Chapter 11 opens by putting Section 10.5 to work. The sine form of the sinusoid is adopted once and for all, and each of its four parameters is given a physical meaning that word problems state in ordinary language. Covers deriving a sinusoid from a physical setup such as a rotating wheel, the distinction between period, ordinary frequency and angular frequency, fitting a sinusoid to measured data by taking the baseline and amplitude from the extremes, and locating the phase from the position of a maximum.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 11 — Applications of Trigonometry
§11.1 Applications of Sinusoids, pp. 881-885
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 881-885 — the pages these objectives are drawn from
Warm-up
You can read the four parameters off a formula and off a graph. A word problem states them in a third way, in ordinary English, and translating is the new skill.
Discussion prompt
A buoy bobs between 2 metres and 6 metres above the seabed, completing a full cycle every 10 seconds. Without writing any formula, name the amplitude, the baseline and the period.
Hint: Two of the three come from the same pair of numbers.
Answer:
The baseline is the average of the extremes, four metres. The amplitude is half their difference, two metres. The period is stated directly, ten seconds.
\[ B = \tfrac{6 + 2}{2} = 4, \qquad A = \tfrac{6 - 2}{2} = 2, \qquad T = 10 \]
Three of the four parameters, from one sentence and no algebra. The fourth, the phase, needs one more piece of information — where in the cycle the clock starts — and that is the part word problems are least explicit about.
Concept
From here on the sinusoid is written in the sine form with time as the variable. The four parameters are unchanged from Section 10.5, but each now carries a physical meaning that an applied problem will describe rather than state.
baseline — The value B about which the sinusoid oscillates, equal to the average of its maximum and minimum. In an application it is the equilibrium or average level of whatever is being modelled.
\[ S(t) = A\sin(\omega t + \phi) + B \]
The book commits to the sine form here for clarity, having stayed neutral in Section 10.5. Nothing is lost: a cosine model can always be rewritten as a sine one with a different phase.
Figure (svg): A sinusoid with its four parameters labelled and their physical meanings written beside each: amplitude as maximum displacement, baseline as the average level, period as the time for one cycle, and phase shift as a head start
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 881-882
Section
Section 1
Concept
Amplitude, baseline, period and phase shift each answer a question an applied problem will pose in words. Recognising which question is being answered is the whole translation step.
The amplitude and baseline together determine the maximum and minimum: the maximum is B plus the amplitude and the minimum is B minus it. Read backwards, that is how a problem's stated extremes become parameters.
| Parameter | Formula | What it measures |
|---|---|---|
| Amplitude | the absolute value of A | the maximum displacement from the baseline |
| Baseline | B | the average level oscillated about |
| Period | two pi over omega | the time for one complete cycle |
| Phase shift | negative phi over omega | how much of a head start the cycle has |
Figure (svg): A sinusoid with its four parameters labelled and their physical meanings written beside each: amplitude as maximum displacement, baseline as the average level, period as the time for one cycle, and phase shift as a head start
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 881-882
Picture it
A word problem never says amplitude. It says how far something swings.
Figure (svg): Two columns pairing each phrase a word problem might use with the parameter it is describing
Making this translation explicitly, in writing, before touching any formula is what turns a wordy problem into a one-line substitution.
Worked example
No formula is written until every parameter has been named.
\[ \text{A tide varies between } 1.2 \text{ and } 5.2 \text{ m, repeating every } 12.4 \text{ h. Find } A, \; B \text{ and } \omega. \]
Take the baseline as the average of the extremes
Why: Add and halve.
\[ B = \frac{5.2 + 1.2}{2} = 3.2 \]
Take the amplitude as half the difference
Why: Subtract and halve.
\[ A = \frac{5.2 - 1.2}{2} = 2 \]
Read the period directly
Why: It is stated in the problem.
\[ T = 12.4 h \]
Convert the period to angular frequency
Why: Two pi over the period.
\[ \omega = 2 \pi / 12.4 = 0.5068 \]
Figure (svg): The solution to Worked example read the parameters from a description shown as a ladder of expressions, one row per legal move
\[ A = 2, \quad B = 3.2, \quad \omega = \frac{2\pi}{12.4} \approx 0.507 \]
Verify: reconstruct the extremes
Why: The maximum is B plus A, which is 5.2, and the minimum is B minus A, which is 1.2. Both match the problem, confirming that the baseline and amplitude were taken the right way round.
Matching
Every applied problem states its parameters in words like these.
Match the pairs
Why: A mean or average level is the baseline; a variation either way is the amplitude; a repeat interval is the period. The location of a maximum does not give the phase directly but determines it, since a sine peaks a quarter period after its cycle starts. Making these four identifications is the entire modelling step.
