8. Read a wave and write its equation

Explain it like I am five

Picture describing ocean swells to a friend: their height, the distance between crests (tops), and what the water does as you start watching. A wave's equation needs the same three clues. Here the midline, the middle line, is y = 0, and ω (omega) is positive.

Example: a wave tops out at 7, bottoms out at −7, repeats every 12 units, and at x = 0 sits in the middle heading down. Height: the amplitude is half the top-to-bottom gap, $7 − (−7) over 2$ = $14 over 2$ = 7. Start: sine rises from the middle; this one falls, so A = −7. Width: the period is T = 12, and one cycle needs ωT = 2π, so ω = $2 π over 12$ = $π over 6$ . The equation is y = −7 sin( $π over 6$ x).

Check at a quarter period, x = 3: the inside is $π over 6$ × 3 = $π over 2$ , so y = −7 × 1 = −7, the bottom. The four starts, with a the amplitude: middle up is a sin, middle down is −a sin, top is a cos, bottom is −a cos.

In plain words

Imagine finding a label for a repeating wallpaper pattern. You measure how tall it is, how wide one repeat is, and where the pattern begins. Those three clues name a sine or cosine wave too. This lesson uses unshifted waves, meaning there is no extra angle added or subtracted inside the function. Their middle, called the midline, is y = 0. You also choose the inside multiplier ω to be positive. Within that family, the start at x = 0 tells you sine or cosine and its sign. The vertical height tells you the amplitude. One complete horizontal repeat tells you the period. You turn those observations into an equation.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Reminder

Midline and maximum/minimum. The middle is halfway between the extremes: for 8 and −8, $8 + (−8) over 2$ = 0, so the midline is y = 0.
Subtract a negative. Subtracting −8 adds 8: 8 − (−8) = 16.
Amplitude and absolute value. If A = −8, amplitude |A| = 8. The sign records reflection.
Sine and cosine at zero. sin 0 = 0 and cos 0 = 1, so −8 sin 0 = 0 while −8 cos 0 = −8.
Quarter-turn values. sin( $π over 2$ ) = 1 and cos π = −1. These locate the first sine peak and halfway cosine extreme.
Solve a one-step equation. If 10ω = 2π, divide both sides by 10 to get ω = $π over 5$ . Substitute back: 10 × $π over 5$ = 2π.
Cancel common factors. $2 π over 6 π$ = $1 over 3$ because the common nonzero factor 2π cancels.
Divide by a fraction. 2π ÷ $7 π over 2$ = 2π × $2 over 7 π$ = $4 over 7$ .
Period and cycle. For y = sin(2x), one complete cycle has width T = π because 2 × π = 2π.
Point plotting. A column with input 5 and output 8 plots the point (5, 8), horizontally first.
Negative input identities. sin(−2x) = −sin(2x) chooses a positive inside multiplier; cos(−2x) = cos(2x) needs no sign change.
Range and intervals. An amplitude of 8 around zero gives −8 ≤ y ≤ 8, written [−8, 8].

Why it works. At zero, sine is zero and cosine is 1. With ω > 0, positive sine initially rises, and negative sine initially falls. Positive cosine starts at its maximum, while negative cosine starts at its minimum. These facts identify the function and sign for the restricted unshifted family. The vertical span contains two amplitudes. A full cycle makes the inside angle advance by 2π, so ωT = 2π. Solving that equation gives ω = . Together, these measurements rebuild the wave.

RuleRule: For nonconstant y = A sin(ωx) or y = A cos(ωx), centered at y = 0 with ω > 0, amplitude a = and ω = . At x = 0: middle up chooses a sin; middle down chooses −a sin; top chooses a cos; bottom chooses −a cos.

The same idea, five ways

Say it

Say: start chooses the wave and sign, height chooses amplitude, width chooses omega.

Write it

For an unshifted sine or cosine wave centered at zero, its start, amplitude, and period determine an equation when ω is chosen positive.

