Recovering the amplitude, period, and vertical and horizontal shifts from a sinusoid's graph, choosing between sine and cosine and deciding whether to reflect, converting a period or frequency into b, modelling circular motion, and fitting a sinusoidal regression to data.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 14 — Trigonometric Graphs, Identities, and Equations
Write Trigonometric Models
Objectives
Five outcomes. From a picture or a table to an equation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-945 — the lesson these objectives are drawn from
Warm-up
Lesson 14.2 graphed a function once you knew a, b, h and k.
Discussion prompt
Now reverse it. A wave has maximum 5 and minimum negative 1, and crosses its midline going upward at x equal to zero, repeating every pi over 2. What is its equation?
Hint: Where is the middle of that range, and how far does it swing?
Answer:
\[ k = \tfrac{5+(-1)}{2} = 2, \quad |a| = \tfrac{5-(-1)}{2} = 3 \]
\[ \text{period } \tfrac{\pi}{2} = \tfrac{2\pi}{b} \;\Longrightarrow\; b = 4 \]
Crossing the midline going up at zero is what a sine does, so h is 0 and the model is y equal to 3 sine 4x plus 2. Four readings, four constants, and the equation follows.
Concept
A sinusoid is the graph of a sine or cosine function. To write its equation, take the mean of the maximum and minimum for k and half their difference for the amplitude, read the period to get b, and choose the function whose starting behaviour matches so that h is zero.
sinusoid — The graph of a sine or cosine function, including any amplitude change, period change, reflection or translation. Every sinusoid can be written as a sine model and equally as a cosine model.
\[ k = \frac{M+m}{2}, \quad |a| = \frac{M-m}{2} \]
Sinusoidal regression uses every data point rather than just the extremes, and is what real measurements usually call for.
Figure (svg): Two columns comparing the four-step method with sinusoidal regression
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-943
Section
Section 1
Concept
The vertical shift k is the mean of the maximum and minimum values, and the amplitude is half their difference. Both come straight from two readings off the graph.
\[ k = \frac{M+m}{2}, \qquad |a| = \frac{M-m}{2} \]
These two formulas work whatever the midline is, which is why they replace reading the amplitude off the axis.
Figure (svg): A sinusoid with its maximum, minimum and midline marked
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-941 — Solve a multi-step problem
Picture it
Example 1: a sinusoid with maximum 5 and minimum negative 1.
Figure (svg): A sinusoid with its maximum, minimum and midline marked
The mean of 5 and negative 1 is 2, the midline. Half their difference is 3, the amplitude. The rest of the graph decides the other two constants.
Worked example
Example 1, Steps 1 and 2.
\[ \text{A sinusoid has } M = 5 \text{ and } m = -1. \text{ Find } k \text{ and } |a|. \]
Read the maximum and minimum
Why: From the highest and lowest points.
\[ M = 5, m = -1 \]
Take their mean
Why: Five plus negative 1, over 2.
\[ k = 2 \]
Take half their difference
Why: Five minus negative 1, over 2.
\[ | a | = 3 \]
Interpret
Why: The wave is centred at 2 and swings 3 either way.
\[ \text{from } -1\text{ to } 5 \]
Figure (svg): A sinusoid with its maximum, minimum and midline marked
\[ k = 2, \quad |a| = 3 \]
Verify: check the two extremes come back
Why: Midline plus amplitude is 2 plus 3, which is 5, and midline minus amplitude is 2 minus 3, which is negative 1. Reconstructing the extremes from k and a is the fastest possible check on both.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-941
Fill the middle
Example 1, Step 2.
Fill in the blanks
k = \frac2___ = \frac______ = ___
Why: Two. The mean of the two extremes is always the level the wave is centred on.