Worked example
A problem will give one of the three. The formula needs a specific one.
\[ \text{A wheel turns at } 15 \text{ revolutions per minute. Find } T, \; f \text{ and } \omega \text{ in seconds.} \]
Convert to revolutions per second
Why: Fifteen per minute is a quarter per second.
\[ f = 0.25\text{ per second} \]
Take the period as the reciprocal
Why: One over the ordinary frequency.
\[ T = 4\text{ seconds} \]
Convert to angular frequency
Why: Multiply the ordinary frequency by two pi.
\[ \omega = 0.5 \pi\text{ per second} \]
Check against the period formula
Why: Two pi over the period gives the same thing.
\[ 2 \pi / 4 = \frac{\pi}{2} \]
Figure (svg): The solution to Worked example the three ways of stating a rate shown as a ladder of expressions, one row per legal move
\[ T = 4 \text{ s}, \quad f = \tfrac{1}{4} \text{ Hz}, \quad \omega = \tfrac{\pi}{2} \text{ s}^{-1} \]
Verify: check the two routes agree
Why: Going by frequency gives two pi times one quarter, which is pi over two. Going by period gives two pi over 4, also pi over two. The two routes must agree, since the period and the frequency are reciprocals.
Trap
\[ 15 \text{ rpm} \;\Longrightarrow\; x(t) = A\sin(15t) + B \]
Put the stated rate straight into the formula as omega
Why: It is a rate and omega is a rate, so the substitution looks natural.
But 15 revolutions per minute is an ordinary frequency, counting cycles, while omega counts radians. The formula would describe a wheel turning about 2.4 times as fast, and in the wrong time unit as well.
\[ \omega = 2\pi f = 2\pi \cdot \tfrac{15}{60} = \tfrac{\pi}{2} \text{ s}^{-1} \]
Convert the rate to radians per unit time before substituting
Why: Multiply by two pi, and fix the time unit at the same time.
This is the same silent error as in Lesson 10.1c, and it has the same tell: check the period the formula implies. Here the correct omega gives a period of 4 seconds, which is 15 revolutions a minute — and the wrong one would have given a period of about 0.42 minutes, or 25 revolutions a minute.
Fill the middle
A quantity varies between 8 and 20, with a period of 6.
Fill in the blanks
B = \frac146 = pi/3, \quad A = \frac______ = ___, \quad \omega = \frac______ = ___
Why: The baseline is the average of the extremes and the amplitude is half their difference. The angular frequency is two pi over the period. Checking: a baseline of 14 with amplitude 6 gives a maximum of 20 and a minimum of 8, matching the description.
Prediction
A sinusoid has amplitude 5 and baseline 12.
Predict first
What are its maximum and minimum?
Correct: 17 and 7.
Why: The maximum is the baseline plus the amplitude, and the minimum is the baseline minus it. So 12 plus 5 is 17 and 12 minus 5 is 7. This is the reverse of how the parameters were found from the data, and running it both ways is the standard check: if the reconstructed extremes do not match the problem, the amplitude and baseline were swapped somewhere.
Socratic
Section 10.5 stayed neutral between the cosine and sine forms. This section chooses the sine.
Discussion prompt
What does committing buy, and is anything lost?
Hint: What has to be agreed before two people can compare their answers?
Answer:
It buys comparability. The same physical situation modelled as a cosine and as a sine gives two correct formulas with different phases, and a reader cannot check one against the other without first converting. Fixing the form removes that friction, which matters more in an applications chapter than in a graphing one.
Nothing is lost, because the two forms are interconvertible: a cosine of theta is a sine of theta plus pi over two, so any cosine model becomes a sine model by adjusting the phase. The choice is a convention, and the book is explicit that it is one — which is the right way to introduce a convention.
Section
Section 2
Concept
The period, the ordinary frequency and the angular frequency all say the same thing about how fast a sinusoid repeats. Which one a problem states varies, and only one of the three goes into the formula.
\[ T = \frac{1}{f} = \frac{2\pi}{\omega}, \qquad \omega = 2\pi f \]
This is exactly the distinction from Lesson 10.1c, where omega was a rate of turning. Nothing has changed except that the rotating object may now be metaphorical.
Figure (svg): The relationship between period, ordinary frequency and angular frequency, shown as a chain with the conversion between each pair
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 881-881
Picture it
One piece of physical information, expressible three ways.
Figure (svg): The relationship between period, ordinary frequency and angular frequency, shown as a chain with the conversion between each pair
The habit worth building: write down all three whenever any one is given. It costs two lines and makes every subsequent substitution safe.
