Anything that rises and falls in a repeating pattern — daylight hours, tides, temperatures — can be built from a sine or cosine curve. The real skill is reading the amplitude, period and shift straight out of a written description.
9 min read3 worked examples
Building a model from a description
A quantity that rises and falls smoothly between a highest and a lowest value, repeating on a fixed cycle, can be modelled using a sine graph — one built entirely from information given in the question, rather than memorised. You just need three things: the largest and smallest values it reaches (which tell you how high the wave sits and how far it swings), and one moment when a maximum or minimum happens (which tells you where the wave is shifted to).
Worked example 1
Setting up the model from given information
The number of hours of daylight in a city, $D$, varies through the year and can be modelled by an equation of the form $D(t) = p+q\sin(30(t-r))°$, where $t$ is the month number ($t=1$ for January, ..., $t=12$ for December). The longest day of the year has 16 hours of daylight and occurs in June ($t=6$); the shortest day has 8 hours and occurs in December ($t=12$). Find the equation for $D(t)$.
MethodUse the max and min to fix the wave's height, then use when the max occurs to fix its shift
1
The curve sits centred on the average of its extreme values: $\dfrac{16+8}{2} = 12$.
Since sin oscillates symmetrically, the "resting level" of D(t) is exactly halfway between its biggest and smallest values.
2
It then swings up and down from that level by $\dfrac{16-8}{2} = 4$.
This is half the total gap between maximum and minimum — how far D(t) departs from its resting level of 12.
3
So far: $D(t) = 12+4\sin(30(t-r))°$ for some shift $r$. $D$ is at its maximum when $\sin(30(t-r))°=1$, i.e. when $30(t-r)=90°$. This happens at $t=6$: $30(6-r)=90 \Rightarrow 6-r=3 \Rightarrow r=3$.
Substitute the month where the maximum is known to occur, then solve for r.
▪ solved
Model: $D(t) = 12+4\sin(30(t-3))°$.
Sense check
The coefficient 30 gives a period of $\dfrac{360°}{30}=12$ months — exactly one year, as expected. Checking December: $D(12)=12+4\sin(30\times9)°=12+4\sin270°=12-4=8$ ✓, matching the given minimum.
Worked example 2
Using the model to predict a value
Using the model $D(t)=12+4\sin(30(t-3))°$ from Worked Example 1, find the number of hours of daylight in September ($t=9$).
MethodSubstitute directly into the model
1
$D(9) = 12+4\sin(30(9-3))° = 12+4\sin(180°)$.
Substitute t = 9 into the model.
2
$\sin180°=0$, so $D(9) = 12+4(0) = 12$.
Evaluate the sine term.
▪ solved
September has 12 hours of daylight — exactly the average of the max and min, as expected for a month roughly midway between the longest and shortest days.
Worked example 3 — Challenge
Solving the model for a target value
Using the same model, find the two months in which there are exactly 14 hours of daylight.
MethodSet the model equal to the target and solve as a trig equation
Answer: $t=4$ and $t=8$, i.e. April and August — symmetric either side of the longest day in June, exactly as the shape of a sine curve predicts.
Common mistake
Forgetting to widen the range for the substituted variable (here $u=30(t-3)$) before solving is the same error seen in the previous lesson — it's just as easy to lose solutions in a modelling context as in a pure equation.
Check yourself
Using the model $D(t)=12+4\sin(30(t-3))°$, find the two months in which there are exactly 10 hours of daylight. (Hint: this time sin(30(t−3))° is negative — one solution will correspond to a value of t below 1, which wraps around to the previous December.)