Worked example
Guided Practice 1 and 2.
\[ \text{Find } k \text{ and } |a| \text{ for a graph with } M = 2, m = -2, \text{ and for one with } M = 1, m = -3. \]
First: take the mean
Why: Two plus negative 2, over 2.
\[ k = 0 \]
First: half the difference
Why: Two minus negative 2, over 2.
\[ | a | = 2 \]
Second: take the mean
Why: One plus negative 3, over 2.
\[ k = -1 \]
Second: half the difference
Why: One minus negative 3, over 2.
\[ | a | = 2 \]
Figure (svg): The solution to Worked example two more graphs shown as a ladder of expressions, one row per algebraic move
\[ k = 0, |a| = 2; \qquad k = -1, |a| = 2 \]
Verify: notice the two share an amplitude
Why: Both swing 2 either side of their midline, yet one is centred on the axis and the other 1 unit below it. Amplitude and midline are genuinely independent readings, and confusing them is what makes reading the amplitude off the axis unreliable.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 942-942
Trap
\[ M = 5, \; m = -1 \]
Read the amplitude off the highest point
Why: The graph reaches 5, so 5 is taken as the amplitude.
\[ |a| = 5 \quad \text{(wrong)} \]
That would make the graph swing from negative 5 to 5 about the axis, which is not what the picture shows at all.
\[ |a| = \frac{M-m}{2} = \frac{5-(-1)}{2} = 3 \]
Take half the difference between the extremes
Why: Amplitude is measured from the midline.
\[ \text{from } 2-3 = -1 \text{ to } 2+3 = 5 \]
Only when the midline happens to be the axis do the maximum and the amplitude coincide.
Fill the middle
Example 1, Step 4.
Fill in the blanks
|a| = \frac3___ = \frac______ = ___
Why: Three. Half the total swing, measured from the midline outward to either extreme.
Matching
Mean and half difference.
Match the pairs
Why: The last is the jump rope of Example 2, where the numbers are inches rather than plain units but the arithmetic is identical. Only the second has its midline on the axis, and only there would the maximum and amplitude agree.
Prediction
Commit before reasoning.
Predict first
For which sinusoids is the amplitude equal to the maximum value?
Correct: Only those whose midline is the x-axis, that is with k equal to zero.
\[ k = 0 \;\Longrightarrow\; M = |a| \]
Why: When k is zero the maximum is 0 plus the amplitude, so the two coincide, which is why the parent graphs of Lesson 14.1 made them look like the same thing. As soon as the wave is lifted or lowered they part company, and the physical models of this lesson are almost all lifted — a height or a temperature is rarely centred on zero. Using the mean and half-difference formulas works in both cases and is therefore the habit worth forming.
Section
Section 2
Concept
Look at what the graph does where the input is zero. Rising through the midline calls for sine, falling through it for negative sine, starting at the maximum for cosine, and starting at the minimum for negative cosine. Choosing well makes h zero.
\[ \sin, \; -\sin, \; \cos, \; -\cos \]
Any sinusoid can be written all four ways with a suitable horizontal shift, but only one of the four needs no shift at all.
Figure (svg): The four possible starting shapes at x equal to zero
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-942 — Decide whether to use sine or cosine
Picture it
The four shapes, each at the origin.
Figure (svg): The four possible starting shapes at x equal to zero
Reading the graph at x equal to zero picks exactly one of the four, and with it both the function and the sign of a.
Worked example
Example 1, Step 3.
\[ \text{A sinusoid crosses its midline } y = 2 \text{ going upward at } x = 0. \text{ Which function?} \]
Look at the value at x equal to 0
Why: It is on the midline.
Look at the direction
Why: Rising.
Conclude the function and sign
Why: Sine with a positive.
\[ a > 0 \]
Set the horizontal shift
Why: No shift is needed.
\[ h = 0 \]
Figure (svg): The four possible starting shapes at x equal to zero
\[ y = a\sin bx+k, \; a > 0 \]
Verify: confirm with the amplitude and midline
Why: Putting the earlier readings in gives y equal to 3 sine 4x plus 2, and at x equal to pi over 8 that is 3 times 1 plus 2, or 5 — the stated maximum in the right place. Substituting a known point is the final check on the whole model.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-941
Sorting
Look at the value and direction at zero.
Sort into buckets
Sort each described behaviour at x equal to 0.
The last two describe the same situation in different words, which is the point: a phrase like at its lowest point when the clock starts is a direct instruction to use a negative cosine.