In math

y = A sin(ωx) or y = A cos(ωx)
A ≠ 0, ω > 0, midline y = 0
a = |A| = $maximum − minimum over 2$
ω = $2 π over T$
Range: −a ≤ y ≤ a, or [−a, a]

Identify a repeating fabric by its starting motif, height, and width.

See it

The graph's start, height, and cycle width determine the coefficients in this restricted family.

The same idea, other ways

As a picture

Read three independent features. Vertical distance gives amplitude, horizontal repeat distance gives period, and the first point with its direction gives the signed sine or cosine start.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

As a recipe

The graph is the cooked result. Its height tells you how much outside multiplier was used. Its cycle width tells you the inside multiplier. Its start tells you which base wave and reflection were used.

With small numbers

A wave reaches 1 and −1, repeats every 2π, and crosses zero going up. Its amplitude is $1 − (−1) over 2$ = 1, its ω is $2 π over 2 π$ = 1, and its equation is y = sin x. Substituting x = $π over 2$ gives 1, matching its first peak.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Why the three clues suffice

Choosing ω > 0 fixes the direction of the input's travel. The start then fixes the family and sign. The extreme heights fix |A|. The full cycle fixes ω because ωT = 2π. No other coefficient remains to choose in y = A sin(ωx) or y = A cos(ωx).

The table pairs each unshifted zero-midline start with its signed equation family.

Start at x = 0, with ω > 0	Equation family
Middle going up	y = a sin(ωx)
Middle going down	y = −a sin(ωx)
Top	y = a cos(ωx)
Bottom	y = −a cos(ωx)

.1Middle going up: positive sine

You see the curve cross y = 0 at x = 0 and rise immediately to the right. In the unshifted family with ω > 0, this is the positive sine start. Imagine stepping onto a swing as it passes through the middle and heads upward. The graph's direction matters because both positive and negative sine equal zero at the start. Its first quarter-cycle height tells them apart.

Rule: With A ≠ 0 and ω > 0, an unshifted zero-midline wave that starts at the middle going up has y = a sin(ωx), where a is its positive amplitude.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

The same idea, five ways

Say it

Say: middle and up means positive sine.

Write it

A positive sine curve starts at zero and first moves upward.

In math

y = a sin(ωx), a > 0, ω > 0
f(0) = 0
f( $T over 4$ ) = a

Begin a ride at the middle while it is climbing.

See it

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Worked exampleMiddle going up gives positive sine

The pictured unshifted wave centered at y = 0 has a maximum of 3, a minimum of −3, a full cycle of 2π, and is at the middle going up at x = 0. Find its amplitude, period, positive ω, and equation. In plain words, use the pictured wave to name the sine or cosine rule that produces it.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 3 and −3. Their middle is y = 0.Their sum is zero, so $3 + (−3) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $3 − (−3) over 2$ = $6 over 2$ = 3.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the middle going up. Choose positive sine, so A = 3.Sine begins at zero. The positive parent first rises, so a positive multiplier gives the observed direction.
Read one complete repeat as T = 2π. Find ω = $2 π over 2 π$ = 1. Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = 3 sin x.The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $π over 2$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 3
Period T = 2π
ω = 1
y = 3 sin x

Check Check by substituting instead of reading the graph. The inside products are 1 × 0 = 0, 1 × = , 1 × π = π, 1 × = , and 1 × 2π = 2π. Thus 3 sin 0 = 0, 3 sin() = 3, 3 sin π = 0, 3 sin() = −3, and 3 sin(2π) = 0. These match all five table columns. The extremes of the equation are 3 and −3, so half their difference returns 3. Also 1 × 2π = 2π verifies both the solved ω and the measured period.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: y = 3 sin x and y = −3 sin x are interchangeable because both start at zero.

The first has y = 3 at , and the second has y = −3 there.

✓ Instead: Use the initial rise to select y = 3 sin x.

Tips and tricks

Tip: Check one quarter period after zero. A positive height confirms positive sine.

