Turns the circle's rotation into a wave. Unrolls the unit circle to produce the sine and cosine graphs, identifies the amplitude, period and midline, and applies Chapter 1's transformations — with the added complication that an inside factor changes the period, so the phase shift must be read after factoring.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 6 — Periodic Functions
§6.1 Graphs of the Sine and Cosine Functions, pp. 716-738
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 716-738 — the pages these objectives are drawn from
Warm-up
The unit circle gave a value for every angle. Plotting those values against the angle is the whole of this section.
Discussion prompt
Imagine a point moving counterclockwise around the unit circle. Describe how its height changes over one full lap.
Hint: Start at the right, go up, across the top, down the left, and back.
Answer:
It starts at height zero, rises to 1 at the top, falls back through zero at the far left, drops to negative 1 at the bottom, and returns to zero.
Plotted against the angle, that is a wave: up, down, and back to where it started. One lap of the circle gives one complete wave.
And because a second lap repeats the same heights, the wave repeats forever. That periodicity is the whole reason these functions describe oscillation, and it comes directly from the circle closing on itself.
Concept
Plotting the moving point's vertical coordinate against the angle produces the sine curve; plotting its horizontal coordinate produces the cosine. Both repeat every full turn because the circle closes.
period — The horizontal length of one complete cycle of a periodic function. For the sine and cosine it is 2 pi, because one full turn of the circle returns the point to its start.
\[ \sin(\theta+2\pi)=\sin\theta, \qquad \cos(\theta+2\pi)=\cos\theta \]
Every feature of the graph is a fact about the circle in disguise. The maximum of 1 is the top of the circle, the zeros are where the point crosses an axis, and the period of 2 pi is the circumference in radians — which is why no separate memorisation is needed.
Figure (svg): The unit circle beside a set of axes, with the height of the moving point traced out horizontally to produce the sine curve
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 716-722
Section
Section 1
Concept
The sine graph plots the moving point's height as the angle increases. The cosine graph plots its horizontal position, which produces the same shape started at a different place.
The last point is worth stating because it means there is really only one curve. The cosine is the sine with a head start of a quarter turn, so anything proved about one transfers to the other with a shift — which is why the section can treat them together.
Figure (svg): The unit circle beside a set of axes, with the height of the moving point traced out horizontally to produce the sine curve
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 716-723
Picture it
The point's height on the left becomes the curve's height on the right.
Figure (svg): The unit circle beside a set of axes, with the height of the moving point traced out horizontally to produce the sine curve
Every feature of the curve corresponds to a position on the circle: the peak to the top, the zeros to the horizontal axis crossings, and the full wave to one complete lap.
Worked example
Four quarter-turns give the shape.
\[ \text{Give the sine's values at } 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2} \text{ and } 2\pi. \]
Start at the right of the circle
Why: Height zero.
\[ \sin 0 = 0 \]
Quarter turn to the top
Why: Height 1.
\[ \sin(\frac{\pi}{2}) = 1 \]
Half turn to the left
Why: Back to height zero.
\[ \sin(\pi) = 0 \]
Three quarters to the bottom, then back
Why: Down to -1, then zero.
\[ -1,\text{ then } 0 \]
Figure (svg): The unit circle beside a set of axes, with the height of the moving point traced out horizontally to produce the sine curve
\[ 0,\;1,\;0,\;-1,\;0 \]
Verify: check the pattern repeats
Why: At 2 pi the value is zero, the same as at 0 — because the point has returned to where it started. Every subsequent lap repeats these same five values, which is exactly what periodicity means and is why the graph continues forever in both directions.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 717-719
Matching
Every graph feature is a position on the circle.
Match the pairs
Why: The sine is the vertical coordinate, so its extremes occur where the point is highest and lowest and its zeros where the point is level with the origin. The period corresponds to a full lap because that is when the point returns to its start.
Worked example
Same shape, different starting point.
\[ \text{Show that } \cos x=\sin\bigl(x+\tfrac{\pi}{2}\bigr). \]
Compare starting values
Why: Cosine starts at 1, sine at 0.
