The graphs of cosine and sine, and the four numbers that describe every transformation of them. Covers periodicity as the property that makes one cycle sufficient, the fundamental cycle and its five quarter marks, the quarter-mark method for graphing any sinusoid without tracking transformations one at a time, the identification of amplitude, period, phase shift and vertical shift from a formula and from a graph, and the recognition that a cosine plus a sine at the same frequency is itself a single sinusoid.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.5 Graphs of the Trigonometric Functions, pp. 790-800
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 790-800 — the pages these objectives are drawn from
Warm-up
You know the cosine and sine of every special angle. Plotting those values is the whole of the first half of this lesson.
Discussion prompt
Write down the cosine of 0, pi over two, pi, three pi over two and two pi. Now imagine those five values plotted as heights above those five positions. What shape are you drawing?
Hint: Just list the five numbers in order and picture the heights.
Answer:
\[ 1, \; 0, \; -1, \; 0, \; 1 \]
High, level, low, level, high again — a wave. Those five points are the peaks, troughs and crossings of one full cycle, and joining them smoothly gives the cosine curve.
The whole of this section is that observation plus one more: because coterminal angles share their cosine, the pattern repeats forever in both directions. Five points determine one cycle, and one cycle determines everything.
Concept
Coterminal angles have the same cosine and the same sine, so adding two pi to the input never changes the output. A function with that property is called periodic, and the smallest positive shift that leaves it unchanged is called its period.
periodic function — A function for which some positive number c satisfies f of t plus c equals f of t for every t in the domain. The smallest such positive number, when it exists, is the period.
\[ \cos(t + 2\pi k) = \cos(t), \qquad \sin(t + 2\pi k) = \sin(t) \]
This is why graphing these functions is tractable at all. Draw one cycle and the rest of the graph is a copy-and-paste operation, in both directions, forever.
Figure (svg): One full cycle of the cosine curve from zero to two pi, with the five quarter marks highlighted at zero, pi over two, pi, three pi over two and two pi
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 790-791
Section
Section 1
Concept
Cosine and sine have identical domains, ranges and periods, and both are continuous and smooth — no gaps, no jumps, no corners. They differ in exactly one listed property.
Evenness and oddness is the only difference, and geometrically it is not much of one: the two graphs are the same curve, slid horizontally by a quarter period.
| Property | cosine | sine |
|---|---|---|
| Domain | all real numbers | all real numbers |
| Range | from -1 to 1 | from -1 to 1 |
| Continuous and smooth | yes | yes |
| Even or odd | even | odd |
| Period | two pi | two pi |
Figure (svg): One full cycle of the cosine curve from zero to two pi, with the five quarter marks highlighted at zero, pi over two, pi, three pi over two and two pi
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 790-792
Picture it
Plotted together, the relationship between cosine and sine is a horizontal slide and nothing more.
Figure (svg): The cosine and sine curves drawn on the same axes over two full periods, showing that the sine is the cosine shifted right by pi over two
That slide is exactly the cofunction identity from Lesson 10.4a, seen rather than computed. Any statement about the sine graph is a statement about the cosine graph moved a quarter period.
Worked example
This is how the graph is actually constructed. Nothing is assumed about its shape.
\[ \text{Plot } y = \cos(x) \text{ over } [0, 2\pi] \text{ using the common values.} \]
Evaluate at the quarter marks
Why: These five give the extremes and the zeros.
\[ 1, 0, -1, 0, 1 \]
Evaluate at the halfway points too
Why: The eighth marks at pi over four and so on fill in the shape.
\[ +- \sqrt{2} / 2\text{ at the eighths} \]
Plot the nine points
Why: Five landmarks plus four intermediate values.
Join them smoothly
Why: The function is continuous and smooth, so no corners are allowed.
Figure (svg): The solution to Worked example build the fundamental cycle from the table shown as a ladder of expressions, one row per legal move
\[ y = \cos(x) \text{ on } [0, 2\pi]: \text{ starts at } 1, \text{ dips to } -1 \text{ at } \pi, \text{ returns to } 1 \]
Verify: check the endpoints agree
Why: The value at zero and the value at two pi are both 1. They must be equal, because the two inputs are coterminal — and that agreement is precisely what lets the cycle tile the line without a jump.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 791-791
Sorting
Periodic means the values repeat under some fixed shift.
Sort into buckets
Sort each function.
Worked example
Periodicity is a computational tool, not just a description of the picture.
\[ \text{Find } \cos\left(\frac{27\pi}{4}\right) \text{ using periodicity.} \]
Subtract whole periods
Why: Two pi written with denominator four is eight pi over four.
\[ 27 \pi / 4 - 8 \pi / 4 = 19 \pi / 4 \]
Subtract again
Why: Still larger than two pi.
\[ 19 \pi / 4 - 8 \pi / 4 = 11 \pi / 4 \]
Subtract once more
Why: Now inside one period.
\[ 11 \pi / 4 - 8 \pi / 4 = 3 \pi / 4 \]
Evaluate the reduced angle
Why: Three pi over four is in quadrant two with reference angle pi over four.
\[ \cos = -\sqrt{2} / 2 \]
Figure (svg): The solution to Worked example use periodicity to evaluate shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{27\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2} \]
Verify: count the periods removed
Why: Three periods is six pi, which is twenty-four pi over four, and twenty-seven minus twenty-four is three. So exactly three whole cycles were removed, and the function's value cannot have changed.
