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Lesson Explainer: Modeling with Trigonometric Functions Mathematics • 10th Grade

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 functionsβ€”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 π‘οŠ¦ around which the modeled quantity oscillates, an amplitude 𝐴 telling us by how much the model oscillates, and a period 𝑇, which is the length of time or distance before the model repeats: 𝑉=𝑏+𝐴360𝑇π‘₯.∘sin

Here, π‘₯ is our variable and 𝑉 is the value that the model is predicting. The coefficient 360𝑇 in the argument to the sine function has the effect of compressing the sine graph to the desired period 𝑇. If we are working in radians, then we will replace 360𝑇 with 2πœ‹π‘‡. The amplitude 𝐴 multiplies the sine function’s amplitude of 1, resulting in a model that oscillates between 𝑏+𝐴 and π‘βˆ’π΄οŠ¦ around the baseline π‘οŠ¦, attaining its maximum and minimum values exactly once per period.

As an example, consider a room whose temperature 𝑋 oscillates by Β±3∘C around a baseline average temperature of 22∘C, with a period of 24 hours. Working in degrees, we have a period coefficient of 36024=15, and a suitable trigonometric model would be 𝑋=22+3(15𝑑),sin where 𝑑 is the time in hours. This model predicts a maximum temperature of 22+3=25 when the sine function reaches its maximum of 1 at 90∘, so 𝑑=9015=6hours, and a minimum temperature of 22βˆ’3=19 when the sine function reaches its minimum of βˆ’1 at 270∘, so 𝑑=27015=18hours.

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, 𝑆, is affected by the tides. It can be represented on a particular day by 𝑆=4(30𝑛)+28sin, where 𝑆 is measured in metres and 𝑛 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 4(30𝑛)+28=24sin for 0≀𝑛<12. Notice, however, that the value of 24 is attained precisely when sin(30𝑛) is at its minimum value of βˆ’1. Furthermore, since 30Γ—12=360, 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 𝑉=𝑏+𝐴360𝑇π‘₯sin will always start at the baseline π‘οŠ¦ at π‘₯=0, which may not be the case in a real-world phenomenon. Therefore, a model may contain an offset π‘Ÿ: 𝑉=𝑏+𝐴360𝑇(π‘₯+π‘Ÿ).sin

This has the effect of shifting the model by π‘Ÿ to the left or right according to whether π‘Ÿ 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 ]βˆ’1,1[ 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 𝑇=12+7ο€»πœ‹12(π‘‘βˆ’10)sin, where 𝑑 is the time after midnight in hours. At what times of the day was the temperature 10∘C? 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 12∘C, the amplitude of its oscillation is 7, and its period is 2πœ‹Γ·πœ‹12=24hours, offset by 10 hours.

We are looking for solutions to the equation 12+7ο€»πœ‹12(π‘‘βˆ’10)=10,sin for 0≀𝑑<24. Let us first rearrange the equation so that we have a pure trigonometric function on the left-hand side: sinο€»πœ‹12(π‘‘βˆ’10)=βˆ’27.

Since βˆ’27 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 sin to both sides of the equation: πœ‹12(π‘‘βˆ’10)=ο€Όβˆ’27.sin

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 πœ‹ in this equation tells us that we are working in radians, so make sure your calculator is set to radians.
  • Second, sinοŠ±οŠ§ο€Όβˆ’27 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 β€œto the nearest minute.” Since 𝑑 is given in hours, this is an accuracy of around two decimal places. We should use four or five decimal places of sinοŠ±οŠ§ο€Όβˆ’27 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 βˆ’0.28975 (point 𝐴 below) for sinοŠ±οŠ§ο€Όβˆ’27; however, you may get 3.43134 (point 𝐡) if your calculator is calibrated to return values in the range [0,2πœ‹[ rather than ο“βˆ’πœ‹2,πœ‹2.

Note that 𝐡=βˆ’π΄+πœ‹=3.43134 and that all solutions to the equation πœ‹12(π‘‘βˆ’10)=ο€Όβˆ’27sin come from the equations πœ‹12(π‘‘βˆ’10)=𝐴+2π‘˜πœ‹οŠ§ and πœ‹12(π‘‘βˆ’10)=𝐡+2π‘˜πœ‹, with integral constants π‘˜οŠ§ and π‘˜οŠ¨. We are interested in solutions in the range 0≀𝑑<24.

We can now rearrange and substitute the values of 𝐴 and 𝐡 in the equations to find 𝑑:𝑑=12πœ‹π΄+10+24π‘˜=12πœ‹(βˆ’0.28975)+10+24π‘˜=8.893236…+24π‘˜,οŠ§οŠ§οŠ§οŠ§π‘‘=12πœ‹π΅+10+24π‘˜=12πœ‹(3.43134)+10+24π‘˜=23.10675+24π‘˜.