Check at a second point
Why: At pi over 2 the cosine is 0.
\[ \sin e\text{ at } \pi\text{ is } 0 \]
Note the pattern
Why: The cosine is a quarter turn ahead.
\[ \text{shift left by } \frac{\pi}{2} \]
State it
Why: Adding inside shifts left.
\[ \cos x = \sin(x + \frac{\pi}{2}) \]
Figure (svg): The solution to Worked example the cosine as a shifted sine shown as a ladder of expressions, one row per legal move
\[ \cos x=\sin\bigl(x+\tfrac{\pi}{2}\bigr) \]
Verify: test at one more value
Why: At x equal to pi the cosine is negative 1, and the sine of three pi over two is also negative 1 — agreeing. The two curves are the same wave with different starting phases, which is why they share every property except where their cycles begin.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 719-723
Trap
\[ \text{the sine and cosine graphs must be memorised separately} \]
Learn each curve's shape and key values independently
Why: Two sets of five values are committed to memory.
Twice as much is memorised as necessary.
They are the same curve. The cosine is the sine shifted left a quarter turn, so knowing one gives the other.
Both come from the same circle, one reading the vertical coordinate and one the horizontal, and those differ only by where the lap is considered to start.
Learn the circle and read both off it. That is one picture rather than two curves, and it also supplies the values at every angle rather than only the five special ones.
Prediction
The sine is the vertical coordinate of a point on the unit circle.
Predict first
What is its range?
Correct: From -1 to 1 inclusive.
Why: A point on a circle of radius 1 is never more than 1 unit from the origin vertically, and it reaches both extremes at the top and bottom. The range is a fact about the circle's radius rather than about the graph, which is why it is the same for the cosine.
Faded example
The cosine is the horizontal coordinate.
Fill in the blanks
\cos 0 = 1, \quad \cos\tfrac-1___ = 0, \quad \cos\pi = ___
Why: At angle zero the point is at the right of the circle, one unit across, so the cosine is 1. At a half turn it is at the left, giving negative 1. The cosine starts at its maximum where the sine starts at zero, which is the quarter-turn offset between them.
Socratic
The number seems arbitrary until the circle is considered.
Discussion prompt
Explain where the period of 2 pi comes from.
Hint: How much angle is one full lap?
Answer:
One full lap of the circle is 2 pi radians, by §5.1's definition — the circumference of a unit circle divided by its radius.
After a full lap the point is back where it started, so every coordinate takes the value it had before. That is exactly what a period is.
So the period is not an extra fact but the circumference in radians. Had angles been measured in degrees the period would be 360, and in fact the number is whatever a full turn measures — which is another reason radians are the natural unit here.
Section
Section 2
Concept
The amplitude is half the distance between the maximum and minimum, and the midline is the horizontal level the curve oscillates about. A coefficient outside sets the first and a constant outside sets the second.
\[ y=A\sin(Bx)+D \]
The half in the second point is the commonest slip. A curve running between 3 and 11 has a total swing of 8 and an amplitude of 4, and reporting 8 gives a curve twice as tall as intended.
Figure (svg): A transformed sine curve with its amplitude, midline and period marked, showing which parameter controls each
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 723-728
Picture it
Amplitude, midline and period, each controlled by one parameter.
Figure (svg): A transformed sine curve with its amplitude, midline and period marked, showing which parameter controls each
The amplitude is measured from the midline to a peak, which is half the total height. Reading it as the full swing doubles the answer.
Worked example
Three parameters, three features.
\[ \text{For } y=3\sin(x)-2, \text{ give the amplitude, midline, maximum and minimum.} \]
Read the amplitude
Why: The size of the outside coefficient.
\[ \text{amplitude } 3 \]
Read the midline
Why: The constant added outside.
\[ \text{midline } y = -2 \]
Find the maximum
Why: Midline plus amplitude.
\[ -2 + 3 = 1 \]
Find the minimum
Why: Midline minus amplitude.
\[ -2 - 3 = -5 \]
Figure (svg): A transformed sine curve with its amplitude, midline and period marked, showing which parameter controls each
\[ A=3, \; \text{midline } y=-2, \; \max=1, \; \min=-5 \]
Verify: check the total swing
Why: From negative 5 to 1 is a total of 6, which is twice the amplitude of 3 — as it must be, since the amplitude is measured from the midline to a peak. That doubling relationship is the check that catches an amplitude reported as the full swing.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 724-726
Faded example
A sinusoid has maximum 12 and minimum 2.