Worked example
Guided Practice 1 and 2.
\[ \text{Write a model for a sinusoid with maximum } 2 \text{ at } x = 0 \text{ and minimum } -2 \text{ at } x = \tfrac{2\pi}{3}; \text{ and one with maximum } 1 \text{ at } x = \tfrac{1}{2} \text{ and minimum } -3 \text{ at } x = \tfrac{3}{2}. \]
First: it starts at the maximum
Why: So a positive cosine.
\[ k = 0, | a | = 2 \]
First: max to min is half a period
Why: Two thirds pi doubled is 4 pi/3.
\[ b = \frac{3}{2} \]
Second: it peaks a quarter period in
Why: So a positive sine with no shift.
\[ k = -1, | a | = 2 \]
Second: max to min is half a period
Why: One doubled is 2.
\[ b = \pi \]
Figure (svg): The solution to Worked example two more choices shown as a ladder of expressions, one row per algebraic move
\[ y = 2\cos\tfrac{3x}{2}; \qquad y = 2\sin\pi x-1 \]
Verify: substitute the given points
Why: The first at x equal to 2 pi over 3 gives 2 cosine pi, which is negative 2 — the stated minimum. The second at x equal to one half gives 2 sine of pi over 2 minus 1, which is 1 — the stated maximum. Both models reproduce the points they were built from.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 942-942
Error analysis
A student models a wave that begins at its lowest point.
Annotate
On: \( y = 36\cos 4\pi t+39 \)
Testing the model at the input the problem describes catches this in one substitution, and it is worth doing before anything else.
Fill the middle
Example 2, Step 3.
Fill in the blanks
\text- \;\Longrightarrow\; a = ___36
Why: Negative 36. Reflecting the cosine puts its starting point at the bottom of the range instead of the top.
Comparison
Fill the blanks. The same wave, two ways.
Comparison matrix
| Question | As a sine | As a cosine |
|---|---|---|
| Starts at | the midline | the maximum |
| Needs h when the graph starts at the maximum | yes, a quarter period | no |
| Amplitude | the same | the same |
| Midline | the same | the same |
Both descriptions are correct for any sinusoid; choosing the one that needs no shift simply saves work and removes a place to make a sign error.
Prediction
Commit before reasoning.
Predict first
A wave starts at its maximum. Could it still be written as a sine model?
Correct: Yes — with a horizontal shift of a quarter period, since sine reaches its maximum a quarter period in.
\[ \sin\!\left(x+\tfrac{\pi}{2}\right) = \cos x \]
Why: Shifting a sine graph left by a quarter period puts its maximum at the start, which is exactly where cosine's is; that is the identity sine of x plus pi over 2 equals cosine of x seen as a graph. So every sinusoid has both a sine form and a cosine form. Choosing the one with h equal to zero is a convenience, not a necessity, and it is why the four-shape check is worth doing first.
Section
Section 3
Concept
The period equals 2 pi over b, so b is 2 pi over the period. If a rate is given instead, the period is its reciprocal, so b is 2 pi times the frequency.
\[ b = \frac{2\pi}{T} = 2\pi f \]
In a graph the period can be measured between consecutive maximums, or as twice the distance from a maximum to the next minimum.
Figure (svg): Converting a period or a rate into the constant b
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-942 — Find the amplitude and period
Picture it
The single equation both routes go through.
Figure (svg): Converting a period or a rate into the constant b
A period of pi over 2 gives b equal to 4; two revolutions a second gives a period of a half and b equal to 4 pi.
Worked example
Example 1, Step 4.
\[ \text{A graph repeats every } \tfrac{\pi}{2}. \text{ Find } b. \]
Write the relationship
Why: Period equals two pi over b.
\[ \frac{\pi}{2} = 2 \pi / b \]
Cross-multiply
Why: B times pi over 2 equals 2 pi.
\[ b \pi = 4 \pi \]
Solve
Why: Divide by pi.
\[ b = 4 \]
Sanity-check
Why: Four cycles fit in the usual 2 pi.
Figure (svg): Converting a period or a rate into the constant b
\[ b = \frac{2\pi}{\pi/2} = 4 \]
Verify: check by going back the other way
Why: Two pi over 4 is pi over 2, the stated period. Reversing the conversion is a two-second check that catches multiplying where you should have divided.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-941
Fill the middle
Example 1, Step 4.
Fill in the blanks
b = \frac4___ = 2\pi \cdot \frac______ = ___
Why: Four. The pi cancels, leaving a whole number, which is what makes periods that are simple multiples of pi so convenient.