.2Middle going down: negative sine

You see the curve cross y = 0 at x = 0 and fall immediately to the right. It starts at the same height as positive sine but travels in the opposite direction. Think of a swing passing through the middle on its way downward. With ω chosen positive, the negative outside multiplier creates that reflection. The amplitude still measures a positive distance from the middle to an extreme.

Rule: With ω > 0, an unshifted zero-midline wave that starts at the middle going down has y = −a sin(ωx), where a > 0 is the amplitude.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

The same idea, five ways

Say it

Say: middle and down means negative sine.

Write it

A negative sine curve begins at zero and first moves below the middle.

In math

y = −a sin(ωx), a > 0, ω > 0
f(0) = 0
f( $T over 4$ ) = −a

Begin a ride at the middle while it is descending.

See it

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Worked exampleMiddle going down gives negative sine

The pictured unshifted wave centered at y = 0 has a maximum of 5, a minimum of −5, a full cycle of 4π, and is at the middle going down at x = 0. Find its amplitude, period, positive ω, and equation. In plain words, use the pictured wave to name the sine or cosine rule that produces it.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 5 and −5. Their middle is y = 0.Their sum is zero, so $5 + (−5) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $5 − (−5) over 2$ = $10 over 2$ = 5.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the middle going down. Choose negative sine, so A = −5.Sine begins at zero. The positive parent first rises, so a negative multiplier gives the observed direction.
Read one complete repeat as T = 4π. Find ω = $2 π over 4 π$ = $1 over 2$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = −5 sin( $1 over 2$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = π. Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 5
Period T = 4π
ω = $1 over 2$
y = −5 sin( $1 over 2$ x)

Check Check by substituting instead of reading the graph. The inside products are × 0 = 0, × π = , × 2π = π, × 3π = , and × 4π = 2π. Thus −5 sin 0 = 0, −5 sin() = −5, −5 sin π = 0, −5 sin() = 5, and −5 sin(2π) = 0. These match all five table columns. The extremes of the equation are 5 and −5, so half their difference returns 5. Also × 4π = 2π verifies both the solved ω and the measured period.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: a downward start means the amplitude is −5.

The start direction determines the sign of A, while amplitude is a distance.

✓ Instead: A = −5 and amplitude = |−5| = 5.

Tips and tricks

Tip: Write both amplitude and A. They have different jobs when the wave is reflected.

.3Top: positive cosine

You see the curve at its highest point when x = 0. It then heads down toward the middle. Think of a ride beginning with its seat at the very top. In the unshifted family, cosine provides this start because cos 0 = 1. Multiplying by a positive amplitude places that first point at the top of the wave. The period changes how far you travel horizontally before returning to that same top.

Rule: With ω > 0, an unshifted zero-midline wave that starts at the top has y = a cos(ωx), where a > 0 is its amplitude.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

The same idea, five ways

Say it

Say: top means positive cosine.

Write it

A positive cosine curve starts at its maximum height.

In math

y = a cos(ωx), a > 0, ω > 0
f(0) = a
f(T) = a

Begin a ride in its highest seat.

See it

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Worked exampleA top start gives positive cosine

The pictured unshifted wave centered at y = 0 has a maximum of 4, a minimum of −4, a full cycle of 6π, and is at the top at x = 0. Find its amplitude, period, positive ω, and equation. In plain words, use the pictured wave to name the sine or cosine rule that produces it.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 4 and −4. Their middle is y = 0.Their sum is zero, so $4 + (−4) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $4 − (−4) over 2$ = $8 over 2$ = 4.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the top. Choose positive cosine, so A = 4.Cosine at zero is 1, so A cos 0 = A. The start is the maximum, 4.
Read one complete repeat as T = 6π. Find ω = $2 π over 6 π$ = $1 over 3$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = 4 cos( $1 over 3$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $3 π over 2$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 4
Period T = 6π
ω = $1 over 3$
y = 4 cos( $1 over 3$ x)

Check Check by substituting instead of reading the graph. The inside products are × 0 = 0, × = , × 3π = π, × = , and × 6π = 2π. Thus 4 cos 0 = 4, 4 cos() = 0, 4 cos π = −4, 4 cos() = 0, and 4 cos(2π) = 4. These match all five table columns. The extremes of the equation are 4 and −4, so half their difference returns 4. Also × 6π = 2π verifies both the solved ω and the measured period.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: the x-distance from the top to the next bottom is the period.