Worked example
The period is stated, so omega comes from one reciprocal.
\[ \text{A pendulum swings with period } 2.4 \text{ s, amplitude } 0.3 \text{ m about a central position. Find } \omega. \]
Identify what was given
Why: The period, directly.
\[ T = 2.4 s \]
Convert to angular frequency
Why: Two pi over the period.
\[ \omega = 2 \pi / 2.4 \]
Simplify
Why: Two pi over 2.4 is five pi over six.
\[ \omega = 5 \pi / 6 \]
Note the other two parameters
Why: Amplitude 0.3, baseline zero since it swings about a central position.
\[ A = 0.3, B = 0 \]
Figure (svg): The solution to Worked example from a period to a model shown as a ladder of expressions, one row per legal move
\[ \omega = \frac{2\pi}{2.4} = \frac{5\pi}{6} \approx 2.618 \text{ s}^{-1} \]
Verify: check the period back
Why: Two pi divided by five pi over six is twelve over five, which is 2.4 seconds. The round trip returns the given period.
Sorting
Only one of the three can go straight into the formula.
Sort into buckets
Sort each statement.
Worked example
Example 11.1.1's rotation data. Read carefully: two revolutions, not one.
\[ \text{A wheel completes } 2 \text{ revolutions in } 2 \text{ min } 7 \text{ s. Find its period and } \omega. \]
Convert the time to seconds
Why: Two minutes and seven seconds.
\[ 127\text{ seconds} \]
Halve it for one revolution
Why: The period is the time for ONE cycle.
\[ T = 63.5 s \]
Convert to angular frequency
Why: Two pi over the period.
\[ \omega = 2 \pi / 63.5 \]
Simplify
Why: Multiplying top and bottom by 2 clears the decimal.
\[ \omega = 4 \pi / 127 \]
Figure (svg): The solution to Worked example two revolutions in a stated time shown as a ladder of expressions, one row per legal move
\[ T = \tfrac{127}{2} \text{ s}, \qquad \omega = \frac{4\pi}{127} \approx 0.0989 \text{ s}^{-1} \]
Verify: check by counting
Why: At this omega, the argument advances by four pi over 127 each second, so after 127 seconds it has advanced by four pi — which is exactly two revolutions. That matches the stated rotation.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 883-883
Error analysis
A student reads that a wheel makes 2 revolutions in 127 seconds.
Annotate
On: \( T = 127 \text{ s} \;\Longrightarrow\; \omega = \frac{2\pi}{127} \)
The definition of period is the time for one complete cycle, and problems frequently state a time for several. Reading the sentence for how many cycles are being described takes a moment and is worth it.
Faded example
A wave has frequency 50 hertz. Find its angular frequency and period.
Fill in the blanks
\omega = 2\pi f = 100 pi \text0.02, \qquad T = \frac______ = ___ \text___
Why: Fifty cycles per second times two pi radians per cycle gives one hundred pi radians per second, about 314. The period is the reciprocal of the frequency, one fiftieth of a second. Checking, two pi over one hundred pi is one fiftieth, so the two routes agree.
Estimation
A model uses omega equal to 0.26 per hour.
Predict first
Roughly what period does that correspond to?
Correct: About 24 hours.
Why: The period is two pi over omega, which is about 6.28 divided by 0.26, roughly 24 hours. That value should be recognisable: a model of a daily cycle will always have omega close to 0.26 per hour, just as a monthly one has omega close to pi over six per month. Learning to recognise a few standard omegas is a genuine time-saver when checking applied models.
Edge cases
The angular frequency omega is required to be positive in the standard form.
Discussion prompt
What would a negative omega describe physically, and how is such a case handled?
Hint: What does the sign of an angular velocity mean, from Lesson 10.1c?
Answer:
A negative angular frequency describes rotation in the opposite direction — clockwise rather than counterclockwise, or a wave travelling the other way.
\[ \sin(-\omega t + \phi) = -\sin(\omega t - \phi) \]
It is handled by the odd identity: the negative comes out of the sine and is absorbed into the amplitude A, which may then be negative without affecting the amplitude, since the amplitude is the absolute value. So a negative omega is always removable, which is why the standard form can insist on a positive one without losing any generality.
Section
Section 3
Concept
For an object moving on a circle, the four parameters map onto four features of the apparatus. Reading each off is more reliable than trying to write the formula in one go.
The last of the four is the one that needs an equation rather than a reading. Substituting t equal to zero and solving for the phase is always available and always works.
Figure (svg): A Ferris wheel of radius sixty-four feet on an eight foot platform, with a rider at the lowest point and the height above the ground marked from the ground up through the platform to the centre and out to the rider
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 882-883
Picture it
Every number in the model appears somewhere in this picture.
Figure (svg): A Ferris wheel of radius sixty-four feet on an eight foot platform, with a rider at the lowest point and the height above the ground marked from the ground up through the platform to the centre and out to the rider
The centre is 72 feet up because the platform is 8 and the radius is 64. That single addition is the whole of the baseline calculation, and it is the step most easily skipped.
