Applications of trigonometric functions
Theory
A periodic model describes a quantity that rises and falls in a repeating cycle — tide height, temperature, or the height of a Ferris wheel car — using \(y=a\sin\!\big(n(t-e)\big)+b\) or the cosine version. The amplitude \(a\), period \(\dfrac{2\pi}{n}\), mean level \(b\) and phase \(e\) are read from the context, after which the model can be evaluated, solved for a given value, or searched for its maximum and minimum.
Many real quantities repeat in a regular cycle. A sinusoidal model \(y=a\sin\!\big(n(t-e)\big)+b\) (or the cosine form) captures this: the graph oscillates evenly above and below a central line. Here \(t\) is time and \(y\) is the modelled quantity — a depth, a temperature, a height.
The amplitude \(a\) is how far the quantity swings above or below its centre, so \(a=\dfrac{\text{max}-\text{min}}{2}\). The mean level (or midline) \(b=\dfrac{\text{max}+\text{min}}{2}\) is the value the curve oscillates about. The period is the time for one full cycle; since the graph repeats every \(\dfrac{2\pi}{n}\), the constant is \(n=\dfrac{2\pi}{\text{period}}\).
The phase \(e\) slides the curve horizontally so it starts in the right place. Because \(\sin\) and \(\cos\) never leave \([-1,1]\), the maximum value is \(b+a\) and the minimum is \(b-a\); a cosine is convenient when the cycle begins at a peak or trough, a sine when it begins at the mean level and is rising.
The general sinusoidal model, with amplitude \(a\), horizontal factor \(n\), phase \(e\) and mean level \(b\):
Amplitude and mean level from the maximum and minimum values:
The period fixes the constant \(n\), and the extreme values follow from \(a\) and \(b\):
How to build and use a periodic model
- Mean and amplitude. From the maximum and minimum, take \(b=\dfrac{\text{max}+\text{min}}{2}\) and \(a=\dfrac{\text{max}-\text{min}}{2}\).
- Find \(n\). Read the period (time for one cycle) and use \(n=\dfrac{2\pi}{\text{period}}\).
- Choose the function and phase. Use cosine if the cycle starts at a peak or trough, sine if it starts at the mean rising; set the phase \(e\) (or \(c\)) so the curve begins in the right place.
- Evaluate. Substitute a time \(t\) into the model to predict the quantity there.
- Solve or optimise. To find when \(y=k\), solve the trig equation over the given domain; the maximum \(b+a\) and minimum \(b-a\) occur where the sine or cosine equals \(\pm 1\).
| \(a\) | \(=\) | \(4\) (coefficient of \(\sin\)) |
| \(b\) | \(=\) | \(7\) (the added constant) |
| \(\text{period}\) | \(=\) | \(\dfrac{2\pi}{\pi/6}=12\) hours |
| \(\text{max}\) | \(=\) | \(7+4=11\) |
| \(\text{min}\) | \(=\) | \(7-4=3\) |
| \(h(2)\) | \(=\) | \(5-4\cos\!\big(\dfrac{\pi}{3}\big)\) |
| \(=\) | \(5-4(\dfrac{1}{2})=3\) |
| \(h_{\max}\) | \(=\) | \(5-4(-1)=9\) |
| \(\dfrac{\pi}{6}t\) | \(=\) | \(\pi\) |
| \(t\) | \(=\) | \(6\) seconds |
| \(b\) | \(=\) | \(\dfrac{25+13}{2}=19\) |
| \(a\) | \(=\) | \(\dfrac{25-13}{2}=6\) |
| \(n\) | \(=\) | \(\dfrac{2\pi}{24}=\dfrac{\pi}{12}\) |
| \(T\) | \(=\) | \(6\cos\!\big(\dfrac{\pi}{12}(t-16)\big)+19\) |
| \(3\sin\!\big(\dfrac{\pi}{6}(t-1)\big)+6\) | \(=\) | \(7.5\) |
| \(\sin\!\big(\dfrac{\pi}{6}(t-1)\big)\) | \(=\) | \(0.5\) |
| \(\dfrac{\pi}{6}(t-1)\) | \(=\) | \(\dfrac{\pi}{6}\;\Rightarrow\; t-1=1\) |
| \(\dfrac{\pi}{6}(t-1)\) | \(=\) | \(\dfrac{5\pi}{6}\;\Rightarrow\; t-1=5\) |
| \(t\) | \(=\) | \(2\) or \(6\) |
Common pitfalls
Frequently asked questions
How do you find the amplitude and mean level of a periodic model?
The amplitude is half the distance between the maximum and minimum, \(a=\dfrac{\text{max}-\text{min}}{2}\); the mean level (midline) is their average, \(b=\dfrac{\text{max}+\text{min}}{2}\). For a tide between \(3\) and \(11\) m the amplitude is \(4\) m and the mean is \(7\) m.
How do you find the period of a sine or cosine model?
The period is the time for one complete cycle. In \(y=a\sin(nt)+b\) the period is \(\dfrac{2\pi}{n}\), so \(n=\dfrac{2\pi}{\text{period}}\). A tide repeating every \(12\) hours has \(n=\dfrac{2\pi}{12}=\dfrac{\pi}{6}\).
How do you find the equation of a trigonometric model from a context?
Take \(b\) as the mean of the max and min, \(a\) as half their difference, and \(n=\dfrac{2\pi}{\text{period}}\). Choose sine or cosine and a phase so the curve starts correctly — for example a cosine with a maximum at a known time gives \(y=a\cos\!\big(n(t-c)\big)+b\).
How do you find when a periodic model reaches a given value?
Set the model equal to the target and solve. Isolate the sine or cosine, take the inverse for a base angle, use the graph's symmetry to find every angle in the domain, then convert back to \(t\). A model usually reaches a value more than once per cycle.
How do you find the maximum and minimum of a trigonometric model and when they occur?
Since \(\sin\) and \(\cos\) lie in \([-1,1]\), the maximum is \(b+a\) and the minimum is \(b-a\). Solve \(\sin=\pm1\) or \(\cos=\pm1\) for \(t\) to find the times. For \(h=5-4\cos\!\big(\dfrac{\pi}{6}t\big)\), the maximum \(9\) occurs when \(\cos=-1\), at \(t=6\).
Should I use sine or cosine to model periodic data?
Either works, since a cosine is a shifted sine. Use cosine when the cycle starts at a peak or trough (no phase shift needed) and sine when it starts at the mean level rising; a phase shift lets either begin anywhere.