Worked example
Example 2, Step 4.
\[ \text{A rope makes } 2 \text{ revolutions per second. Find } b. \]
Convert the rate to a period
Why: Two turns a second means half a second each.
\[ T = 0.5 \]
Write the relationship
Why: Half equals two pi over b.
\[ 0.5 = 2 \pi / b \]
Solve
Why: B equals 2 pi over 0.5.
\[ b = 4 \pi \]
Check
Why: Two pi over 4 pi is one half.
Figure (svg): The solution to Worked example b from a rate shown as a ladder of expressions, one row per algebraic move
\[ b = 2\pi f = 2\pi(2) = 4\pi \]
Verify: notice the two routes agree
Why: Going through the period gives 2 pi over 0.5, and going through the frequency gives 2 pi times 2, and both are 4 pi. Period and frequency being reciprocals is exactly what makes the two formulas the same formula.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 942-942
Trap
\[ \text{maximum at } x = \tfrac{1}{2}, \text{ minimum at } x = \tfrac{3}{2} \]
Take the gap as the period
Why: One unit between the two extremes.
\[ T = 1 \;\Longrightarrow\; b = 2\pi \quad \text{(wrong)} \]
A maximum and the next minimum are half a period apart, not a whole one. The period here is 2.
\[ T = 2\!\left(\tfrac{3}{2}-\tfrac{1}{2}\right) = 2 \]
Double the gap between consecutive extremes
Why: Or measure between consecutive maximums.
\[ b = \frac{2\pi}{2} = \pi \]
Measuring maximum to maximum avoids the doubling entirely, and is the safer reading when the graph shows enough of the curve.
Fill the middle
Guided Practice 2.
Fill in the blanks
\text2 \tfrac______, \text___ \tfrac______ \;\Longrightarrow\; T = 2(1) = ___
Why: Two. Consecutive extremes are half a period apart, so the gap has to be doubled before converting to b.
Matching
Period or frequency, one equation.
Match the pairs
Why: The last needed the doubling step: max to min is 2 pi over 3, so the period is 4 pi over 3 and b is 2 pi divided by that, which is three halves. Missing the doubling would have given b equal to 3.
Prediction
Commit before reasoning.
Predict first
On a graph, which measurement gives the period most reliably?
Correct: From one maximum to the next maximum, since that is a full cycle.
\[ \text{peak to peak} = T; \quad \text{peak to trough} = \tfrac{T}{2} \]
Why: Peak to peak is exactly one period with no factor to remember. Maximum to minimum is half a period and midline crossing to the next crossing is also half, so both need doubling and both invite the error. When the graph shows two peaks, use them; when it does not, double carefully and write the doubling down as its own step.
Section
Section 4
Concept
For an object moving in a circle, the highest and lowest points give the midline and amplitude, the rate of rotation gives b, and where the object starts chooses between sine and cosine and fixes the sign of a.
\[ h = -36\cos 4\pi t+39 \]
Stating the starting position is what removes the need for a horizontal shift, which is why problems are usually worded that way.
Figure (svg): A jump rope's height modelled as a reflected cosine wave
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 942-942 — Model circular motion
Picture it
Example 2: a rope swinging between 3 and 75 inches, twice a second.
Figure (svg): A jump rope's height modelled as a reflected cosine wave
The rope starts at the bottom, so the model is a reflected cosine, and its midline of 39 inches is the height of the rope's centre of rotation.
Worked example
Example 2.
\[ \text{A rope's highest point is } 75 \text{ in and lowest } 3 \text{ in, at } 2 \text{ revolutions per second, starting at the lowest point.} \]
Find k and the amplitude
Why: Mean and half difference of 75 and 3.
\[ k = 39, | a | = 36 \]
Choose the function
Why: It starts at the minimum.
Set the sign of a
Why: Reflected.
\[ a = -36 \]
Find b from the rate
Why: Two turns a second means period 0.5.
\[ b = 4 \pi \]
Figure (svg): A jump rope's height modelled as a reflected cosine wave
\[ h = -36\cos 4\pi t+39 \]
Verify: test the model at t equal to 0
Why: Negative 36 times cosine 0 plus 39 is negative 36 plus 39, which is 3 — the stated lowest point. Testing at the moment the problem describes is the single most valuable check on a model.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 942-942
Fill the middle
Example 2, Step 2.
Fill in the blanks
k = \frac39___ = \frac______ = ___
Why: Thirty-nine inches, which is the height of the centre about which the rope rotates.