Fill in the blanks
\text5 = \frac7___ = ___, \qquad \text___ = \frac______ = ___
Why: Half the difference gives the amplitude and the average gives the midline. Checking: 7 plus 5 is 12 and 7 minus 5 is 2, recovering both extremes. Half the difference and the average are the same two formulas that find a radius and a centre.
Worked example
The maximum and minimum determine both.
\[ \text{A sinusoid runs between } 4 \text{ and } 10. \text{ Find its amplitude and midline.} \]
Find the total swing
Why: Maximum minus minimum.
\[ 10 - 4 = 6 \]
Halve it for the amplitude
Why: The amplitude is half the swing.
\[ \text{amplitude } 3 \]
Average them for the midline
Why: The midpoint of the two.
\[ \frac{10 + 4}{2} = 7 \]
Check
Why: Seven plus 3 is 10; 7 minus 3 is 4.
Figure (svg): The solution to Worked example find the parameters from a graph shown as a ladder of expressions, one row per legal move
\[ A=3, \qquad \text{midline } y=7 \]
Verify: note the two formulas
Why: The amplitude is half the difference and the midline is the average — which is exactly how you would find the radius and centre of an interval. That is the right way to think of it: the curve oscillates within an interval, and these are its half-width and its centre.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 726-728
Error analysis
A student reads a sinusoid's amplitude from its extremes.
Annotate
On: \( \max=5, \min=-3 \;\Longrightarrow\; \text{amplitude } = 8 \)
Amplitude is half the swing, measured from the midline. Computing the midline first and then the distance to a peak makes the halving automatic rather than something to remember.
Prediction
The coefficient outside a sine is negative 4 rather than 4.
Predict first
What changes?
Correct: The curve reflects, but the amplitude is still 4.
Why: Amplitude is a distance and is always positive, so it is the size of the coefficient. The negative sign reflects the curve over its midline, turning peaks into troughs — a visible change in shape that leaves the amplitude alone.
Sorting
Each feature has one parameter.
Sort into buckets
Sort each feature by which parameter sets it.
Explain it to yourself
The amplitude is half the total swing.
Discussion prompt
Explain why, in terms of where it is measured from.
Hint: What is the amplitude a distance from?
Answer:
The amplitude is the distance from the midline to a peak, not from the trough to the peak. The midline sits halfway between them.
So the total swing spans two amplitudes: one down from the midline to the trough and one up to the peak. Halving the swing recovers one of them.
Thinking of the curve as oscillating within an interval makes this natural: the amplitude is the interval's half-width and the midline is its centre, exactly as a radius relates to a diameter.
Section
Section 3
Concept
An inside coefficient compresses the graph horizontally, so it shortens the period. The period is a full turn divided by that coefficient.
\[ \text{period}=\frac{2\pi}{|B|} \]
Reading the compression as a period change is the one genuinely new thing here. For a non-periodic function a horizontal compression is hard to see; for a wave it is immediately visible as more cycles in the same space.
Figure (svg): Two sine curves on the same axes, one with the standard period and one compressed by a factor inside the function, showing the period halving
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 728-732
Picture it
The solid curve completes two waves where the dashed one completes one.
Figure (svg): Two sine curves on the same axes, one with the standard period and one compressed by a factor inside the function, showing the period halving
The inside factor of 2 halves the period, which is §1.5's compression by one half seen through a periodic lens.
Worked example
Divide a full turn by the coefficient.
\[ \text{Find the period of } y=\sin(3x). \]
Identify the inside coefficient
Why: It multiplies x.
\[ B = 3 \]
Apply the formula
Why: Full turn over B.
\[ 2 \pi / 3 \]
Interpret
Why: Three waves fit in a full turn.
Check
Why: Three times the period is 2 pi.
Figure (svg): Two sine curves on the same axes, one with the standard period and one compressed by a factor inside the function, showing the period halving
\[ \text{period}=\frac{2\pi}{3} \]
Verify: count the waves
Why: Three periods of 2 pi over 3 make exactly 2 pi, so three complete waves fit in one full turn — which is what a coefficient of 3 does. Multiplying the period by B and getting a full turn back is the reliable check.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 729-730
Sorting
Compare the coefficient against 1.