The bottom is only halfway through the cycle. The direction and height have not both repeated.

✓ Instead: Measure top to next top. Top to bottom is .

Tips and tricks

Tip: Measure between consecutive peaks. Skipping a peak counts more than one period.

.4Bottom: negative cosine

You see the curve at its lowest point when x = 0. It then moves up toward the middle. Think of starting a ride in its lowest seat. Negative cosine supplies this start because cos 0 = 1 and a negative outside multiplier turns that output into a negative height. You still find the positive amplitude from half the total vertical span. Keep the negative sign only in A and in the equation.

Rule: With ω > 0, an unshifted zero-midline wave that starts at the bottom has y = −a cos(ωx), where a > 0 is the amplitude.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

The same idea, five ways

Say it

Say: bottom means negative cosine.

Write it

A negative cosine curve begins at its minimum height.

In math

y = −a cos(ωx), a > 0, ω > 0
f(0) = −a
f( $T over 2$ ) = a

Begin a ride in its lowest seat.

See it

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Worked exampleA bottom start gives negative cosine

The pictured unshifted wave centered at y = 0 has a maximum of 7, a minimum of −7, a full cycle of 9, and is at the bottom at x = 0. Find its amplitude, period, positive ω, and equation. In plain words, use the pictured wave to name the sine or cosine rule that produces it.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 7 and −7. Their middle is y = 0.Their sum is zero, so $7 + (−7) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $7 − (−7) over 2$ = $14 over 2$ = 7.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the bottom. Choose negative cosine, so A = −7.Cosine at zero is 1, so A cos 0 = A. The start is the minimum, −7.
Read one complete repeat as T = 9. Find ω = $2 π over 9$ = $2 π over 9$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = −7 cos( $2 π over 9$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $9 over 4$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 7
Period T = 9
ω = $2 π over 9$
y = −7 cos( $2 π over 9$ x)

Check Check by substituting instead of reading the graph. The inside products are × 0 = 0, × = , × = π, × = , and × 9 = 2π. Thus −7 cos 0 = −7, −7 cos() = 0, −7 cos π = 7, −7 cos() = 0, and −7 cos(2π) = −7. These match all five table columns. The extremes of the equation are 7 and −7, so half their difference returns 7. Also × 9 = 2π verifies both the solved ω and the measured period.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: a bottom start can use y = −7 sin(ωx).

Sine at zero is zero, so the equation would begin at the middle rather than −7.

✓ Instead: Use negative cosine; −7 cos 0 = −7.

Tips and tricks

Tip: Substitute x = 0 first. Cosine returns A, and sine returns zero.

Strategy: step by step

1. Confirm that the midline is y = 0 and that the problem uses an unshifted sine or cosine wave. This is the family the four-start method identifies.
2. Find the amplitude from half the vertical span. It is positive for a changing wave.
3. At x = 0, choose sine or cosine and the sign of A from the starting height and direction. Set |A| equal to the amplitude.
4. Measure the horizontal distance from one point to the next matching point with the same direction. That distance is T.
5. Solve ωT = 2π by dividing both sides by T. Choose ω = $2 π over T$ > 0 and write the equation.
6. Substitute the five quarter-period inputs into the equation. Compare all five outputs with the graph, and verify ωT = 2π.

Strategy

Strategy: write an equation from a sine or cosine graph

Is this an unshifted sine or cosine wave with midline y = 0?

YesContinue with the four-start method and choose ω > 0.