Trap
A student concludes: since cosine is periodic and cosine is a smooth wave, every periodic function is a smooth wave.
But the definition of periodic says only that the values repeat. It says nothing whatever about the shape of what repeats.
A constant function repeats its value under every shift, so it satisfies the definition of periodic — and it has no period at all, because there is no smallest positive shift that works.
A square wave and a sawtooth are periodic with corners and jumps. What makes cosine and sine special is not periodicity but that they are continuous and smooth as well, which is why Theorem 10.22 lists those separately.
The book is careful here for a reason: periodicity buys you repetition and nothing else. Every other pleasant property of these graphs has to be established on its own.
Prediction
The cosine of 0.3 is about 0.955.
Predict first
What is the cosine of 0.3 plus 100 pi?
Correct: About 0.955.
Why: One hundred pi is fifty whole periods, since each period is two pi. Adding a whole number of periods leaves the value untouched, so the answer is identical. Notice that no computation of the cosine was needed at all — periodicity answers the question completely, and this is why reducing an angle into one revolution is always the correct first move.
Two truths and a lie
Rule out the statements that are true. The survivor is the false one.
Eliminate the wrong options
One of these statements about periodicity is wrong.
Survives elimination: A
Why: Statement A is false, and a constant function is the counterexample: it repeats under every positive shift, so there is no smallest one and the period does not exist. The definition is careful to say the smallest such number if it exists, precisely to accommodate this case. It is a small point but it is the reason Theorem 10.22 lists continuity and smoothness separately from periodicity.
Socratic
The whole strategy of this section is to draw one cycle and stop.
Discussion prompt
Explain precisely why that is legitimate, and what would have to be true for it to fail.
Hint: What does the graph look like on the interval from two pi to four pi?
Answer:
Every input outside one cycle differs from an input inside it by a whole number of periods, and periodicity says the value at those two inputs is identical. So the graph over any other interval of length two pi is an exact translated copy of the fundamental cycle.
It would fail if the function were not exactly periodic — if, say, each cycle were slightly taller than the last. That happens in real damped oscillations, where the amplitude decays, and there one cycle genuinely does not determine the rest. The pure sinusoid is the idealisation in which it does.
Section
Section 2
Concept
To graph one cycle of any sinusoid, set its argument — the expression inside the cosine or sine — equal to each of the five quarter marks and solve for x. That gives the five x-coordinates directly, in one step each.
The period is then simply the last x-value minus the first, which is a free check on the arithmetic and requires no separate formula.
Figure (svg): A table showing the quarter-mark method: the argument of the cosine is set equal to each of the five standard values and solved for x, producing the five x-coordinates where the key points sit
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 793-793
Picture it
Five equations, each a one-line solve. The transformations never have to be applied in order.
Figure (svg): A table showing the quarter-mark method: the argument of the cosine is set equal to each of the five standard values and solved for x, producing the five x-coordinates where the key points sit
The alternative — shift, then stretch, then shift again, tracking each point through each step — is far more error-prone, because the order of the operations matters and is easy to get wrong.
Worked example
Example 10.5.1, part 1. Five equations, then five substitutions.
\[ \text{Graph one cycle of } f(x) = 3\cos\left(\frac{\pi x - \pi}{2}\right) + 1 \text{ and state the period.} \]
Identify the argument
Why: Everything inside the cosine.
\[ \frac{\pi x - \pi}{2} \]
Set it equal to each quarter mark and solve
Why: Five short linear equations.
\[ x = 1, 2, 3, 4, 5 \]
Substitute each x back into f
Why: The amplitude 3 and shift 1 turn the cosine values into heights.
\[ 4, 1, -2, 1, 4 \]
Read the period off the endpoints
Why: The cycle runs from x equal to 1 to x equal to 5.
\[ \text{period } = 4 \]
Figure (svg): One cycle of three cosine of the quantity pi x minus pi over two, plus one, running from x equal to one to x equal to five with a maximum of four and a minimum of negative two
\[ \text{points } (1,4), (2,1), (3,-2), (4,1), (5,4); \quad \text{period } 4 \]
Verify: check with the period formula
Why: Written in standard form the coefficient of x inside the cosine is pi over two, so the period is two pi divided by pi over two, which is 4. The quarter-mark method and the formula agree, and the first gave the graph as well.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 793-793
Fill the middle
Find the x-values for one cycle of the cosine of two x minus pi over three.
Fill in the blanks
2x - \fracpi/67 pi/6 = 0 \;\Longrightarrow\; x = ___ \qquad 2x - \frac______ = 2\pi \;\Longrightarrow\; x = ___
Why: The first equation gives two x equal to pi over three, so x is pi over six — that is where the cycle begins, and it is the phase shift. The last gives two x equal to two pi plus pi over three, which is seven pi over three, so x is seven pi over six. The period is the difference, seven pi over six minus pi over six, which is pi — matching two pi over the coefficient 2.