Sort into buckets
Sort each function.
Worked example
Run the formula backwards.
\[ \text{A sinusoid has period } 8. \text{ Find its horizontal coefficient.} \]
Write the formula
Why: Period equals 2 pi over B.
\[ 8 = 2 \pi / B \]
Multiply up
Why: Clearing the fraction.
\[ 8 B = 2 \pi \]
Solve
Why: Divide by 8.
\[ B = \frac{\pi}{4} \]
Check
Why: 2 pi over pi over 4 is 8.
Figure (svg): The solution to Worked example find B from a period shown as a ladder of expressions, one row per legal move
\[ B=\frac{\pi}{4} \]
Verify: sanity-check the size
Why: A period of 8 is longer than the standard 2 pi, which is about 6.28, so B should be less than 1 — and pi over 4 is about 0.785. A stretched wave needs a small coefficient, and checking that direction catches an inverted formula.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 730-732
Trap
\[ y=\sin(4x) \;\Longrightarrow\; \text{period}=2\pi\cdot 4=8\pi \]
Combine the coefficient with the standard period
Why: The two numbers are multiplied together.
The period is reported as longer than the standard one.
A coefficient above 1 compresses, so the period gets shorter, not longer. The correct period is 2 pi over 4, which is pi over 2.
Four complete waves then fit in a full turn, which is what multiplying the input by 4 does.
Check the direction against the compression. A bigger inside coefficient means a faster wave and a shorter period, which is §1.5's reversal appearing again.
Faded example
For the function cosine of 6x.
Fill in the blanks
\text6 = \frac3___} = \frac______}
Why: Two pi over 6 simplifies to pi over 3. Six complete waves fit in a full turn, which is what the coefficient of 6 produces. Multiplying the period by 6 recovers 2 pi, confirming it.
Prediction
A function has an inside coefficient of 4.
Predict first
How many complete cycles fit between 0 and 2 pi?
Correct: Four.
Why: The period is 2 pi over 4, so four of them fit into 2 pi. The inside coefficient counts the cycles per full turn directly, which is the most useful reading of it — and it is the reciprocal of the period relationship.
Explain it to yourself
The period formula divides rather than multiplies.
Discussion prompt
Explain why, using §1.5's inside rule.
Hint: What does an inside factor do to a graph?
Answer:
An inside factor compresses horizontally by its reciprocal, which §1.5 established: multiplying the input by B squeezes the picture towards the vertical axis by a factor of B.
Squeezing a repeating pattern means each cycle occupies less horizontal space, so the period shrinks by that same factor — which is a division.
So the formula is not a new fact about waves. It is §1.5's horizontal compression, made visible because a periodic function's compression shows up as a countable change in how many cycles fit, where an ordinary function's would be hard to see at all.
Section
Section 4
Concept
A horizontal shift can only be read once the inside is written as the coefficient times a bracketed difference. Reading the constant directly overstates the shift by the coefficient.
\[ y=A\sin\bigl(B(x-h)\bigr)+D \;\Longrightarrow\; \text{shift } h \]
This is the same error §1.5 warned about, and it bites harder here because trigonometric arguments are so often written unfactored. The habit worth building is to factor as the very first step, before any parameter is read.
Figure (svg): A sinusoid with a horizontal shift, showing that the shift must be read after factoring the horizontal coefficient out of the argument
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 732-736
Picture it
The same function written two ways, and only one of them displays the shift.
Figure (svg): A sinusoid with a horizontal shift, showing that the shift must be read after factoring the horizontal coefficient out of the argument
The unfactored form suggests a shift of pi and the factored form shows the true shift of pi over 2. The factor of 2 is exactly the horizontal coefficient.
Worked example
The factoring is compulsory.
\[ \text{Find the phase shift of } y=\sin(2x-\pi). \]
Factor the coefficient out
Why: Two comes out of both terms.
\[ \sin(2(x - \frac{\pi}{2})) \]
Read the shift
Why: The bracketed subtraction.
\[ \frac{\pi}{2} \]
Note the direction
Why: A minus means right.
\[ \text{right } \frac{\pi}{2} \]
Note the period is unaffected
Why: Still 2 pi over 2.