Worked example
Guided Practice 3.
\[ \text{How does the model change if the lowest point is } 5 \text{ in and the highest is } 70 \text{ in?} \]
Recompute k
Why: Mean of 70 and 5.
\[ k = 37.5 \]
Recompute the amplitude
Why: Half of 70 minus 5.
\[ | a | = 32.5 \]
Keep the rest
Why: Same rate, same starting point.
\[ b = 4 \pi,\text{ still reflected} \]
Write the model
Why: Substitute the new constants.
\[ h = -32.5 \cos(4 \pi t) + 37.5 \]
Figure (svg): The solution to Worked example adjust the model shown as a ladder of expressions, one row per algebraic move
\[ h = -32.5\cos 4\pi t+37.5 \]
Verify: check both extremes
Why: Midline plus amplitude is 37.5 plus 32.5, which is 70, and midline minus amplitude is 5. Both stated heights come back, and note that both k and a changed — raising the floor by 2 inches and lowering the ceiling by 5 shifts the centre and shrinks the swing at once.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 942-942
Error analysis
A student models a rope swinging between 3 and 75 inches.
Annotate
On: \( h = -72\cos 4\pi t+39 \)
The wrong model would swing from negative 33 to 111 inches, so a rope passing 33 inches below the ground. Checking the extremes catches it at once.
Fill the middle
Example 2, check.
Fill in the blanks
h(0) = -36\cos 0+39 = -36+39 = 3
Why: Three inches, the stated lowest point. That confirms both the reflection and the midline in a single substitution.
Matching
In a circular-motion model.
Match the pairs
Why: Every constant in the model is something you could measure directly, which is what separates a model from a curve merely fitted to points. If a constant has no physical meaning, that is worth investigating.
Ranking
Smallest first.
Put in order
Why: The values are 3, 36, 39, 39 and 75 inches, with the height at one eighth of a second equal to the midline because that is a quarter period in. The amplitude of 36 is a distance rather than a height, and it sits just below the midline numerically by coincidence.
Section
Section 5
Concept
When data is given as a table rather than as a clean maximum and minimum, a graphing calculator's sinusoidal regression fits a model using all the points at once, returning it in the form a sine of bx plus c, plus d.
\[ K = 23.9\sin(0.533t-2.69)+82.4 \]
The constant d plays the role of k, and the c inside the bracket carries the horizontal shift in the unfactored form.
Figure (svg): A sinusoidal regression fitted to monthly energy data
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 943-943 — Use sinusoidal regression
Picture it
Example 3: monthly energy use at Cape Canaveral.
Figure (svg): A sinusoidal regression fitted to monthly energy data
No data point lies exactly on the curve, and none has to. The regression balances all of them, which is what makes it right for measurements.
Worked example
Example 3.
\[ \text{A regression on monthly energy data returns } a = 23.9, b = 0.533, c = -2.69, d = 82.4. \text{ Write and interpret the model.} \]
Write the model
Why: In the calculator's form.
\[ K = 23.9 \sin(0.533 t - 2.69) + 82.4 \]
Interpret d
Why: The midline.
\[ \text{average use about } 82, 400\text{ kWh} \]
Interpret a
Why: The amplitude.
\[ \text{swings } 23, 900\text{ either way} \]
Interpret b
Why: Two pi over b is the period.
\[ \text{about } 11.8\text{ months} \]
Figure (svg): A sinusoidal regression fitted to monthly energy data
\[ T = \frac{2\pi}{0.533} \approx 11.8 \]
Verify: check the model against a data point
Why: At t equal to 1 the model gives about 62.5, and the recorded value was 61.9 — within one percent. A regression is not expected to pass through any point exactly, so agreement to within the natural scatter of the data is what success looks like.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 943-943
Fill the middle
Example 3.
Fill in the blanks
T = \frac11.8___ \approx ___ \text___
Why: About 11.8 months. The regression was not told the cycle was annual, and it recovered very nearly twelve months from the data alone.