Taking π‘˜=π‘˜=0 gives us the solutions 𝑑=8.89327 and 𝑑=23.10675 lying within 0≀𝑑<24, as desired. However, there is a final point of caution before giving our answer. The question says, β€œGive 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 β€œ8 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 β€œ0.89327 of an hour” into minutes, using 8.893236…×60=53.59416 and similarly 0.10675Γ—60=6.405. At last, we have times of 𝑑=08:54 and 𝑑=23:06 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 𝑑=𝑑+0.3ο€Ό2πœ‹10π‘‘οˆοŠ¦sin, where π‘‘οŠ¦ is the depth of the lake on a quiet day and 𝑑 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 𝑦=π‘‘οŠ¦, the depth of the lake on a calm day. We are not given the value of π‘‘οŠ¦, 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 π‘‘οŠ¦. We should immediately convert this to 0.8 m to match the dimensions of 𝑑 and π‘‘οŠ¦. 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 π‘‘οŠ¦.

On the other hand, we can read the magnitude of the waves’ oscillation off from the equation 𝑑=𝑑+0.3ο€Ό2πœ‹10π‘‘οˆοŠ¦sin. Since the maximum value of the sine function is 1, the height 𝑑 of the water reaches a maximum of 0.3 m above π‘‘οŠ¦. Since Nabil’s feet are 0.2 m above π‘‘οŠ¦, the water reaches them when 0.3ο€Ό2πœ‹10π‘‘οˆsin 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 0≀π‘₯<2πœ‹ the function sin(π‘₯) is greater than or equal to 23.

Well, sin(π‘₯)=23 has two solutions for 0≀π‘₯<2πœ‹: π‘₯=0.7297radians and π‘₯=2.4119radians. The distance between π‘₯ and π‘₯ is 2.4119βˆ’0.7297=1.6822. Dividing by the full period of 2πœ‹, we have 1.6822Γ·2πœ‹=0.2677, and multiplying by 100 for a percentage gives a final answer of 26.8% to one decimal place.

The examples we have considered so far have involved models of the form 𝑇=π‘˜+𝐴(𝑓(π‘₯)),sin where π‘˜ and 𝐴 are constants and the argument 𝑓(π‘₯) is some expression in the parameter π‘₯. In particular, they have contained only one trigonometric function. We turn our attention now to models of the form 𝑇=π‘˜+𝐴(π‘₯)+𝐡(π‘₯),sincos with π‘˜, 𝐴, and 𝐡 being constants, containing a linear combination of sine and cosine.

Our approach to problems involving such models will be to rewrite the expression 𝐴(π‘₯)+𝐡(π‘₯)sincos in a form containing only one trigonometric function. This is always possible using the harmonic identities.

Identities: The Harmonic Identities

For all 𝐴, 𝐡, and π‘₯, we have the following identities: 𝐴(π‘₯)+𝐡(π‘₯)≑𝑅(π‘₯+𝛼),𝐴(π‘₯)βˆ’π΅(π‘₯)≑𝑅(π‘₯βˆ’π›Ό),𝐴(π‘₯)+𝐡(π‘₯)≑𝑅(π‘₯βˆ’π›Ό),𝐴(π‘₯)βˆ’π΅(π‘₯)≑𝑅(π‘₯+𝛼),sincossinsincossincossincoscossincos where 𝑅 and 𝛼 are given by 𝑅=√𝐴+𝐡 and 𝛼=ο€Όπ΅π΄οˆ.tan

Let us prove the first of these identities: 𝐴(π‘₯)+𝐡(π‘₯)≑𝑅(π‘₯+𝛼).sincossin

Starting with the expression on the right-hand side and applying the angle sum identity, we have 𝑅(π‘₯+𝛼)≑𝑅((π‘₯)(𝛼)+(π‘₯)(𝛼)).sinsincoscossin

Expanding the bracket on the right and equating coefficients with 𝐴(π‘₯)+𝐡(π‘₯)sincos yields the two equations 𝑅(𝛼)=𝐴cos and 𝑅(𝛼)=𝐡.sin

Taking the quotient, 𝑅(𝛼)𝑅(𝛼)=(𝛼)=𝐡𝐴sincostan and so 𝛼=ο€Όπ΅π΄οˆ.tan

To find 𝑅, we square 𝑅(𝛼)cos and 𝑅(𝛼)sin and add them together: 𝐴+𝐡=(𝑅(𝛼))+(𝑅(𝛼))=𝑅(𝛼)+𝑅(𝛼)=𝑅(𝛼)+(𝛼).cossincossincossin

And then we use the identity cossin(π‘₯)+(π‘₯)≑1 to get 𝐴+𝐡=𝑅, and so 𝑅=√𝐴+𝐡.