Figure (svg): A sinusoid with a horizontal shift, showing that the shift must be read after factoring the horizontal coefficient out of the argument
\[ \text{shift } \tfrac{\pi}{2} \text{ right} \]
Verify: check where the wave starts
Why: A sine normally starts a cycle at zero going up. Setting the argument to zero gives 2x equal to pi, so x is pi over 2 — the cycle starts there, confirming the shift. Finding the input that makes the argument vanish is the reliable check.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 733-735
Faded example
For the argument 5x minus 10.
Fill in the blanks
5x - 10 = 5(x - 2), \textright___ \text______
Why: Factoring 5 out gives 5 times the quantity x minus 2, so the shift is 2 to the right rather than 10. The horizontal coefficient divided the shift by 5, which is what factoring reveals and what reading the constant directly misses.
Worked example
Factor first, then everything reads off.
\[ \text{Describe } y=4\cos(3x+\pi)-1. \]
Factor the inside
Why: Three out of both terms.
\[ 3(x + \frac{\pi}{3}) \]
Read the amplitude and midline
Why: From the outside numbers.
\[ A = 4, D = -1 \]
Read the period
Why: Full turn over 3.
\[ 2 \pi / 3 \]
Read the shift
Why: Plus inside means left.
\[ \text{left } \frac{\pi}{3} \]
Figure (svg): The solution to Worked example read all four parameters shown as a ladder of expressions, one row per legal move
\[ A=4, \; T=\tfrac{2\pi}{3}, \; \text{shift } \tfrac{\pi}{3} \text{ left}, \; D=-1 \]
Verify: check the maximum's location
Why: A cosine peaks where its argument is zero, which is at x equal to negative pi over 3 — the shift, as predicted. And the peak's height is the midline plus the amplitude, negative 1 plus 4, which is 3. Both checks confirm the parameters were read from the factored form correctly.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 735-736
Error analysis
A student finds the phase shift of a compressed sinusoid.
Annotate
On: \( y=\sin(4x-\pi) \;\Longrightarrow\; \text{shift } \pi \text{ right} \)
Factor the horizontal coefficient out before reading any shift. The shift is always divided by that coefficient, and the unfactored constant overstates it every time.
Prediction
A sinusoid is shifted horizontally.
Predict first
Which of its features change?
Correct: Only where the cycle starts.
Why: A horizontal shift slides the whole curve sideways without changing its shape, so the amplitude, period and midline are all unaffected. Only the starting point of each cycle moves, which is what a phase shift means.
Sorting
Inside changes are reversed, as always.
Sort into buckets
Sort each factored argument.
Step zero
You are given a sinusoid with a compressed and shifted argument.
Discussion prompt
What do you do before reading any parameter, and why?
Hint: Which parameter is misread if you skip it?
Answer:
Factor the horizontal coefficient out of the entire argument, so the inside reads as B times a bracketed difference.
Without that, the phase shift is misread — the constant in an unfactored argument is the shift multiplied by B, so it overstates the shift by exactly that factor.
The amplitude, midline and period can all be read without factoring, so the shift is the only casualty. But it is a silent one: the resulting sketch has the right shape in the wrong place, and nothing about it looks wrong.
Section
Section 5
Concept
A sinusoid is sketched by drawing the midline, marking the amplitude above and below it, and dividing one period into quarters to place the five key points.
The quarter-period division is what makes the sketch accurate without plotting. A sine's five key points over one cycle are midline, maximum, midline, minimum, midline — and a cosine's are maximum, midline, minimum, midline, maximum.
Figure (svg): A transformed sine curve with its amplitude, midline and period marked, showing which parameter controls each
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 736-738
Picture it
The midline and amplitude frame the picture before any point is plotted.
Figure (svg): A transformed sine curve with its amplitude, midline and period marked, showing which parameter controls each
With the midline and the two extremes drawn, the five key points across one period determine the curve completely.
Worked example
Frame first, then the five points.
\[ \text{Sketch } y=2\sin(2x)+1 \text{ over one period.} \]
Draw the midline and extremes
Why: Midline 1, amplitude 2.
\[ \text{runs from } -1\text{ to } 3 \]
Find the period
Why: Two pi over 2.