Worked example
Example 10.5.1, part 2. The x-values come out in decreasing order, which is fine.
\[ \text{Graph one cycle of } g(x) = \tfrac{1}{2}\sin(\pi - 2x) + \tfrac{3}{2} \text{ and state the period.} \]
Set the argument equal to each quarter mark
Why: The argument is pi minus two x.
\[ \pi - 2 x = 0, \frac{\pi}{2}, \pi, 3 \pi / 2, 2 \pi \]
Solve each for x
Why: Because the coefficient of x is negative, the x-values decrease.
\[ \frac{\pi}{2}, \frac{\pi}{4}, 0, -\frac{\pi}{4}, -\frac{\pi}{2} \]
Substitute back to get the heights
Why: Amplitude one half, midline three halves.
\[ \frac{3}{2}, 2, \frac{3}{2}, 1, \frac{3}{2} \]
Read the period off the interval covered
Why: From negative pi over two to pi over two.
\[ \text{period } = \pi \]
Figure (svg): The solution to Worked example an argument with a negative coefficient shown as a ladder of expressions, one row per legal move
\[ \text{cycle on } \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], \quad \text{period } \pi \]
Verify: check with the period formula
Why: The coefficient of x inside the sine has absolute value 2, so the period is two pi over 2, which is pi. That matches the interval length, and note that the negative coefficient affected the direction the marks were traversed but not the period.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 793-794
Error analysis
A student graphs two sine of three x plus one.
Annotate
On: \( 2\sin(3x) + 1 = 0, \; \tfrac{\pi}{2}, \; \pi, \; \tfrac{3\pi}{2}, \; 2\pi \)
The quarter marks describe positions around one turn of the circle, so they belong wherever the angle lives — inside the function. Anything outside the function is a transformation of the output and has nothing to do with them.
Ranking
Graph one cycle of a sinusoid by the quarter-mark method.
Put in order
Why: Identifying the argument first is what prevents the commonest error, since the quarter marks belong inside the function. Solving gives the horizontal positions and substituting gives the heights, and only then is there anything to plot. Note that the period falls out for free as the difference between the first and last x-values.
Prediction
A sinusoid has argument four x plus pi.
Predict first
How long will one cycle be?
Correct: pi over 2.
Why: Setting the argument to zero gives x equal to negative pi over four, and setting it to two pi gives x equal to pi over four. The difference is pi over two. Equivalently, the period is two pi divided by the coefficient of x, which is two pi over 4. The coefficient inside the function compresses the graph horizontally, and a larger coefficient means a shorter period — the two are inversely related, which is worth internalising.
Explain it to yourself
The alternative method is to apply the transformations one at a time: shift, stretch, shift again.
Discussion prompt
Explain why the quarter-mark method is more reliable, and identify the specific thing that goes wrong with the transformation approach.
Hint: Does it matter whether you stretch first or shift first?
Answer:
The transformation approach requires the operations to be done in the right order, and horizontal shifting and horizontal stretching do not commute. Shifting right by pi then compressing by 2 is not the same as compressing by 2 then shifting right by pi — the second gives a shift of pi over two.
The quarter-mark method never sequences anything. It solves one equation per point, and each solve automatically accounts for every horizontal transformation at once, in the correct combination, because the argument already contains them all. That is why it is worth learning even though it looks less conceptual.
Section
Section 3
Concept
Every sinusoid can be written in one standard form, and four numbers read off that form describe it completely: how tall it is, how long a cycle takes, where the cycle starts, and how high the midline sits.
Theorem 10.23 — For omega positive, the sinusoid has period two pi over omega, amplitude the absolute value of A, phase shift negative phi over omega, and vertical shift B.
\[ C(x) = A\cos(\omega x + \phi) + B, \qquad S(x) = A\sin(\omega x + \phi) + B \]
The parameter omega is called the angular frequency and counts how many cycles fit into an interval of length two pi. It must be positive for the theorem as stated, and the even-odd identities can always make it so.
Figure (svg): A general sinusoid with its four defining quantities marked: the amplitude measured from the midline to the peak, the period measured between successive peaks, the vertical shift locating the midline, and the phase shift locating where the cycle begins
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 794-795
Picture it
Each parameter is a measurable feature of the picture, not just a letter in a formula.
Figure (svg): A general sinusoid with its four defining quantities marked: the amplitude measured from the midline to the peak, the period measured between successive peaks, the vertical shift locating the midline, and the phase shift locating where the cycle begins
Reading them off a graph and reading them off a formula are the two directions of the same skill, and the next section does the first.