NoUse the later transformation method for a displaced midline or general phase shift. The four-start shortcut does not determine those extra changes.

↓

At x = 0, is the curve on the middle line?

YesChoose sine. Rising to the right gives positive A; falling gives negative A.

NoChoose cosine if the start is an extreme. A top start gives positive A; a bottom start gives negative A.

↓

Do the proposed period endpoints have the same height and direction?

YesCheck that they are consecutive matching positions. Their horizontal distance is T.

NoContinue to the next matching position. A zero crossing going down does not match a zero crossing going up.

↓

Do all five substituted points match the graph?

YesThe equation fits the start, extremes, and complete period.

NoRecheck the sign, read one full period, and divide 2π by T again.

Confirm the family and y = 0 midline. Choose positive ω.
Measure amplitude, read the starting position and direction, and choose signed A with sine or cosine.
Measure one complete repeat T and calculate ω = $2 π over T$ .
Write the equation and check all five quarter-period points.

Worked exampleRead a bottom-start graph with a plain-number period

The pictured unshifted wave centered at y = 0 has a maximum of 8, a minimum of −8, a full cycle of 10, and is at the bottom at x = 0. Find its amplitude, period, positive ω, and equation. In plain words, use the pictured wave to name the sine or cosine rule that produces it.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 8 and −8. Their middle is y = 0.Their sum is zero, so $8 + (−8) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $8 − (−8) over 2$ = $16 over 2$ = 8.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the bottom. Choose negative cosine, so A = −8.Cosine at zero is 1, so A cos 0 = A. The start is the minimum, −8.
Read one complete repeat as T = 10. Find ω = $2 π over 10$ = $π over 5$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = −8 cos( $π over 5$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $5 over 2$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 8
Period T = 10
ω = $π over 5$
y = −8 cos( $π over 5$ x)

Check Check by substituting instead of reading the graph. The inside products are × 0 = 0, × = , × 5 = π, × = , and × 10 = 2π. Thus −8 cos 0 = −8, −8 cos() = 0, −8 cos π = 8, −8 cos() = 0, and −8 cos(2π) = −8. These match all five table columns. The extremes of the equation are 8 and −8, so half their difference returns 8. Also × 10 = 2π verifies both the solved ω and the measured period.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: read a taller middle-up graph

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 3 and −3. Their middle is y = 0.Their sum is zero, so $3 + (−3) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $3 − (−3) over 2$ = $6 over 2$ = 3.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the middle going up. Choose positive sine, so A = 3.Sine begins at zero. The positive parent first rises, so a positive multiplier gives the observed direction.
Read one complete repeat as T = 2π. Find ω = $2 π over 2 π$ = 1. Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = 3 sin x.The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $π over 2$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 3
Period T = 2π
ω = 1
y = 3 sin x

Rung 2Rung 2: read a reflected wave with a long period

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 5 and −5. Their middle is y = 0.Their sum is zero, so $5 + (−5) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $5 − (−5) over 2$ = $10 over 2$ = 5.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the middle going down. Choose negative sine, so A = −5.Sine begins at zero. The positive parent first rises, so a negative multiplier gives the observed direction.
Read one complete repeat as T = 4π. Find ω = $2 π over 4 π$ = $1 over 2$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = −5 sin( $1 over 2$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = π. Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 5
Period T = 4π
ω = $1 over 2$
y = −5 sin( $1 over 2$ x)

Rung 3Rung 3: bottom start and a plain-number period

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 8 and −8. Their middle is y = 0.Their sum is zero, so $8 + (−8) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $8 − (−8) over 2$ = $16 over 2$ = 8.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the bottom. Choose negative cosine, so A = −8.Cosine at zero is 1, so A cos 0 = A. The start is the minimum, −8.
Read one complete repeat as T = 10. Find ω = $2 π over 10$ = $π over 5$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = −8 cos( $π over 5$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $5 over 2$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 8
Period T = 10
ω = $π over 5$
y = −8 cos( $π over 5$ x)

Rung 4Rung 4: bottom start and a fractional period

The pictured unshifted wave centered at y = 0 has a maximum of 4, a minimum of −4, a full cycle of $7 π over 2$ , and is at the bottom at x = 0. Find its amplitude, period, positive ω, and equation. In plain words, use the pictured wave to name the sine or cosine rule that produces it.