Worked example
Example 11.1.1. A wheel of diameter 128 feet on an 8 foot platform, two revolutions in 127 seconds, rider starting at the bottom.
\[ \text{Find } h(t), \text{ the height above the ground } t \text{ seconds after passing the lowest point.} \]
Find the amplitude from the radius
Why: The diameter is 128, so the radius is 64.
\[ A = 64 \]
Find the baseline from the centre's height
Why: Eight feet of platform plus 64 feet of radius.
\[ B = 72 \]
Find omega from the rotation rate
Why: Two revolutions in 127 seconds gives a period of 63.5.
\[ \omega = 4 \pi / 127 \]
Find the phase from the starting position
Why: At t equal to zero the height is 8, the lowest point, so the sine must be negative one.
\[ \phi = -\frac{\pi}{2} \]
Figure (svg): The solution to Worked example height on a Ferris wheel shown as a ladder of expressions, one row per legal move
\[ h(t) = 64\sin\left(\frac{4\pi}{127}t - \frac{\pi}{2}\right) + 72 \]
Verify: check three positions
Why: At t equal to zero the sine of negative pi over two is negative one, giving 72 minus 64, which is 8 feet — the bottom, as required. At a quarter period, about 15.9 seconds, the argument is zero and the height is 72, the centre. At half a period the height is 136, the top, matching the stated overall height.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 883-884
Ranking
Derive a sinusoid from a description of circular motion.
Put in order
Why: The first three are readings from the description and can be done in any order among themselves, but all three must precede the phase, since imposing a starting condition requires the other parameters to be in place. The final check is what catches an omitted platform height or a halved period, and it costs three substitutions.
Worked example
The same wheel, timed from a different moment. Only the phase changes.
\[ \text{Find } h(t) \text{ if the clock starts when the rider is level with the centre and rising.} \]
Note that three parameters are unchanged
Why: The apparatus has not changed, only the clock.
Impose the new starting condition
Why: At t equal to zero the height is 72, the baseline.
\[ \sin(\phi) = 0 \]
Use the second condition, that it is rising
Why: A sine at phase zero is increasing; at phase pi it is decreasing.
\[ \phi = 0 \]
Write the model
Why: No phase shift at all.
\[ h(t) = 64 \sin(4 \pi t / 127) + 72 \]
Figure (svg): The solution to Worked example a different starting position shown as a ladder of expressions, one row per legal move
\[ h(t) = 64\sin\left(\frac{4\pi}{127}t\right) + 72 \]
Verify: check the height is rising at t = 0
Why: Just after zero the argument is small and positive, so the sine is small and positive and the height exceeds 72. The rider is rising, as required. Note that the height condition alone did not determine the phase — phase pi would also give 72 — and the second condition was genuinely needed.
Trap
\[ h(t) = 64\sin\left(\tfrac{4\pi}{127}t - \tfrac{\pi}{2}\right) + 64 \]
Take the baseline as the radius, since that is where the centre is above the bottom of the wheel
Why: The radius is the natural number in view and it does locate the centre relative to the wheel.
But the height is measured from the ground, and the wheel sits on an 8 foot platform. The centre is 72 feet above the ground, not 64.
\[ B = 8 + 64 = 72 \]
Measure the baseline from the reference level the question uses
Why: Here that is the ground, so both the platform and the radius contribute.
The check is to evaluate at the lowest point: the model must give the physically correct minimum. With B equal to 72 the minimum is 8, matching the platform height. With B equal to 64 it would be zero, which would put the rider at ground level.
Faded example
A wheel of radius 10 m has its centre 12 m above the ground and turns once every 40 seconds. A rider starts at the top.
Fill in the blanks
A = 10, \; B = 12, \; \omega = pi/20, \quad h(0) = 22 \;\Longrightarrow\; \sin\phi = 1 \;\Longrightarrow\; \phi = pi/2
Why: Two pi over 40 gives pi over twenty. Starting at the top means the height is the maximum, 22, so the sine must equal 1 and the phase is pi over two. The model is 10 sine of the quantity pi t over twenty plus pi over two, plus 12 — which at t equal to zero gives 10 plus 12, namely 22.
Prediction
A rider on a Ferris wheel starts at the lowest point.
Predict first
What must the phase be, in the sine form?
Correct: -pi/2.
Why: Starting at the lowest point means the height is at its minimum, so the sine must equal negative one at t equal to zero. The sine is negative one at negative pi over two, which is therefore the phase. Note that three pi over two would also give a sine of negative one and is equally correct — the phase is determined only up to a multiple of two pi, and negative pi over two is simply the tidiest representative.
Missing information
A problem says: a wheel of radius 15 metres turns once a minute. Find the height of a rider as a function of time.
Discussion prompt
Name everything missing and say what each would supply.
Hint: Three of the four parameters are determined. Which, and what is left?
Answer:
The amplitude is 15 from the radius and omega is two pi over 60 from the rotation rate. Two things are missing.
The height of the centre above the ground is unstated, so the baseline is unknown — the model cannot say whether the lowest point is at ground level or ten metres up. And the starting position is unstated, so the phase is unknown.
