-e)+b\) describes a quantity that rises and falls regularly. The amplitude \(|a|\) is how far it swings from the centre, the mean \(b\) is that centre value, and the period \(\dfrac{2\pi}{n}\) is the time for one full cycle.
The maximum is \(b+|a|\) and the minimum is \(b-|a|\), so the range is \([\,b-|a|,\ b+|a|\,]\). These translate directly to, say, highest and lowest tide.
To find when a value \(y_0\) occurs, set the model equal to \(y_0\) and solve the trigonometric equation over the stated time domain, using the reference angle and symmetry.
For a model \(y=a\sin n(t-e)+b\):
How to work with a sinusoidal model
- Identify the parameters: read \(a\), \(n\) and \(b\) from the rule (or a graph).
- Interpret: amplitude \(|a|\), period \(\dfrac{2\pi}{n}\), mean \(b\), and extremes \(b\pm|a|\).
- Evaluate: to find the value at a time, substitute \(t\) and use exact special-angle values.
- Solve: to find a time, set the model equal to the target and solve the trig equation over the stated domain.
Amplitude and mean — from the rule:
| \(|a|\) | \(=\) | \(3\) |
| \(b\) | \(=\) | \(4\) |
Period — \(n=\dfrac{\pi}{6}\):
| \(\text{period}\) | \(=\) | \(\dfrac{2\pi}{\pi/6}\) |
| \(=\) | \(12\) |
Highest and lowest — \(b\pm|a|\):
| \(\text{max}\) | \(=\) | \(4+3=7\) |
| \(\text{min}\) | \(=\) | \(4-3=1\) |
Amplitude \(3\) m, period \(12\) h, mean \(4\) m; highest \(7\) m, lowest \(1\) m.
Substitute \(t=2\):
| \(d\) | \(=\) | \(4+3\sin\!\left(\dfrac{\pi}{6}\times 2\right)\) |
| \(=\) | \(4+3\sin\dfrac{\pi}{3}\) |
Use \(\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}\):
| \(=\) | \(4+3\times\dfrac{\sqrt3}{2}\) | |
| \(=\) | \(4+\dfrac{3\sqrt3}{2}\) |
Depth \(=4+\dfrac{3\sqrt3}{2}\) m \(\approx 6.60\) m.
Maximum — \(H\) is greatest when \(\cos\dfrac{\pi}{10}t=-1\):
| \(H_{\max}\) | \(=\) | \(11-9(-1)\) |
| \(=\) | \(20\) |
Time — solve \(\cos\dfrac{\pi}{10}t=-1\):
| \(\dfrac{\pi}{10}t\) | \(=\) | \(\pi\) |
| \(t\) | \(=\) | \(10\) |
Maximum height \(20\) m, first reached at \(t=10\) s.
Set the model equal to \(5.5\) and isolate sine:
| \(4+3\sin\dfrac{\pi}{6}t\) | \(=\) | \(5.5\) |
| \(3\sin\dfrac{\pi}{6}t\) | \(=\) | \(1.5\) |
| \(\sin\dfrac{\pi}{6}t\) | \(=\) | \(\dfrac12\) |
Solve — let \(u=\dfrac{\pi}{6}t\); over \(0\le t\le 12\), \(u\) runs over \([0,2\pi]\):
| \(u\) | \(=\) | \(\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\) |
Back-substitute \(t=\dfrac{6u}{\pi}\):
| \(t\) | \(=\) | \(1,\ 5\) |
Depth is \(5.5\) m at \(t=1\) h and \(t=5\) h.
Common pitfalls
Frequently asked questions
What do the parts of y = a sin n(t - e) + b mean in a model?
\(|a|\) is the amplitude (the swing), \(b\) is the mean or centre value, \(\dfrac{2\pi}{n}\) is the period, and \(e\) shifts the cycle in time.
How do I find the maximum and minimum of a model?
The maximum is \(b+|a|\) and the minimum is \(b-|a|\): the mean value plus or minus the amplitude.
How do I find the period of a tide or temperature model?
Use \(\dfrac{2\pi}{n}\). For example \(n=\dfrac{\pi}{6}\) gives a period of \(12\) hours.
How do I find when a model reaches a certain value?
Set the model equal to that value, isolate the sine or cosine, then solve the trig equation over the stated time domain.
Why might there be two answers for the time?
Because the quantity passes through most values twice each cycle — once rising and once falling — so there are two times per period.