Worked example
Guided Practice 4.
\[ \text{Monthly average temperatures run from } 29 \text{ in January to } 74 \text{ in July. Write a sine model.} \]
Find the midline and amplitude
Why: Mean and half difference of 74 and 29.
\[ k = 51.5, | a | = 22.5 \]
Find b from the period
Why: Twelve months.
\[ b = 2 \pi / 12 =\text{ about } 0.524 \]
Place the peak
Why: Sine peaks a quarter period after h; July is t = 7.
\[ h = 7 - 3 = 4 \]
Write the model
Why: Substitute all four.
\[ T = 22.5 \sin(0.524(t - 4)) + 51.5 \]
Figure (svg): The solution to Worked example model Boston temperatures shown as a ladder of expressions, one row per algebraic move
\[ T \approx 22.5\sin(0.524t-2.09)+51.5 \]
Verify: compare with what a regression would return
Why: A calculator fitting all twelve points returns coefficients close to these but not identical, because it balances every month rather than only the two extremes. The four-step method and the regression agree to within about a degree here, which is reassuring for both.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 943-943
Trap
\[ K(1) = 62.5 \text{ but the data says } 61.9 \]
Conclude the regression is wrong
Why: The model missed a data point.
\[ \text{reject the model} \quad \text{(wrong)} \]
A regression is a best fit, not an interpolation. Real measurements scatter, and a curve through every point would be fitting the noise.
\[ \text{compare the misses with the scatter} \]
Judge the fit by how the residuals look overall
Why: Small and unpatterned is good.
\[ \text{misses of under } 1\% \text{ across the year} \]
The textbook's own verdict is that the model appears to be a good fit, which is exactly the right standard.
Matching
In a sine of bx plus c, plus d.
Match the pairs
Why: The calculator's c is not h itself: factoring b out of bx plus c shows the shift is negative c over b. For this model that is 2.69 over 0.533, about 5.05 months, which puts the annual low in early January.
Comparison
Fill the blanks. Two ways to get a model.
Comparison matrix
| Question | Four-step method | Sinusoidal regression |
|---|---|---|
| Data used | the maximum and minimum only | every data point |
| Tools | arithmetic | a graphing calculator |
| Best for | a stated highest and lowest value | a table of measurements |
| Passes through every point? | through both extremes | usually none exactly |
The hand method is exact about the two points it uses and silent about the rest; the regression is approximate about every point and therefore more trustworthy when the data is noisy.
Prediction
Commit before reasoning.
Predict first
A curve could be found passing exactly through all twelve monthly measurements. Why is the regression preferred?
Correct: Because the measurements contain noise, and a curve through every point would model the noise rather than the underlying cycle.
\[ \text{four constants} \ll \text{twelve data points} \]
Why: Monthly averages vary from year to year for reasons that have nothing to do with the seasons, so insisting on passing through this year's exact figures would build those accidents into the model. A four-constant sinusoid cannot chase every wiggle, and that limitation is a feature: it forces the model to capture only the repeating pattern. Fitting the noise is called overfitting, and a model that does it predicts next year worse, not better.
Comparison
Fill the blanks. Four constants, four questions.
Comparison matrix
| Constant | Question to ask | Answer comes from |
|---|---|---|
| k | what level is it centred on? | the mean of the maximum and minimum |
| a | how far does it swing? | half the difference of the extremes |
| b | how often does it repeat? | 2 pi divided by the period |
| h | where does a cycle begin? | usually zero, if the function is chosen well |
Choosing among sine, negative sine, cosine and negative cosine is what usually makes h zero, which is why that choice is worth making before anything else.
Pattern
Extremes, function, period.
Test the finished model at the input the problem describes. One substitution catches nearly every error.
OpenStax Algebra and Trigonometry 2e, §8.1 Graphs of the Sine and Cosine Functions §8.1
Check
Mean and half difference.
Check your understanding
A sinusoid has maximum 5 and minimum -1. What are its midline and amplitude?
Answer: A
Why: The mean of 5 and -1 is 2, and half their difference is 3.
Check
Match the behaviour at the start.
Check your understanding
A rope is at its lowest point when t = 0, with maximum 75 and minimum 3 inches. Which model?
Answer: A
Why: Starting at the minimum calls for a negative cosine, with amplitude 36 and midline 39.
Check
Consecutive extremes are half a period apart.
Check your understanding
A graph has a maximum at x = 1/2 and the next minimum at x = 3/2. What is b?
Answer: A
Why: The gap of 1 is half a period, so the period is 2 and b is 2 pi over 2.