Worked example
The function from the earlier example, now analysed rather than plotted.
\[ \text{Find the four parameters of } f(x) = 3\cos\left(\frac{\pi x - \pi}{2}\right) + 1. \]
Rewrite the argument in standard form
Why: Split the fraction so the coefficient of x is visible.
\[ (\frac{\pi}{2}) x + (-\frac{\pi}{2}) \]
Read off A, omega, phi and B
Why: Matching against A cosine of omega x plus phi, plus B.
\[ A = 3, w = \frac{\pi}{2}, \phi = -\frac{\pi}{2}, B = 1 \]
Compute the period and amplitude
Why: Two pi over omega, and the absolute value of A.
\[ \text{period } 4,\text{ amplitude } 3 \]
Compute the phase shift
Why: Negative phi over omega, not negative phi.
\[ \text{shift } = 1,\text{ to the right} \]
Figure (svg): The solution to Worked example read the four parameters off a formula shown as a ladder of expressions, one row per legal move
\[ |A| = 3, \quad \tfrac{2\pi}{\omega} = 4, \quad -\tfrac{\phi}{\omega} = 1, \quad B = 1 \]
Verify: compare with the earlier graph
Why: The graph ran from x equal to 1 to x equal to 5, so it started at 1 and had length 4 — matching the phase shift and period. Its maximum was 4 and minimum negative 2, an amplitude of 3 about a midline of 1. Every parameter matches the picture.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 795-796
Matching
Each of the four changes the graph in exactly one way.
Match the pairs
Why: A scales vertically, so it sets the amplitude; omega scales horizontally, so it sets the period as two pi over omega; phi translates horizontally, but the translation is negative phi over omega because the horizontal scaling applies to it too; and B translates vertically. Note the asymmetry: B is the vertical shift directly, while phi is not the horizontal shift directly.
Worked example
The theorem requires a positive omega, so a negative coefficient must be dealt with before reading anything off.
\[ \text{Find the four parameters of } g(x) = \tfrac{1}{2}\sin(\pi - 2x) + \tfrac{3}{2}. \]
Factor out the negative from the argument
Why: Pi minus two x is the negative of two x minus pi.
\[ \sin(-(2 x - \pi)) \]
Apply the odd identity
Why: Sine is odd, so the minus sign comes out front and joins the amplitude.
\[ -(\frac{1}{2}) \sin(2 x - \pi) + \frac{3}{2} \]
Read off the parameters
Why: Now omega is positive.
\[ A = -\frac{1}{2}, w = 2, \phi = -\pi, B = \frac{3}{2} \]
Compute the four quantities
Why: Amplitude is the absolute value, so the negative A does not reduce it.
Figure (svg): The solution to Worked example make omega positive first shown as a ladder of expressions, one row per legal move
\[ |A| = \tfrac{1}{2}, \quad \text{period } \pi, \quad \text{shift } \tfrac{\pi}{2} \text{ right}, \quad B = \tfrac{3}{2} \]
Verify: reconcile with the earlier graph
Why: The graph was drawn on the interval from negative pi over two to pi over two, so it appears to start at negative pi over two rather than pi over two. But the theorem is detecting a different cycle — another complete cycle begins at pi over two — and both are correct descriptions of the same graph. A sinusoid has infinitely many cycles and the phase shift names one of them.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 796-796
Trap
\[ y = \cos(2x - \pi) \quad\Longrightarrow\quad \text{phase shift } = \pi \]
Take the constant inside the argument, with its sign flipped, as the shift
Why: For a coefficient of 1 that is correct, so the habit is reinforced early and then fails silently.
The cycle actually begins where the argument is zero, which is at two x equal to pi, so x equal to pi over two — half of what was claimed.
\[ -\frac{\phi}{\omega} = -\frac{-\pi}{2} = \frac{\pi}{2} \]
Divide by omega as well as negating
Why: The horizontal compression by omega compresses the shift too.
The quarter-mark method never makes this mistake, because setting the argument to zero and solving does the division automatically. If you are ever unsure of a phase shift, solve for where the argument is zero and you have it.
Faded example
Find the four parameters of five sine of the quantity three x plus pi over two, minus two.
Fill in the blanks
A = 5, \; \omega = 3, \; \phi = \tfrac2 pi/3-pi/6, \; B = -2 \;\Longrightarrow\; \text___ = ___, \; \text___ = ___
Why: The period is two pi over three. The phase shift is negative phi over omega, which is negative pi over two divided by 3, giving negative pi over six — a shift of pi over six to the left. The amplitude is 5 and the midline sits at negative 2, so the graph oscillates between negative 7 and 3.
Prediction
A sinusoid has amplitude 3 and vertical shift negative 1.
Predict first
What are its maximum and minimum values?
Correct: 2 and -4.
Why: The midline sits at negative 1 and the wave rises and falls 3 units from it, so the maximum is negative one plus three, which is 2, and the minimum is negative one minus three, which is negative 4. Reading these off is the fastest check on any sinusoid problem: the maximum and minimum must average to the vertical shift and differ by twice the amplitude.
Edge cases
The theorem requires omega to be positive.
Discussion prompt
What would go wrong if omega were allowed to be negative, and how do the even-odd identities rescue the situation?
Hint: Try computing the period with a negative omega.
Answer:
The period formula would give a negative period, which is meaningless — a cycle has a length, and lengths are positive. The phase shift formula would also change sign spuriously.
\[ \cos(-\omega x + \phi) = \cos(\omega x - \phi), \qquad \sin(-\omega x + \phi) = -\sin(\omega x - \phi) \]
The even-odd identities let any negative omega be turned positive: for cosine the minus sign simply disappears, and for sine it comes out front and is absorbed into A. Since the amplitude is the absolute value of A, absorbing a minus sign there costs nothing — which is exactly what the second worked example did.