The graph's start, height, and cycle width determine the coefficients in this restricted family.

Substitute each input to check the reconstructed curve's five heights.

Confirm that the highest and lowest heights are 4 and −4. Their middle is y = 0.Their sum is zero, so $4 + (−4) over 2$ = 0. This checks the zero-midline requirement. The question already specifies an unshifted wave.
Find the amplitude: $4 − (−4) over 2$ = $8 over 2$ = 4.The distance from bottom to top contains two equal amplitudes.
At x = 0 the graph is at the bottom. Choose negative cosine, so A = −4.Cosine at zero is 1, so A cos 0 = A. The start is the minimum, −4.
Read one complete repeat as T = $7 π over 2$ . Find ω = 2π ÷ $7 π over 2$ = 2π × $2 over 7 π$ = $4 over 7$ . Choose ω > 0.A full cycle advances the inside angle by 2π, so ωT = 2π and dividing both sides by T finds the inside multiplier.
Write y = −4 cos( $4 over 7$ x).The outside multiplier gives the correct heights and start; the inside multiplier gives the observed period.
For a second check, split the period into quarters: $T over 4$ = $7 π over 8$ . Use the five-column picture to compare the equation with the graph.Checking the start, both extremes, the middle crossings, and the end tests more than the starting height alone.

Answer

Amplitude = 4
Period T = $7 π over 2$
ω = $4 over 7$
y = −4 cos( $4 over 7$ x)

Check Check by substituting instead of reading the graph. The inside products are × 0 = 0, × = , × = π, × = , and × = 2π. Thus −4 cos 0 = −4, −4 cos() = 0, −4 cos π = 4, −4 cos() = 0, and −4 cos(2π) = −4. These match all five table columns. The extremes of the equation are 4 and −4, so half their difference returns 4. Also × = 2π verifies both the solved ω and the measured period.

Rung 5Rung 5: measure the period from the picture

Use the labeled graph of an unshifted sine or cosine wave centered at y = 0. Read its highest and lowest heights, choose the function and its sign, measure one complete repeat, and write its equation with positive ω. In plain words, use the picture's labels to find the wave's height and width before choosing its rule.

Read two consecutive top inputs, then subtract them to measure a full repeat.

These computed check columns match the graph after its period is measured.

