In this explainer, we will learn how to solve problems that are modeled using trigonometric functions.
Many real-world phenomena, both natural and artificial, exhibit some form of periodicity. That is to say that they repeat after some period. Examples might include the horizontal displacement of a moving pendulum over time, the change in time of depth of water at a point due to tidal forces, or the change in magnetic field intensity across a distance. Such phenomena can often be modeled using trigonometric functions—usually some combination of sine and cosine.
A typical trigonometric model will predict the value of some measurable quantity by an equation in some variable (usually time or distance from a fixed point) involving a combination of the sine and cosine functions. Such a model has three important features: a baseline value around which the modeled quantity oscillates, an amplitude telling us by how much the model oscillates, and a period , which is the length of time or distance before the model repeats:
Here, is our variable and is the value that the model is predicting. The coefficient in the argument to the sine function has the effect of compressing the sine graph to the desired period . If we are working in radians, then we will replace with . The amplitude multiplies the sine function’s amplitude of 1, resulting in a model that oscillates between and around the baseline , attaining its maximum and minimum values exactly once per period.
As an example, consider a room whose temperature oscillates by around a baseline average temperature of , with a period of 24 hours. Working in degrees, we have a period coefficient of , and a suitable trigonometric model would be where is the time in hours. This model predicts a maximum temperature of when the sine function reaches its maximum of 1 at , so , and a minimum temperature of when the sine function reaches its minimum of at , so .
Let us look at an example in which we need to calculate how many times in a given period a particular value is attained.
Example 1: Solving a Real-World Problem Modeled by a Trigonometric Function
The depth of the water in a fishing port, , is affected by the tides. It can be represented on a particular day by , where is measured in metres and is the time elapsed, in hours, after midnight. How many times is the depth of the water exactly 24 metres in the period 00:00–12:00?
We are asked to find the number of solutions to the equation for . Notice, however, that the value of 24 is attained precisely when is at its minimum value of . Furthermore, since , we can see that the period of this model is 12 hours. We know that the sine function reaches its minimum exactly once per period. Therefore, the depth of the water is (predicted to be) exactly 24 metres once in the period 00:00–12:00.
Note that a model of the form will always start at the baseline at , which may not be the case in a real-world phenomenon. Therefore, a model may contain an offset :
This has the effect of shifting the model by to the left or right according to whether is positive or negative.
An important point to remember about the sine function is that while it attains its maximum and minimum once per period, it takes every other value in the interval twice per period, as we will see in our next example.
Example 2: Solving a Real-World Problem Modeled by a Trigonometric Function
The outside temperature (in degrees Celsius) on a certain day was modeled with , where is the time after midnight in hours. At what times of the day was the temperature ? Give your answer to the nearest minute using a 24-hour format.
We can read off from the equation that this model has a baseline temperature of , the amplitude of its oscillation is 7, and its period is , offset by 10 hours.
We are looking for solutions to the equation for . Let us first rearrange the equation so that we have a pure trigonometric function on the left-hand side:
Since is neither a maximum nor a minimum of the sine function, we know that we are looking for two solutions in a 24-hour period. We now apply to both sides of the equation:
It is now time to get our calculators out. Before we calculate anything though, there are a few things to watch out for.
- First, the presence of in this equation tells us that we are working in radians, so make sure your calculator is set to radians.
- Second, is almost certainly an irrational number, so whatever decimal value we write down will only be an approximation. We need to be careful that the approximation we use moving forward in the calculation will guarantee an appropriate degree of accuracy in our final answer. Notice that the question asks for our answer to be given “to the nearest minute.” Since is given in hours, this is an accuracy of around two decimal places. We should use four or five decimal places of at this stage of the calculation to make sure that rounding errors do not compound to give us a wrong answer in the end.
Your calculator will probably return a value of (point below) for ; however, you may get 3.43134 (point ) if your calculator is calibrated to return values in the range rather than .
Note that and that all solutions to the equation come from the equations and with integral constants and . We are interested in solutions in the range .
We can now rearrange and substitute the values of and in the equations to find :
Taking gives us the solutions and lying within , as desired. However, there is a final point of caution before giving our answer. The question says, “Give your answer to the nearest minute using a 24-hour format.” The decimal 8.89327 is certainly not in 24-hour format. Indeed, it may be read as “8 and 0.89327 hours” (after midnight). It is very important to remember in questions like this that there are 60 minutes in an hour, not 100! To convert to the 24-hour clock format, we need to convert the “0.89327 of an hour” into minutes, using and similarly . At last, we have times of and to the nearest minute.
Certain problems involving trigonometric models may require some work of interpretation before they can be answered. Often, a diagram or sketch is useful.