Section
Section 4
Concept
Given a graph, the four parameters can be measured off it directly. Two of them come from the vertical extremes and two from the horizontal geometry of one cycle.
The answer is never unique. Any cycle can be chosen as the starting one, and a cosine fit and a sine fit will differ in phase — the book's own example produces two different-looking formulas for the same curve, both correct.
Figure (svg): A sinusoid drawn on a grid with its range and one full cycle marked, showing how the amplitude comes from half the vertical spread and the vertical shift from the average of the two extremes
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 796-797
Picture it
The two vertical parameters come from the same pair of numbers, combined two different ways.
Figure (svg): A sinusoid drawn on a grid with its range and one full cycle marked, showing how the amplitude comes from half the vertical spread and the vertical shift from the average of the two extremes
Half the difference gives the amplitude and half the sum gives the midline. If the two extremes are symmetric about zero, the vertical shift is zero and only the amplitude survives.
Worked example
Example 10.5.2, part 1. One cycle runs from negative one to five, with range from negative three halves to five halves.
\[ \text{Find } C(x) = A\cos(\omega x + \phi) + B \text{ matching a cycle on } [-1, 5] \text{ with range } \left[-\tfrac{3}{2}, \tfrac{5}{2}\right]. \]
Find the period and hence omega
Why: The cycle length is five minus negative one, which is 6.
\[ \text{period } 6,\text{ so } w = \frac{\pi}{3} \]
Find the phase shift and hence phi
Why: The cycle begins at negative one, so negative phi over omega is negative one.
\[ \phi = \frac{\pi}{3} \]
Find the amplitude
Why: Half the difference of the two extremes.
\[ A = \frac{\frac{5}{2} + \frac{3}{2}}{2} = 2 \]
Find the vertical shift
Why: The average of the two extremes.
\[ B = \frac{\frac{5}{2} - \frac{3}{2}}{2} = \frac{1}{2} \]
Figure (svg): The solution to Worked example fit a cosine shown as a ladder of expressions, one row per legal move
\[ C(x) = 2\cos\left(\frac{\pi}{3}x + \frac{\pi}{3}\right) + \frac{1}{2} \]
Verify: test at the start of the cycle
Why: At x equal to negative one the argument is negative pi over three plus pi over three, which is zero, and the cosine of zero is 1. So the value is two plus one half, which is five halves — the maximum, exactly where a cosine cycle should start.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 797-797
Fill the middle
A sinusoid oscillates between a maximum of 11 and a minimum of negative 3.
Fill in the blanks
A = \frac74 = ___, \qquad B = \frac______ = ___
Why: The vertical spread is 14, so the amplitude is 7, and the midline sits at the average of the two extremes, which is 4. Checking: a midline of 4 with amplitude 7 gives a maximum of 11 and a minimum of negative 3, matching. The two formulas are half the difference and half the sum, and they are worth memorising as a pair since neither works alone.
Worked example
Example 10.5.2, part 2. Three parameters are unchanged; only the phase differs.
\[ \text{Find } S(x) = A\sin(\omega x + \phi) + B \text{ for the same graph.} \]
Carry over the three unchanged parameters
Why: Period, amplitude and vertical shift describe the curve regardless of which function is used.
\[ w = \frac{\pi}{3}, A = 2, B = \frac{1}{2} \]
Find where a sine cycle begins
Why: A sine cycle starts at the midline going upward, which happens at x equal to seven halves.
\[ \text{shift } = \frac{7}{2} \]
Compute phi from the shift
Why: Phi is negative omega times the shift.
\[ \phi = -(\frac{\pi}{3}) (\frac{7}{2}) = -7 \pi / 6 \]
Assemble the formula
Why: Same amplitude, period and midline; new phase.
\[ S(x) = 2 \sin(\pi x / 3 - 7 \pi / 6) + \frac{1}{2} \]
Figure (svg): The solution to Worked example fit a sine to the same graph shown as a ladder of expressions, one row per legal move
\[ S(x) = 2\sin\left(\frac{\pi}{3}x - \frac{7\pi}{6}\right) + \frac{1}{2} \]
Verify: check the two formulas agree
Why: At x equal to negative one the sine version gives the argument negative pi over three minus seven pi over six, which is negative three pi over two, whose sine is 1. So the value is five halves — matching the cosine version. Two very different-looking formulas, the same curve.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 797-797
Error analysis
A graph oscillates between a maximum of 7 and a minimum of 1. A student reads off the parameters.
Annotate
On: \( A = 7, \qquad B = 0 \)
The two vertical parameters must be found together, from the same pair of numbers. Half the difference is the amplitude and half the sum is the midline, and reading either one alone will get the other wrong.
Prediction
Two students fit a formula to the same sinusoid graph, one using a cosine and one using a sine.
Predict first
Which of their four parameters will differ?
Correct: Only the phase.
Why: The amplitude, period and vertical shift describe the shape and position of the curve itself and do not depend on which function is used to express it. Only the phase differs, because a cosine cycle starts at a maximum while a sine cycle starts at the midline going up — a quarter period apart. This is the cofunction identity again, showing up as a difference between two correct answers.