Read the height labels at the top and bottom points: the highest output is 11 and the lowest output is −11. Their middle is (11 + (−11)) ÷ 2 = 0.A point's second coordinate is its output height. The average of the two extremes locates the midline and confirms that this wave is centered at y = 0.
Find the full height gap: 11 − (−11) = 11 + 11 = 22. Halve it: amplitude = 22 ÷ 2 = 11.Subtracting the bottom height from the top height measures both equal middle-to-extreme distances together. Amplitude measures only one of them.
At x = 0, the graph is at its top. Choose positive cosine, with A = 11.Cosine at zero is 1, so 11 cos 0 = 11 gives the observed top. Sine would give output 0 at zero, which would miss the starting point.
Read the horizontal labels of two consecutive tops: x = 0 and x = 14. Subtract their input positions: T = 14 − 0 = 14. The picture also shows the previous top at x = −14. Using that pair gives T = 0 − (−14) = 14, so subtracting the left input from the right input works even when the first top is not at zero.The period is the horizontal distance between consecutive matching positions in the wave. Top to next top includes both the descent to the bottom and the return to the top.
Reject 7 as the period. The distance from the top at x = 0 to the bottom at x = 7 is 7 − 0 = 7, which is half of 14.The bottom has output −11, while the start has output 11. These positions have different heights, so they do not mark a complete repeat.
Set 14ω = 2π to find the positive inside multiplier. Divide both sides by 14: ω = $2 π over 14$ = $π over 7$ . Substitute back: 14 × $π over 7$ = 2π.The measured input distance 14 must advance the inside angle by one full turn. Division isolates the missing multiplier, and substitution checks that it gives the intended full turn.
Write y = 11 cos( $π over 7$ x).Positive cosine supplies the top start, 11 supplies the amplitude, and $π over 7$ supplies the measured cycle width.
Build the check inputs from the measured period. One quarter is $14 over 4$ = $7 over 2$ . Two quarters give $7 over 2$ + $7 over 2$ = 7. Three quarters give 3 × $7 over 2$ = $21 over 2$ . Four quarters give 4 × $7 over 2$ = 14.Four equal horizontal gaps span one complete cycle. Calculating the multiples gives the five input columns from the starting input 0 to the ending input 14.
Substitute the five inputs into the inside angle. The products are $π over 7$ × 0 = 0, $π over 7$ × $7 over 2$ = $π over 2$ , $π over 7$ × 7 = π, $π over 7$ × $21 over 2$ = $3 π over 2$ , and $π over 7$ × 14 = 2π.These products identify the circle angle reached at each graph input. They are the five quarter-turn angles whose cosine values are already known.
Evaluate each output: 11 cos 0 = 11, 11 cos( $π over 2$ ) = 0, 11 cos π = −11, 11 cos( $3 π over 2$ ) = 0, and 11 cos(2π) = 11. Compare them with the graph and the check table.Cosine reads the circle's horizontal coordinate. Multiplying its parent outputs by 11 gives the graph's observed top, middle, bottom, middle, and top.

Answer

Amplitude = 11
Period T = 14
ω = $π over 7$
y = 11 cos( $π over 7$ x)
The top-to-bottom distance 7 is half a cycle.

Check Measure the same full repeat with the middle crossings: the curve crosses y = 0 going down at x = and going up at x = . Their distance is − = = 7. These opposite-direction crossings are half a cycle apart, so doubling that distance gives T = 14 again. The equation's extreme heights 11 and −11 give (11 − (−11)) ÷ 2 = 11, matching the height read from the picture.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: for a wave with maximum 8 and minimum −8, use amplitude 16.

Sixteen is the entire bottom-to-top span, which contains two amplitudes.

✓ Instead: a = = 8. Its heights are 8 and −8.

✗ Not this: Counterexample: a wave crossing zero going up has period equal to the distance to the next zero crossing.

The next crossing goes down and occurs halfway through a sine cycle.

✓ Instead: Measure to the next zero crossing going up, or double the distance to the downward crossing.

✗ Not this: Counterexample: use the four-start shortcut on y = sin(x + ).

There is an extra angle inside, so the wave is horizontally shifted. At x = 0 its height is sin() = , which is neither its middle nor an extreme.

✓ Instead: General phase-shift problems belong to the later transformation method. Here use the family with no extra inside angle.

✗ Not this: Counterexample: from a downward sine start conclude that ω must be negative.

We choose ω > 0 and put the reflection in A. Negative ω would make the same curve possible with another sign choice.

✓ Instead: For example, sin(−2x) = −sin(2x). In this lesson write A = −1 and ω = 2.