This is the standard shape of an incomplete modelling problem: the parameters describing the apparatus are given and the parameters describing its placement and timing are not. Both kinds are needed, and a problem that omits either is genuinely unanswerable rather than merely hard.
Section
Section 4
Concept
Given a table of measurements that look periodic, three parameters come directly from the data and the fourth comes from where the maximum falls. The book is candid that this is not an exact science.
A calculator's regression feature will produce a different and usually slightly better fit, because it minimises total error rather than matching the two extreme points exactly. Both are legitimate; the hand method is transparent and the regression is optimal.
Figure (svg): Twelve monthly daylight measurements for Fairbanks plotted as points, with a fitted sinusoid through them, showing the maximum in June and the minimum in December
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 884-885
Picture it
Twelve monthly measurements and the sinusoid fitted to them by hand.
Figure (svg): Twelve monthly daylight measurements for Fairbanks plotted as points, with a fitted sinusoid through them, showing the maximum in June and the minimum in December
The fit is good but not perfect, and it could not be: a hand fit passes exactly through the two extremes and accepts whatever error that leaves elsewhere.
Worked example
Example 11.1.2. Hours of daylight in Fairbanks, monthly, with a maximum of 21.8 and a minimum of 3.3.
\[ \text{Fit } H(t) = A\sin(\omega t + \phi) + B \text{ to the twelve monthly values.} \]
Take the baseline as the average of the extremes
Why: Add 21.8 and 3.3 and halve.
\[ B = 12.55 \]
Take the amplitude as half the difference
Why: Subtract and halve.
\[ A = 9.25 \]
Take the period from the context
Why: Daylight repeats annually, and t is measured in months.
\[ T = 12, \omega = \frac{\pi}{6} \]
Find the phase from the maximum's position
Why: The maximum is at month 6, and a sine peaks a quarter period after its cycle starts.
\[ \phi = -\frac{\pi}{2} \]
Figure (svg): The solution to Worked example fit the daylight data shown as a ladder of expressions, one row per legal move
\[ H(t) = 9.25\sin\left(\frac{\pi}{6}t - \frac{\pi}{2}\right) + 12.55 \]
Verify: test at the two extremes
Why: At t equal to 6 the argument is pi minus pi over two, which is pi over two, so the sine is 1 and the value is 21.8 — the observed maximum exactly. At t equal to 12 the argument is two pi minus pi over two, so the sine is negative one and the value is 3.3, the observed minimum. The fit is exact at both extremes by construction.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 884-885
Fill the middle
Monthly temperatures range from a minimum of negative 4 to a maximum of 26 degrees.
Fill in the blanks
B = \frac1115 = ___, \qquad A = \frac______ = ___
Why: The baseline is 11 and the amplitude is 15. Note that the minimum being negative does not complicate anything: the average of 26 and negative 4 is 11, and half their difference is 15. Checking, 11 plus 15 is 26 and 11 minus 15 is negative 4, so both extremes are reproduced.
Worked example
A hand fit is exact where it was built and approximate elsewhere. Checking that is part of the modelling.
\[ \text{Compare the model's value at } t = 3 \text{ with the observed } 12.4 \text{ hours.} \]
Substitute t equal to 3
Why: The argument is pi over two minus pi over two, which is zero.
\[ \text{argument } = 0 \]
Evaluate
Why: The sine of zero is zero, so the value is the baseline.
\[ H(3) = 12.55 \]
Compare with the observation
Why: The measured value was 12.4 hours.
\[ \text{error } 0.15\text{ hours} \]
Interpret
Why: Nine minutes out over a twelve-hour value, about one percent.
Figure (svg): The solution to Worked example assess the fit away from the extremes shown as a ladder of expressions, one row per legal move
\[ H(3) = 12.55, \quad \text{observed } 12.4, \quad \text{error } 0.15 \text{ h} \]
Verify: say why an error is expected
Why: The hand method forced the curve through the maximum and minimum exactly, so the remaining error is distributed across the other ten points. A regression would spread the error more evenly, fitting the extremes slightly less well and the middle slightly better. Neither is wrong; they optimise different things.
Error analysis
A student fits a sinusoid to data ranging from 3.3 to 21.8.
Annotate
On: \( A = 21.8, \qquad B = 0 \)
The two vertical parameters must be found together from the same pair of numbers, half the sum and half the difference. Taking either one alone will get the other wrong, and the sanity check is whether the model's minimum is physically possible.
Prediction
You fit a sinusoid by hand to twelve data points, forcing it through the maximum and minimum.
Predict first
Where will the fit be worst?
Correct: At the points in between.
Why: The method uses only the two extreme values, so the curve passes through them exactly and the other ten points contribute nothing to the fit. Whatever error exists is therefore concentrated away from the extremes. A regression uses all twelve and distributes the error more evenly, which usually makes it slightly worse at the extremes and better in between — a genuine trade rather than a strict improvement.