Missing information
A problem shows a sinusoid graph and asks for its formula. A student produces one, and a classmate produces a different one.
Discussion prompt
Can they both be right? Explain what makes the answer non-unique and what would have to be added to make it unique.
Hint: How many cycles does the graph contain, and how many could be chosen as the starting one?
Answer:
They can both be right. The graph contains infinitely many cycles and any of them may be taken as the starting one, so the phase can differ by any whole number of periods. Choosing a sine rather than a cosine shifts it by another quarter period, and choosing a negative amplitude shifts it by a half period.
To make the answer unique the problem would have to specify the function, require the amplitude to be positive, and restrict the phase to a stated interval such as from zero to two pi. Textbooks rarely do all three, which is why the back of the book often shows one of several correct answers and why checking your formula against the graph matters more than matching the printed answer.
Real world
Daily high temperatures in a temperate city average about 10 degrees over the year, peak near 22 degrees in mid-July, and bottom out near negative 2 degrees in mid-January.
Discussion prompt
Fit a sinusoid modelling the temperature as a function of the day of the year, and say what each of the four parameters means physically.
Hint: Find the amplitude and midline from the two extremes, and the period from the fact that the pattern repeats annually.
Answer:
\[ A = \frac{22 - (-2)}{2} = 12, \qquad B = \frac{22 + (-2)}{2} = 10 \]
\[ \text{period} = 365 \;\Longrightarrow\; \omega = \frac{2\pi}{365}, \qquad \text{peak at day } 196 \]
\[ T(d) = 12\cos\left(\frac{2\pi}{365}(d - 196)\right) + 10 \]
The amplitude of 12 is how far the temperature swings either side of average — the severity of the seasons. The vertical shift of 10 is the annual mean temperature. The period of 365 is the year. The phase shift of 196 days locates midsummer.
This is a genuinely useful model and it is how climate normals are often summarised. Its limitations are equally instructive: real temperature data is noisy around this curve, and the fit degrades near the poles where the seasonal shape stops being sinusoidal.
Section
Section 5
Concept
Expanding the standard form with the sum identity shows that any sinusoid is a combination of a cosine and a sine at the same frequency. Reading that backwards, any such combination must itself be a single sinusoid.
\[ A\cos(\omega x + \phi) = A\cos\phi \cdot \cos(\omega x) - A\sin\phi \cdot \sin(\omega x) \]
Section 11.1 uses this constantly, because it is what turns a physical superposition of two effects into a single wave with a definite amplitude and phase.
Figure (svg): A curve that is the cosine of two x minus root three times the sine of two x, shown to be a single sinusoid of amplitude two
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 798-799
Picture it
The curve looks like it was built from two waves. It is one wave, and the picture shows it.
Figure (svg): A curve that is the cosine of two x minus root three times the sine of two x, shown to be a single sinusoid of amplitude two
Reading the amplitude off the graph gives 2, and the period is pi. Both match the single-sinusoid form exactly, which is the graphical confirmation of the algebra.
Worked example
Example 10.5.3, part 1. Match coefficients, then eliminate phi.
\[ \text{Write } f(x) = \cos(2x) - \sqrt{3}\sin(2x) \text{ as } A\cos(\omega x + \phi). \]
Expand the target form and match frequencies
Why: Both terms have argument 2x, so omega is 2 and B is 0.
\[ w = 2, B = 0 \]
Equate the two coefficients
Why: The cosine coefficient gives one equation, the sine coefficient the other.
\[ A \cos \phi = 1, A \sin \phi = \sqrt{3} \]
Square and add to eliminate phi
Why: The Pythagorean identity collapses the left side to A squared.
\[ A ^{2} = 1 + 3 = 4,\text{ so } A = 2 \]
Find phi from the signs
Why: Cosine of phi is one half and sine of phi is root three over two.
\[ \phi = \frac{\pi}{3} \]
Figure (svg): The solution to Worked example rewrite as a single cosine shown as a ladder of expressions, one row per legal move
\[ f(x) = 2\cos\left(2x + \frac{\pi}{3}\right) \]
Verify: expand it back
Why: Two cosine of two x plus pi over three expands to two times cosine 2x times one half minus two times sine 2x times root three over two, which is cosine 2x minus root three sine 2x. That is the original function.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 798-798
Sorting
The test is whether the arguments match.
Sort into buckets
Sort each function.
Worked example
Example 10.5.3, part 2. The same amplitude, a different phase.
\[ \text{Write the same } f(x) \text{ as } A\sin(\omega x + \phi). \]
Expand the sine form and match
Why: Now the sine coefficient pairs with cosine of phi and the cosine coefficient with sine of phi.
\[ A \sin \phi = 1, A \cos \phi = -\sqrt{3} \]
Square and add
Why: The same elimination gives the same amplitude.
\[ A ^{2} = 4,\text{ so } A = 2 \]
Find phi from the signs
Why: Sine of phi is one half and cosine of phi is negative root three over two, so phi is in quadrant two.
\[ \phi = 5 \pi / 6 \]
Assemble
Why: Same amplitude and frequency, new phase.
\[ f(x) = 2 \sin(2 x + 5 \pi / 6) \]
Figure (svg): The solution to Worked example rewrite the same function as a single sine shown as a ladder of expressions, one row per legal move
\[ f(x) = 2\sin\left(2x + \frac{5\pi}{6}\right) \]
Verify: compare the two phases
Why: The cosine version had phase pi over three and the sine version has five pi over six. Their difference is five pi over six minus two pi over six, which is pi over two — exactly a quarter period in the argument, as the cofunction identity requires.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 799-799
Trap
\[ g(x) = \cos(2x) - \sqrt{3}\sin(3x) \quad\Longrightarrow\quad g(x) = 2\cos(2x + \phi) \]
Apply the coefficient-matching method regardless of the arguments
Why: The two coefficients are the same as in the previous example, so the method looks applicable.