Tips and tricks

Tip: Know cold: The graph input uses radians unless a degree sign is written. Memory cue: π is a half turn; 2π is a full turn.
Tip: Know cold: Amplitude is |A|. Memory cue: amplitude is a distance, so drop the coefficient's direction sign.
Tip: Know cold: T = $2 π over |ω|$ for A ≠ 0 and ω ≠ 0. Memory cue: faster inside, shorter outside distance.
Tip: Know cold: With A > 0 and ω > 0, sine starts at the center rising and cosine starts at the crest. Memory cue: sine at center, cosine at crest.
Tip: Know cold: Sine is odd with origin symmetry; cosine is even with y-axis symmetry. Memory cue: odd changes both point signs; even keeps equal heights at opposite inputs.
Tip: Memory cue: center means sine, crest means cosine. Then choose the sign from the first direction or extreme.
Tip: Amplitude is the positive height measure. Write signed A on a separate line.
Tip: Plain-number periods still produce radian inside angles. If T = 10, ω = $π over 5$ , because $π over 5$ × 10 = 2π.
Tip: Rebuild the equation from three graph measurements rather than memorize individual worksheet curves.
Cheat sheet tip: Keep the four-start rule, a = $maximum − minimum over 2$ , ω = $2 π over T$ , and a note that the shortcut uses unshifted y = 0 waves.
Tip: Understand, then rebuild it when needed: Rebuild exact sine values from the circle and the two special triangles. Rebuild the five key points by dividing T into four gaps. Rebuild amplitude from half the vertical span. Rebuild ω from ωT = 2π. Rebuild a cofunction argument by subtracting from one quarter turn. Do not memorize individual transformed graphs or long value tables; each can be reconstructed from these ideas.
Tip: Put on the cheat sheet: Keep the amplitude, period, range and inverse-period formulas, the four starts, the five-point pattern pictures, and the six cofunction identities. Look these up while studying; practice rebuilding them for a closed-book exam.
Tip: One-page cheat sheet: For y = A sin(ωx) or y = A cos(ωx), A ≠ 0 and ω ≠ 0.
Amplitude = |A| = $maximum − minimum over 2$ .
Domain = (−∞, ∞); range = [−|A|, |A|]; midline y = 0.
T = $2 π over |ω|$ , quarter period = $T over 4$ , |ω| = $2 π over T$ .
Choose ω > 0 when reading a graph. At x = 0: center rising means positive sine; center falling means negative sine; top means positive cosine; bottom means negative cosine.
Use the sine and cosine key-point table pictures, not a memorized set of transformed curves.
Sine: sin(−x) = −sin x. Cosine: cos(−x) = cos x.
Both repeat after a full inside turn of 2π.
sin x = cos( $π over 2$ − x); cos x = sin( $π over 2$ − x).
tan x = cot( $π over 2$ − x); cot x = tan( $π over 2$ − x).
sec x = csc( $π over 2$ − x); csc x = sec( $π over 2$ − x).
Use 90° in place of $π over 2$ for degree arguments; values must be defined.
cos x = sin(x + $π over 2$ ): cosine is sine shifted $π over 2$ left.
This four-start graph method uses an unshifted wave with midline y = 0.

Trap. Trap: The four-start method assumes an unshifted wave centered at y = 0 and a positive ω. A wave with another midline or an arbitrary horizontal shift needs the later transformation method.

Keep in mind

The amplitude is half the gap, not the whole gap: from −7 to 7 the gap is 14 and the amplitude is 7.
Measure the period between two points that match in height and direction, such as top to next top, because top to bottom is only half a period.
ω comes from the period through ω = $2 π over T$ , never the period itself: a period of 30 gives ω = $2 π over 30$ = $π over 15$ , not 30.
The four starts work only for waves centered on y = 0 with no sideways shift; a shifted wave needs the transformation method of 5-6.

Memory hookLook at x = 0: middle going up is sin, middle going down is −sin, top is cos, bottom is −cos.

Flash cards: say the answer out loud, then flip

What is the midline of a wave?

The level line halfway between its top and bottom
here y = 0.

At x = 0 the wave is at its top. Which family?

y = a cos(ωx), with a > 0

At x = 0 the wave is in the middle, heading up. Which family?

y = a sin(ωx), with a > 0

Top 9, bottom −9, starts at the bottom, period 10π. Write the equation.

y = −9 cos(x), since ω = =

A period is 16. Find the positive ω.

A wave goes from a top to the next bottom in 5 units. Is the period 5?

No. Top to bottom is half a cycle, so the period is 10.