Comparison
Fill the blanks from memory. Two methods, two different goals.
Comparison matrix
| Hand fit | Calculator regression | |
|---|---|---|
| Uses | the two extreme values | all the data points |
| Exact at | the maximum and minimum | generally nowhere |
| Minimises | nothing; it matches two points | total squared error |
| Transparent | yes; every step is visible | no; the method is internal |
Neither is better in the abstract. The hand fit is checkable and explains itself; the regression is optimal by a stated criterion. Knowing which you used, and why, is what matters.
Explain it to yourself
The book says that fitting a sinusoid to data manually is not an exact science.
Discussion prompt
Explain what is inexact about it. Is the inexactness in the method, or in the data, or in the assumption?
Hint: Would perfect data make the method exact?
Answer:
Some of it is the method: choosing to match the two extremes exactly is one choice among many, and a different choice gives a different curve.
But the deeper inexactness is in the assumption. Real daylight hours are not exactly sinusoidal — the underlying astronomy produces a shape close to a sinusoid but not identical to one — so no sinusoid fits perfectly however it is chosen.
That distinction matters when reporting results. The model is an approximation to reality, not merely an approximation to the data, and quoting it to four decimal places would claim a precision the model does not have.
Section
Section 5
Concept
Amplitude, baseline and period are all read off. The phase is not: it must be deduced from where in the cycle something happens, and the quarter-period relationship is what converts that.
The substitution method is worth preferring when the given information is a value rather than an extreme, because it needs no reasoning about where in the cycle you are.
Figure (svg): A diagram showing that a sine curve reaches its maximum a quarter period after the start of its cycle, so knowing where the maximum occurs locates the phase shift
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 885-885
Picture it
A sine peaks a quarter of the way through its cycle, and that single fact converts a maximum's position into a phase.
Figure (svg): A diagram showing that a sine curve reaches its maximum a quarter period after the start of its cycle, so knowing where the maximum occurs locates the phase shift
A cosine would peak at the very start of its cycle instead, which is exactly the quarter-period difference between the two forms and the reason the phases differ between a sine model and a cosine model of the same situation.
Worked example
The daylight data again. The maximum is in month 6 and the period is 12.
\[ \text{Find } \phi \text{ for } H(t) = 9.25\sin\left(\tfrac{\pi}{6}t + \phi\right) + 12.55 \text{ with a maximum at } t = 6. \]
Find where the cycle starts
Why: A sine peaks a quarter period after starting, and a quarter of 12 is 3.
\[ \text{cycle starts at } t = 3 \]
That start is the phase shift
Why: By definition the phase shift is where the cycle begins.
\[ \text{phase shift } = 3 \]
Convert to phi
Why: The phase shift is negative phi over omega.
\[ 3 = -\frac{\phi}{\frac{\pi}{6}} \]
Solve
Why: Multiply through.
\[ \phi = -\frac{\pi}{2} \]
Figure (svg): The solution to Worked example phase from the maximum's position shown as a ladder of expressions, one row per legal move
\[ -\frac{\phi}{\omega} = 3 \;\Longrightarrow\; \phi = -3\omega = -3 \cdot \frac{\pi}{6} = -\frac{\pi}{2} \]
Verify: check the maximum lands at t = 6
Why: At t equal to 6 the argument is pi minus pi over two, which is pi over two, and the sine of pi over two is 1 — the maximum. The phase does put the peak where the data says it is.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 885-885
Prediction
A sine model has period 20 and its maximum occurs at t equal to 8.
Predict first
Where does its cycle begin?
Correct: t = 3.
Why: A sine peaks a quarter period after its cycle starts, and a quarter of 20 is 5. So the cycle began 5 units before the peak, at t equal to 3. That is the phase shift, and phi is then negative omega times 3. The commonest slip is to add rather than subtract, giving 13, which would put the peak a quarter period before the cycle started.
Worked example
The alternative route, which needs no reasoning about quarter periods.
\[ \text{A model } 5\sin(2t + \phi) + 9 \text{ has value } 14 \text{ at } t = 0. \text{ Find } \phi. \]
Substitute the known point
Why: Put t equal to zero and the value equal to 14.
\[ 5 \sin(\phi) + 9 = 14 \]
Isolate the sine
Why: Subtract 9 and divide by 5.
\[ \sin(\phi) = 1 \]
Solve
Why: The sine equals 1 only at pi over two, plus multiples of two pi.
\[ \phi = \frac{\pi}{2} \]
Note what this means
Why: The value 14 is the maximum, so t equal to zero is a peak.
Figure (svg): The solution to Worked example phase by substitution shown as a ladder of expressions, one row per legal move
\[ \phi = \frac{\pi}{2} \]
Verify: confirm 14 is the maximum
Why: The baseline is 9 and the amplitude 5, so the maximum is 14. The substitution forced the sine to 1, which only happens at a peak — consistent, and a useful check that the given value was attainable at all. A value of 15 would have given a sine of 1.2 and revealed the problem as impossible.