Divide into quarters
Why: Each quarter is pi over 4.
\[ \text{at } 0, \frac{\pi}{4}, \frac{\pi}{2}, 3 \pi / 4, \pi \]
Place the five key points
Why: Sine pattern: mid, max, mid, min, mid.
\[ 1, 3, 1, -1, 1 \]
Figure (svg): A transformed sine curve with its amplitude, midline and period marked, showing which parameter controls each
\[ \text{midline } 1, \; A=2, \; T=\pi \]
Verify: check the quarter points
Why: At pi over 4 the argument is pi over 2, where the sine is 1, giving 2 plus 1 which is 3 — the maximum, as the pattern predicts. Checking one quarter point confirms both the period and the key-point pattern at once.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 736-737
Ranking
Frame the picture before plotting anything.
Put in order
Why: The midline and amplitude frame the vertical extent, then the period and shift frame the horizontal one. Quartering locates the five key points inside that frame, and joining them is last. Plotting before framing means guessing the scale.
Worked example
Each stated feature gives one parameter.
\[ \text{A wave oscillates between } 6 \text{ and } 14 \text{ with period } 4, \text{ starting at its maximum.} \]
Find the midline and amplitude
Why: Average and half difference.
\[ \text{midline } 10, A = 4 \]
Find the coefficient
Why: Full turn over the period.
\[ B = 2 \pi / 4 = \frac{\pi}{2} \]
Choose sine or cosine
Why: Starting at a maximum is the cosine's pattern.
Assemble
Why: No shift needed.
\[ 4 \cos(\pi x / 2) + 10 \]
Figure (svg): The solution to Worked example write a formula from a description shown as a ladder of expressions, one row per legal move
\[ y=4\cos\bigl(\tfrac{\pi}{2}x\bigr)+10 \]
Verify: check the start and the extremes
Why: At x equal to zero the cosine is 1, giving 4 plus 10 which is 14 — the maximum, as required. And the values run from 6 to 14, matching. Choosing the cosine avoided needing a phase shift, which is why the choice between the two functions is worth making deliberately.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 737-738
Trap
\[ \text{a wave starting at its maximum, modelled as } A\sin(Bx)+D \]
Choose the sine as the default function
Why: The sine is used because it is the more familiar of the two.
A phase shift is then needed to make the wave start at a peak.
A cosine starts at its maximum naturally, so choosing it removes the need for any phase shift at all.
The sine version is not wrong — it needs a shift of a quarter period — but it is more work and more places to make a sign error.
Choose the function that matches the starting behaviour. Starting at a maximum suggests a cosine; starting on the midline going up suggests a sine.
Prediction
A standard sine curve over one period.
Predict first
What is the pattern of its five key points?
Correct: Midline, maximum, midline, minimum, midline.
Why: A sine starts on the midline rising, peaks at a quarter period, returns to the midline at half, troughs at three quarters, and returns at the full period. The cosine's pattern is the second option, which is why choosing the right function can avoid a phase shift.
Faded example
A wave has period 10.
Fill in the blanks
B = \frac105} = \frac______}
Why: Two pi over 10 simplifies to pi over 5. A period longer than the standard 2 pi requires a coefficient below 1, and pi over 5 is about 0.63 — consistent. Checking the direction is what catches an inverted formula.
Real world
Anything that oscillates regularly is modelled by one of these.
Discussion prompt
Daylight hours vary between about 8 and 16 over a year. What are the model's four parameters?
Hint: What are the extremes, the cycle length, and where does the cycle start?
Answer:
The midline is 12 hours, the average of 8 and 16, and the amplitude is 4, half their difference.
The period is one year, so the coefficient is a full turn divided by 365 days — or by 12 if the input is months.
The phase shift places the maximum at the summer solstice, which is where the choice of starting point and of sine or cosine gets decided. A cosine centred on the solstice needs no shift at all, which is usually the tidiest formulation — and it is exactly the choice this section's last worked example made.
Comparison
Fill the blanks from memory. Each controls one feature, and only one of them needs factoring first.
Comparison matrix
| controls | read it by | |
|---|---|---|
| A | the amplitude | taking its size, ignoring the sign |
| B | the period | dividing a full turn by it |
| h | the phase shift | factoring B out first, then reading |
| D | the midline | reading it directly |
Only the third row requires any preparation, and it is the one most often misread. The other three read straight off the formula as written.