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 𝑇=(25+0.4(36π‘₯)+0.3(36π‘₯))sincos∘, where π‘₯ is the number of minutes since the start of the period and the angles are measured in degrees. By writing 0.4(36π‘₯)+0.3(36π‘₯)sincos in the form 𝑅(π‘₯+π‘Ž)sin, 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 25+0.4(36π‘₯)+0.3(36π‘₯)sincos. To do this, we will convert the linear combination 0.4(36π‘₯)+0.3(36π‘₯)sincos of trigonometric functions into a single trigonometric function using the harmonic identity 𝐴(π‘₯)+𝐡(π‘₯)≑𝑅(π‘₯+𝛼),sincossin where 𝛼=ο€Όπ΅π΄οˆtan and 𝑅=√𝐴+𝐡.

In this example, we have 𝐴=0.4 and 𝐡=0.3, so 𝛼=ο€Ό0.30.4=36.8699,tan∘ to four decimal places, and 𝑅=√𝐴+𝐡=√0.4+0.3=0.5.

By the harmonic identity then, we have 0.4(36π‘₯)+0.3(36π‘₯)≑0.5(36π‘₯+36.8699),sincossin and in our model we have 𝑇=25+0.4(36π‘₯)+0.3(36π‘₯)=25+0.5(36π‘₯+36.8699).sincossin

We can see that the minimum value of 𝑇 is attained when sin(36π‘₯+36.8699)=βˆ’1, and so the minimum is 𝑇=24.5. Recall that the sine function reaches its minimum of βˆ’1 at 270∘; so, to find the minute when the aquarium is at its minimum temperature, we need to solve 36π‘₯+36.8699=27036π‘₯=233.1301π‘₯=6.4758.

We conclude that 𝑇 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 𝑉=2βˆ’0.5ο€Όπœ‹π‘‘12+1.2ο€Όπœ‹π‘‘12οˆβ‹…cossinms, where 𝑑 is the time in hours after 00:00 and the angles are given in radians. By expressing 𝑉(𝑑) in the form 2βˆ’π‘…ο€Όπœ‹π‘‘12+π›Όοˆcos, find the time in any 24-hour period that the velocity of the liquid is predicted to be above 3 mβ‹…sβˆ’1. Give your answer to the nearest hour.


Here, we have a trigonometric model 𝑉=2βˆ’0.5ο€Όπœ‹π‘‘12+1.2ο€Όπœ‹π‘‘12cossin given as a linear combination of sine and cosine (plus a constant). The question is asking us to find when the value of 𝑉 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 𝐴(π‘₯)βˆ’π΅(π‘₯)≑𝑅(π‘₯+𝛼).cossincos

Write 𝑉=2βˆ’0.5ο€Όπœ‹π‘‘12+1.2ο€Όπœ‹π‘‘12=2βˆ’ο€Ό0.5ο€Όπœ‹π‘‘12οˆβˆ’1.2ο€Όπœ‹π‘‘12cossincossin so that we have 𝐴=0.5 and 𝐡=1.2. Now, 𝛼=ο€Όπ΅π΄οˆ=ο€Ό1.20.5=1.176,tantan to three decimal places, and 𝑅=√0.5+1.2=√1.69=1.3.

Our model can now be written in the form 𝑉=2βˆ’ο€Ό0.5ο€Όπœ‹π‘‘12οˆβˆ’1.2ο€Όπœ‹π‘‘12=2βˆ’1.3ο€Όπœ‹π‘‘12+1.176.cossincos

So, 𝑉 is greater than 3 precisely when 1.3ο€Όπœ‹π‘‘12+1.176<βˆ’1ο€Όπœ‹π‘‘12+1.176<βˆ’11.3.coscos

Two solutions to cos(π‘₯)=βˆ’11.3 are cosοŠ±οŠ§ο€Όβˆ’11.3=2.448 and 2πœ‹βˆ’2.448=3.835.

We are interested in the values of 𝑑 given by the inequality 2.448β‰€πœ‹π‘‘12+1.176≀3.8351.272β‰€πœ‹π‘‘12≀2.6594.859≀𝑑≀10.157.

So, the number of hours that 𝑉 is greater than 3 is 10.157βˆ’4.859=5.298 or 5 hours to the nearest hour.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • We understand how to model periodic phenomena with trigonometric functions. We can use the baseline π‘οŠ¦, amplitude 𝐴, period 𝑇, and offset π‘Ÿ to produce and interpret models of the form 𝑉=𝑏+𝐴360𝑇(π‘₯+π‘Ÿ).sin
  • We can use the inverse functions sin and cos to find values or ranges of values of a parameter π‘₯ in such models for specified values of 𝑉.
  • 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.

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