Trap
\[ \text{cycle starts at } t = 3, \; \omega = \tfrac{\pi}{6} \;\Longrightarrow\; \phi = 3 \]
Take the phase shift as the phase
Why: The two names are almost the same and the distinction is easy to lose.
But the phase shift is negative phi over omega, not phi. With phi equal to 3 the shift would be negative 3 divided by pi over six, about negative 5.7 — nowhere near the intended 3.
\[ -\frac{\phi}{\omega} = 3 \;\Longrightarrow\; \phi = -3\omega = -\frac{\pi}{2} \]
Solve the relationship rather than reading it as an equality
Why: The phase and the phase shift differ by a factor of negative omega.
This is the same distinction as in Lesson 10.5a and it costs marks in exactly the same way. The safest route is substitution: put a known point into the model and solve for phi directly, which cannot confuse the two because phi is the only unknown.
Faded example
A model has omega equal to pi over 4 and its cycle begins at t equal to 2.
Fill in the blanks
-\fracpi/4-pi/2 = 2 \;\Longrightarrow\; \phi = -2 \cdot ___ = ___
Why: Phi is negative omega times the phase shift, so negative two times pi over four, which is negative pi over two. Checking: at t equal to 2 the argument is pi over two minus pi over two, which is zero, so the sine is zero and rising — exactly the start of a cycle.
Discrimination
Two routes to the phase, suited to different given information.
Sort into buckets
Sort each piece of given information by which method it suits.
Real world
A tide table records high water at 03:12 and the next high water at 15:37, with a range from 0.8 m to 4.6 m above chart datum.
Discussion prompt
Build a complete sinusoidal model of the depth as a function of hours after midnight, naming each parameter as you find it.
Hint: The time between consecutive high waters is one period.
Answer:
\[ T = 15\text{:}37 - 03\text{:}12 = 12\text{ h }25\text{ min} = 12.417 \text{ h} \;\Longrightarrow\; \omega = \frac{2\pi}{12.417} \approx 0.506 \]
\[ B = \tfrac{4.6 + 0.8}{2} = 2.7, \qquad A = \tfrac{4.6 - 0.8}{2} = 1.9 \]
The maximum is at t equal to 3.2 hours, and a sine peaks a quarter period later than its start, so the cycle began at 3.2 minus 3.104, about 0.096 hours. Then phi is negative omega times that, about negative 0.049.
\[ d(t) = 1.9\sin\big(0.506t - 0.049\big) + 2.7 \]
Every parameter came from the table: the two high water times gave the period, the range gave the amplitude and baseline, and the position of high water gave the phase. This is exactly how a published tide table is generated in reverse, and the same four readings would let you extend it to any day.
Comparison
Fill the blanks from memory. The same four parameters, found two different ways.
Comparison matrix
| From a physical setup | From data | |
|---|---|---|
| Amplitude | the radius or swing | half the range of the data |
| Baseline | the height of the centre | the average of the extremes |
| Period | from the stated rotation rate | the repeat interval |
| Phase | from the starting position | from where the maximum falls |
| Accuracy | exact, if the model is right | approximate |
The last row is the real difference. A physical derivation is exact given its assumptions; a data fit is an approximation to measurements that were themselves approximate.
Pattern
Whether the sinusoid comes from a setup or from data, the same five moves cover it.
Writing all three of period, ordinary frequency and angular frequency down whenever any one is given costs two lines and removes the commonest silent error in the chapter.
OpenStax Algebra and Trigonometry 2e, §8.1 Graphs of the Sine and Cosine Functions §8.1
Check
Parameters from a description.
Check your understanding
A quantity oscillates between 40 and 60 with a period of 8. What are its amplitude and baseline?
Answer: C
Why: The baseline is the average of the extremes, half of 100, which is 50. The amplitude is half the difference, half of 20, which is 10. Checking, 50 plus 10 is 60 and 50 minus 10 is 40, reproducing both extremes.
Check
Rates. Which kind were you given?
Check your understanding
A wheel turns at 20 revolutions per minute. What is omega, in radians per second?
Answer: C
Why: Twenty revolutions per minute is one third of a revolution per second. Multiplying by two pi radians per revolution gives two pi over three radians per second, about 2.09. Checking, the period is two pi divided by that, which is 3 seconds — and 20 revolutions a minute is indeed one every 3 seconds.
Check
The phase. Count quarter periods.
Check your understanding
A sine model has period 12 and reaches its maximum at t equal to 7. Where does its cycle begin?
Answer: A
Why: A sine peaks a quarter period after its cycle begins, and a quarter of 12 is 3. So the cycle began 3 units before the peak, at t equal to 4. That value is the phase shift, and phi would then be negative omega times 4.