Pattern
Five steps, and the first is the one that must not be skipped.
Step 1 exists only for step 5, but it has to happen first because the factoring changes what the argument looks like. Doing it before anything else makes the whole reading mechanical.
OpenStax Algebra and Trigonometry 2e, §8.1 Graphs of the Sine and Cosine Functions §8.1
Check
Divide, do not multiply.
Check your understanding
What is the period of the function sine of 5x?
Answer: A
Why: The period is a full turn divided by the inside coefficient, so it is 2 pi over 5. Five complete waves fit into a full turn, which is what multiplying the input by 5 achieves.
Check
Half the swing.
Check your understanding
A sinusoid runs between -3 and 7. What is its amplitude?
Answer: A
Why: The total swing is 10 and the amplitude is half of it, which is 5. The midline is at 2, the average of the two extremes, and 2 plus 5 gives 7 while 2 minus 5 gives negative 3.
Check
Factor first.
Check your understanding
What is the phase shift of the function sine of the quantity 3x minus 6?
Answer: A
Why: Factoring the 3 out gives 3 times the quantity x minus 2, so the shift is 2 to the right. Checking: the argument vanishes at x equal to 2, which is where the cycle starts.
Real world
Every alternating current, sound wave and tide is a sinusoid, and the four parameters have physical names.
Discussion prompt
A mains voltage is described as 230 volts at 50 hertz. Which parameters do those numbers give?
Hint: What does hertz measure, and how does it relate to the period?
Answer:
Fifty hertz means 50 cycles per second, so the period is one fiftieth of a second — and the coefficient B is a full turn divided by that, about 314.
The 230 volts is a kind of average rather than the amplitude directly; the actual peak is higher, about 325 volts, because the quoted figure is a root-mean-square value. But it is the amplitude parameter that the peak corresponds to.
The midline is zero, since the voltage alternates symmetrically about it, and the phase matters when combining two supplies — which is exactly why three-phase power uses three sinusoids shifted by a third of a period each. Every one of this section's parameters has a name in the engineering.
Commit first
State your confidence along with your answer.
Predict first
An inside coefficient of 3 is applied to a sine. What happens to the period?
Correct: It is divided by 3.
Why: An inside factor compresses the graph horizontally by its reciprocal, so each cycle occupies a third of the space and three cycles fit where one did. The amplitude is set by the outside coefficient and is entirely unaffected by anything inside.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the sine graph is a wave, starting from the unit circle.
Hint: What is being plotted against what?
Answer:
Imagine a point moving round the unit circle and plot its height against the angle. It starts level with the centre, rises to the top, comes back down through the middle, drops to the bottom, and returns.
That up-down-up motion, drawn out horizontally, is the wave. Nothing is being approximated: the curve's height at each angle is literally the point's height.
And because the circle closes, a second lap repeats the same heights exactly. A good explanation stresses that the periodicity is not an extra property but a direct consequence of going round a closed path — which is why every oscillating thing in nature ends up described by these functions.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth causes errors that are invisible in the finished sketch, since the curve has the right shape in the wrong place. The second costs marks for a reason that is purely about where the measurement starts from.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw the unit circle beside a set of axes and trace the point's height across to produce one full sine wave, marking which circle positions give the peak, the trough and the zeros. Then write a fully transformed sinusoid, factor its argument, and label the amplitude, period, phase shift and midline, noting which of the four required the factoring.
If your traced wave's peak lines up with the top of the circle, and your phase shift was read from a factored bracket, both halves of the section are on the page.
Recap
Five things, and the first explains why the rest look the way they do.
| if you remember one thing | it should be this |
|---|---|
| about the shape | the graph is the circle unrolled, so periodicity is automatic |
| about amplitude | it is half the total swing, measured from the midline |
| about the period | divide a full turn by B; a bigger B means a faster wave |
| about the shift | factor B out first, or the shift is overstated by B |
Section 6.2 graphs the other four functions, where the asymptotes from §5.3's undefined points become the dominant visual feature.
OpenStax, Precalculus, §6.1 Graphs of the Sine and Cosine Functions §6.1, pp. 716-738 — everything on these slides traces back here
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