Example 3: Solving a Real-World Problem Modeled by a Trigonometric Function
Nabil sits on a pier, his feet dangling 60 cm below the pier. The pier is usually 80 cm above the lake. But this is a windy day, and waves make the depth of the lake oscillate. The depth of the lake under the pier is given, in metres, by , where is the depth of the lake on a quiet day and is the time in seconds. For what fraction of the time are Nabil’s feet touching the water? Express your answer as a percentage correct to one decimal place.
A fair amount of information is given in the question. A sketch will certainly be useful to work out what is going on.
The waves on the lake are oscillating about the central line of , the depth of the lake on a calm day. We are not given the value of , but that does not matter. What is important is the relative distance between the lake surface and the height of the pier. We are told that the pier is 80 cm above . We should immediately convert this to 0.8 m to match the dimensions of and . Similarly, we are told that Nabil’s feet hang 60 cm below the pier. Together with the height of the pier above the water, this tells us that Nabil’s feet are 0.2 m above .
On the other hand, we can read the magnitude of the waves’ oscillation off from the equation . Since the maximum value of the sine function is 1, the height of the water reaches a maximum of 0.3 m above . Since Nabil’s feet are 0.2 m above , the water reaches them when is between 0.2 and 0.3.
Notice that 0.2 is two-thirds of 0.3. The question is asking us for what fraction of a full period the function is greater than or equal to .
Well, has two solutions for : and . The distance between and is . Dividing by the full period of , we have , and multiplying by 100 for a percentage gives a final answer of to one decimal place.
The examples we have considered so far have involved models of the form where and are constants and the argument is some expression in the parameter . In particular, they have contained only one trigonometric function. We turn our attention now to models of the form with , , and being constants, containing a linear combination of sine and cosine.
Our approach to problems involving such models will be to rewrite the expression in a form containing only one trigonometric function. This is always possible using the harmonic identities.
Identities: The Harmonic Identities
For all , , and , we have the following identities: where and are given by and
Let us prove the first of these identities:
Starting with the expression on the right-hand side and applying the angle sum identity, we have
Expanding the bracket on the right and equating coefficients with yields the two equations and
Taking the quotient, and so
To find , we square and and add them together:
And then we use the identity to get and so
The proofs of the other identities are similar and left as an exercise for the interested reader.
Let us see an example of using these identities to solve a problem involving a trigonometric model.
Example 4: Solving a Real-World Trigonometric Problem by Simplifying a Trigonometric Function
Farida models the temperature of her tropical fish tank in a ten-minute period by the equation , where is the number of minutes since the start of the period and the angles are measured in degrees. By writing in the form , calculate the minimum temperature this model predicts for the tank in a ten-minute period and the minute during which this temperature is predicted to occur.
The question wants us to find, first of all, the minimum value of the function . To do this, we will convert the linear combination of trigonometric functions into a single trigonometric function using the harmonic identity where and .
In this example, we have and , so to four decimal places, and
By the harmonic identity then, we have and in our model we have
We can see that the minimum value of is attained when , and so the minimum is . Recall that the sine function reaches its minimum of at ; so, to find the minute when the aquarium is at its minimum temperature, we need to solve
We conclude that reaches its minimum during the 6th minute.
In our last example, we will further practice using the harmonic identities.
Example 5: Solving a Real-World Trigonometric Problem by Simplifying a Trigonometric Function
The velocity of a liquid leaving an outflow pipe over a 24-hour period is modeled by the equation , where is the time in hours after 00:00 and the angles are given in radians. By expressing in the form , find the time in any 24-hour period that the velocity of the liquid is predicted to be above 3 m⋅s−1. Give your answer to the nearest hour.
Here, we have a trigonometric model given as a linear combination of sine and cosine (plus a constant). The question is asking us to find when the value of is greater than 3. The first step is to convert our linear combination of sine and cosine into a single trigonometric function. We will use the harmonic identity
Write so that we have and . Now, to three decimal places, and
Our model can now be written in the form
So, is greater than 3 precisely when
Two solutions to are and .
We are interested in the values of given by the inequality
So, the number of hours that is greater than 3 is or 5 hours to the nearest hour.
Let us finish by recapping a few important concepts from this explainer.
- We understand how to model periodic phenomena with trigonometric functions. We can use the baseline , amplitude , period , and offset to produce and interpret models of the form
- We can use the inverse functions and to find values or ranges of values of a parameter in such models for specified values of .
- We can interpret complicated real-world examples using a sketch or diagram.
- We can rewrite linear combinations of sine and cosine as a single trigonometric function using the harmonic identities to solve more complicated problems involving trigonometric models.