But the two terms have different arguments, two x and three x. The expansion of a single sinusoid produces a cosine and a sine with the SAME argument, so no single sinusoid can equal this.
The function g is genuinely not a sinusoid. Its graph is not a simple wave: it has period two pi, the least common multiple of the two individual periods, and a lumpy shape within each cycle.
The check to run first, before any coefficient matching: do the two arguments match exactly? If they do, the sum is a sinusoid at that frequency. If they do not, it is a genuinely more complicated object, and analysing it is what Fourier series are for.
Faded example
Write three cosine x plus four sine x as a single cosine.
Fill in the blanks
A\cos\phi = 3, \; A\sin\phi = -4 \;\Longrightarrow\; A^2 = 9 + 16 = 25 \;\Longrightarrow\; A = 5
Why: Squaring and adding eliminates phi via the Pythagorean identity, leaving A squared as the sum of the squares of the two coefficients — which is just the distance formula. The amplitude is 5. The phase then satisfies cosine of phi equal to three fifths and sine of phi equal to negative four fifths, putting phi in quadrant four. Note that the sine coefficient picks up a minus sign because the cosine expansion has a minus in it.
Prediction
You combine a cosine of amplitude 5 and a sine of amplitude 12, both at the same frequency.
Predict first
What is the amplitude of the resulting single sinusoid?
Correct: 13.
Why: The amplitude is the square root of the sum of the squares, which is the root of 25 plus 144, namely the root of 169, which is 13. Five, twelve and thirteen are a Pythagorean triple. The formula is exactly the distance formula, because the two coefficients behave like perpendicular components — which is the deep reason the combination works at all, and the reason 17, the plain sum, is too large.
Socratic
The technique eliminates phi in one step by squaring both coefficient equations and adding them.
Discussion prompt
Explain why that works, and what geometric fact it corresponds to.
Hint: What identity connects the squares of a cosine and a sine?
Answer:
\[ (A\cos\phi)^2 + (A\sin\phi)^2 = A^2(\cos^2\phi + \sin^2\phi) = A^2 \]
Factoring out A squared leaves the Pythagorean identity, which is 1, so phi vanishes completely and A squared is exposed. It works because phi appears only through its cosine and sine, and squaring and adding is precisely the combination that identity annihilates.
Geometrically, the two coefficients are the components of a vector of length A at angle phi, and the computation is the distance formula recovering that length. That is not a coincidence: this whole technique is vector addition of two perpendicular oscillations, and Section 11.8 will make the connection explicit.
Comparison
Fill the blanks from memory. The left column reads a formula, the right reads a graph.
Comparison matrix
| Quantity | From the formula | From the graph |
|---|---|---|
| Amplitude | the absolute value of A | half the difference of max and min |
| Vertical shift | B | the average of max and min |
| Period | two pi over omega | the length of one cycle |
| Phase shift | negative phi over omega | where a cycle begins |
| Angular frequency | omega | two pi over the period |
Notice that the phase shift is the only one needing a division by omega. That single asymmetry accounts for most of the errors in this lesson.
Pattern
Whether you are graphing a formula, analysing one, or fitting one to a picture, the same five moves cover it.
If a formula contains both a cosine and a sine, check whether their arguments match before anything else. If they do, combine into a single sinusoid first; if they do not, none of this applies.
OpenStax Algebra and Trigonometry 2e, §8.1 Graphs of the Sine and Cosine Functions §8.1
Check
Reading parameters. Watch the division by omega.
Check your understanding
What is the phase shift of y equals 4 cosine of the quantity 3 x minus pi, plus 2?
Answer: B
Why: Here phi is negative pi and omega is 3, so the phase shift is negative phi over omega, which is pi over 3, a shift to the right. Equivalently, setting the argument to zero gives three x equal to pi, so x equals pi over three — where the cycle begins.
Check
Fitting a graph. Use both extremes together.
Check your understanding
A sinusoid oscillates between a maximum of 9 and a minimum of 1. What are its amplitude and vertical shift?
Answer: C
Why: The amplitude is half the difference, which is half of 8, namely 4. The vertical shift is the average of the two extremes, which is half of 10, namely 5. Checking, a midline of 5 with amplitude 4 gives a maximum of 9 and a minimum of 1.
Check
A hidden sinusoid. Check the frequencies first.
Check your understanding
What is the amplitude of y equals 8 cosine of 5x plus 6 sine of 5x?