Real world
A hospital monitors a patient's blood pressure, which oscillates with each heartbeat between a diastolic minimum of 80 and a systolic maximum of 120 millimetres of mercury, at a pulse of 72 beats per minute.
Discussion prompt
Build a sinusoidal model of the pressure, and then explain why clinicians quote the two extremes rather than the amplitude and baseline the model uses.
Hint: Convert the pulse to an ordinary frequency first, then to omega.
Answer:
\[ B = \tfrac{120 + 80}{2} = 100, \qquad A = \tfrac{120 - 80}{2} = 20 \]
\[ f = \tfrac{72}{60} = 1.2 \text{ Hz} \;\Longrightarrow\; \omega = 2.4\pi \text{ s}^{-1} \]
\[ P(t) = 20\sin(2.4\pi t + \phi) + 100 \]
Clinicians quote 120 over 80 rather than an amplitude of 20 about a baseline of 100 because the two extremes are what is measured and what carries the clinical meaning. Systolic pressure is what the arteries must withstand at peak; diastolic is what perfuses the heart muscle itself between beats.
The model's parameters are exactly the same information rearranged — the mean pressure of 100 and the pulse pressure of 40, which is twice the amplitude, are both quantities clinicians do use in other contexts. Which pair you quote depends on what the number is for, and the arithmetic connecting them is the same half-sum and half-difference as everywhere else in this lesson.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
A problem states that a wheel makes 4 revolutions in 10 seconds. What is the period?
Correct: 2.5 seconds.
\[ T = \frac{10}{4} = 2.5 \text{ s}, \qquad f = \frac{4}{10} = 0.4 \text{ Hz}, \qquad \omega = 0.8\pi \]
Why: The period is the time for one complete cycle, so it is the total time divided by the number of revolutions: 10 divided by 4, which is 2.5 seconds. The distractor 0.4 is the ordinary frequency, 4 revolutions per 10 seconds, which is the reciprocal — a genuinely useful quantity but not the period. Taking the stated 10 seconds as the period is the commonest error and would make the model run four times too slowly.
Explain it
They can graph a sinusoid from a formula but freeze when a word problem describes one instead.
Discussion prompt
In no more than five sentences, give them a procedure for turning any such description into a formula.
Hint: How many numbers do they need, and where does each come from?
Answer:
A usable answer: you need exactly four numbers and each has one place to look. The highest and lowest values give you two at once — their average is the baseline and half their difference is the amplitude. The time for one full cycle gives the period, and two pi divided by that is omega.
The fourth, the phase, is the only one needing thought: find some moment where you know what is happening, substitute it into the formula, and solve for the phase. Then check by putting the highest and lowest moments back in — if the formula does not reproduce them, something is wrong and you will find it in seconds.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The vertical parameters are fixed by computing half the sum and half the difference together and then reconstructing the extremes as a check. Rate conversion is fixed by writing all three of period, frequency and omega whenever any one is given. Physical derivations are fixed by reading three parameters off the apparatus and then imposing the starting condition for the fourth. The phase is fixed by counting quarter periods, or more safely by substituting a known point and solving. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw a general sinusoid with a baseline well above the axis, and label all four parameters on the picture with both their formula and their physical meaning. Underneath, write the three-way conversion between period, ordinary frequency and angular frequency, and beside it note which of the three a problem most often states. In the middle left, draw a Ferris wheel on a platform and derive the height model completely, labelling where each of the four parameters came from. In the middle right, sketch a dozen data points that look sinusoidal and write the three formulas that extract the baseline, amplitude and period from them. In the bottom left, draw one cycle of a sine and mark the quarter-period point where it peaks, writing the rule that converts a maximum's position into a phase shift. In the bottom right, work one complete example of your own: invent a physical situation, state its four facts in words, and produce the model. Finally, circle the one parameter that data never gives you directly, and write the two ways of getting it.
The circled parameter is the phase, and the two ways are counting quarter periods from a named landmark, or substituting a known point into the model and solving for phi. The second is the safer of the two.
Recap
Five things, and the second is where the silent errors live.
| If the question says | Your first move is |
|---|---|
| It varies between two values | Half the sum is B, half the difference is A |
| It repeats every so often | That is the period; omega is two pi over it |
| So many revolutions per minute | Multiply by two pi and fix the time unit |
| It starts at the lowest point | Substitute t = 0 and solve for the phase |
| The maximum occurs at this time | Count back a quarter period for the cycle start |
Sinusoids so far have modelled things that genuinely go round. The next lesson models something that does not: an object bouncing on a spring, whose motion turns out to be sinusoidal for a reason that has nothing to do with circles — and which then extends to damped and forced motion, where the amplitude itself varies.
Stitz & Zeager, College Trigonometry, Ch. 11 Applications of Trigonometry — §11.1 Applications of Sinusoids §11.1, pp. 881-885 — everything on these slides traces back here
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