Real world
In a city the sun rises at 4:31 a.m. on the longest day, day 172, and at 7:44 a.m. on the shortest, day 355. Sunrise times are close to sinusoidal across the year.
Discussion prompt
Write a model for the sunrise time in hours after midnight as a function of the day number.
Hint: Convert both times to decimal hours first.
Answer:
\[ 4\!:\!31 = 4.52, \quad 7\!:\!44 = 7.73 \]
\[ k = \tfrac{7.73+4.52}{2} = 6.125, \quad |a| = \tfrac{7.73-4.52}{2} = 1.605 \]
\[ b = \tfrac{2\pi}{365} \approx 0.0172 \]
\[ S(d) = -1.605\cos\!\left(\tfrac{2\pi}{365}(d-355)\right)+6.125 \]
Sunrise is modelled as about 1.605 cosine units either side of 6:08 a.m., with the latest sunrise on day 355.
The choice of a negative cosine shifted to day 355 is the interesting move. Day 355 is the MAXIMUM sunrise time, so an ordinary cosine shifted there would work; using a negative cosine shifted to day 172, the minimum, would work equally well and is arguably more natural. Both describe the same curve, which is the point of the four-shape check: whichever extreme you know the date of, there is a form that needs no extra arithmetic. Notice also that the two given days are 183 apart, almost exactly half the 365-day period, which is a good sign the sinusoidal assumption is reasonable.
Commit first
Answer, then rate your confidence honestly.
Predict first
A graph has a maximum at x equal to 2 and its next minimum at x equal to 5. Is the period 3?
Correct: No — consecutive extremes are half a period apart, so the period is 6.
\[ T = 2(5-2) = 6 \;\Longrightarrow\; b = \tfrac{\pi}{3} \]
Why: A full cycle runs maximum, midline, minimum, midline, maximum, so a maximum and the next minimum are only halfway through it. Doubling the gap of 3 gives a period of 6 and b equal to pi over 3. Taking the gap as the period is the single most common error in writing a model from a graph, and it halves every prediction the model makes about timing. Measuring peak to peak instead avoids it entirely.
Explain it
They can graph a sine wave from an equation but have never gone the other way.
Discussion prompt
In four sentences or fewer, explain how to find the midline and amplitude from a graph.
Hint: Talk about the highest and lowest points.
Answer:
Find the highest point and the lowest point. The midline is halfway between them, which is their average.
The amplitude is how far the graph reaches above or below that midline, which is half the distance between the two extremes. Those two numbers tell you where the wave sits and how big it is.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the extremes, the mean gives k and half the difference gives the amplitude. For the shapes, ask what the graph does at the start: midline rising, midline falling, at the top, or at the bottom. For periods, measure peak to peak whenever you can, and double explicitly when you cannot. For regressions, d is the midline, a is the amplitude, and 2 pi over b is the period.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a modelling page. Top left: sketch a sinusoid of your own with a midline well away from the axis, mark its maximum and minimum, and compute k and the amplitude from them, showing both formulas. Top right: draw all four starting shapes small, label what each does at the origin, and beside each write a phrase from a word problem that would select it. Middle: take one graph and find b twice, once by measuring peak to peak and once by measuring peak to trough and doubling, checking that the two agree. Bottom left: write a circular-motion model of your own from an invented highest point, lowest point, rotation rate and starting position, then test it at the starting time. Bottom right: sketch a scatter plot of eight invented data points and a fitted sinusoid through them, and note beside it why the curve should not pass through every point.
If any period on your page came from a maximum-to-minimum measurement without doubling, every timing the model predicts is out by a factor of two.
Recap
Five things, and equations recovered from pictures and data.
| If you see | Then |
|---|---|
| A maximum and a minimum | Their mean is k and half their difference the amplitude |
| A graph rising through its midline at zero | Use a positive sine with h equal to zero |
| A graph starting at its lowest point | Use a negative cosine |
| A rate rather than a period | Take the reciprocal, or use b equal to 2 pi times the rate |
| A gap between consecutive extremes | Double it to get the period |
| A table of measurements | Use a sinusoidal regression |
Lesson 14.6 returns to identities with two new ones: formulas for the sine and cosine of a sum or difference of angles.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.5 Write Trigonometric Models §14.5, pp. 941-945 — everything on these slides traces back here
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