Answer: C
Why: Both terms have argument 5x, so this is a single sinusoid. The amplitude is the square root of the sum of the squares of the coefficients, which is the root of 64 plus 36, namely the root of 100, which is 10. Six, eight and ten are a Pythagorean triple.
Real world
The tide at a harbour is high at 3 a.m. at 5.2 metres and low at 9:15 a.m. at 1.4 metres, and the pattern repeats every 12 hours 25 minutes.
Discussion prompt
Model the depth as a sinusoid in hours after midnight, and use it to say when in the morning the depth first reaches 4 metres — the minimum a particular vessel needs.
Hint: Find the four parameters first, then solve the equation the model produces.
Answer:
\[ A = \frac{5.2 - 1.4}{2} = 1.9, \qquad B = \frac{5.2 + 1.4}{2} = 3.3 \]
\[ \text{period } = 12.417 \;\Longrightarrow\; \omega = \frac{2\pi}{12.417} \approx 0.506 \]
\[ D(t) = 1.9\cos\big(0.506(t - 3)\big) + 3.3 \]
The high tide at 3 a.m. sets the phase, since a cosine cycle begins at a maximum. Solving D of t equal to 4 gives the cosine of the argument equal to about 0.368, so the argument is about plus or minus 1.194, giving t about 0.64 and t about 5.36.
So the depth is above 4 metres from about 00:38 until about 05:22. This is exactly how tide tables are computed, though real predictions sum several sinusoids at different frequencies — one for the moon, one for the sun, and more — because a single sinusoid captures the dominant term but not all of it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two sinusoids have the same amplitude, period and vertical shift but different phase shifts. How do their graphs compare?
Correct: They are the same curve, one slid horizontally.
\[ \cos\left(x - \tfrac{\pi}{2}\right) = \sin(x) \]
Why: The phase shift is precisely a horizontal translation, and the other three parameters fix the shape entirely. So two sinusoids agreeing on those three are identical curves in different horizontal positions. This is why cosine and sine are the same curve: they agree on amplitude, period and midline and differ only by a quarter-period shift. The third option is a special case rather than the general answer — a phase difference of exactly half a period does produce a reflection in the midline, but most phase differences do not.
Explain it
They have been taught to graph transformations by shifting, then stretching, then shifting again, and keep getting the horizontal shift wrong.
Discussion prompt
In no more than five sentences, teach them the quarter-mark method and explain why it avoids the problem they are having.
Hint: What are they actually getting wrong, and what does the method never have to do?
Answer:
A usable answer: their problem is that horizontal shifting and horizontal stretching do not commute, so the order matters and the shift gets scaled or not scaled depending on when you apply it. Instead, look at what is inside the cosine and set it equal to each of zero, pi over two, pi, three pi over two and two pi. Solve each little equation for x and you have the five key positions immediately.
Then put each of those x-values back into the whole formula to get the heights, plot the five points, and join them into a wave. You never apply a transformation at all, so there is no order to get wrong — the algebra handles all the horizontal work at once.
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 phase shift is fixed by solving for where the argument is zero, which does the division automatically. The quarter-mark method is fixed by always identifying the argument first, in writing. Reading a graph is fixed by taking half the difference and half the sum of the extremes as a pair rather than separately. Hidden sinusoids are fixed by checking that the arguments match before doing anything, then squaring and adding the coefficients. 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
Across the top of the page, draw two full cycles of the cosine curve and, on the same axes in a different colour, two full cycles of the sine curve, marking the quarter-period shift between them and writing the identity that expresses it. Below, draw a general sinusoid with a midline well above the x-axis, and label all four parameters on the picture: amplitude with a vertical bracket, period with a horizontal bracket, vertical shift as the height of the midline, and phase shift as the position where a cycle begins. Beside it write each parameter's formula in terms of A, omega, phi and B, and circle the one that requires a division. In the middle of the page, work the quarter-mark method completely for the function two sine of the quantity three x minus pi, showing all five equations and all five points. In the bottom left, write the two formulas for reading amplitude and vertical shift off a graph. In the bottom right, convert five cosine x minus twelve sine x into a single cosine, showing the squaring-and-adding step. Finally, write one sentence saying what test decides whether a sum of a cosine and a sine is a sinusoid at all.
The test is whether the two arguments are identical. If they are, the sum is a single sinusoid at that frequency; if they are not, it is periodic but not sinusoidal, and no amount of coefficient matching will make it one.
Recap
Five things, and the second one is the method to use for the rest of the chapter.
| If the question says | Your first move is |
|---|---|
| Graph one cycle | Identify the argument, then set it to the five quarter marks |
| State the period | Two pi over the coefficient of x inside the function |
| State the phase shift | Solve for where the argument is zero |
| Find a formula from this graph | Half the difference and half the sum of the extremes |
| A cosine plus a sine appears | Check the arguments match, then square and add |
Cosine and sine are the well-behaved pair: bounded, continuous, defined everywhere. The next two lessons graph the other four functions, and every one of them has vertical asymptotes where a denominator vanishes — which changes the shape of the picture completely.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.5 Graphs of the Trigonometric Functions §10.5, pp. 790-800 — everything on these slides traces back here
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