7. Sketch a complete cycle with five key points

Explain it like I am five

Picture drawing one hill and one valley on a strip of wallpaper. You need only five landmark dots and a smooth curve through them. These landmarks, the key points, split one cycle (one full repeat of the wave) into four equal steps.

Example: y = 6 cos(πx). A = 6, so the amplitude, the middle-to-top height, is 6. ω = π, so the period T, the width of one cycle, is $2 π over π$ = 2. Each step is a quarter period: $T over 4$ = $2 over 4$ = $1 over 2$ . The inputs are 0, $1 over 2$ , 1, $3 over 2$ , 2. Cosine's heights go top, middle, bottom, middle, top: 1, 0, −1, 0, 1. Multiply each by A = 6: 6, 0, −6, 0, 6. The key points are (0, 6), ( $1 over 2$ , 0), (1, −6), ( $3 over 2$ , 0) and (2, 6).

Join them with a smooth, rounded curve. Why these five: at those inputs the inside, πx, equals 0, $π over 2$ , π, $3 π over 2$ and 2π, the quarter turns where cosine is 1, 0, −1, 0, 1. To draw more, copy the cycle every 2 units.

In plain words

Imagine drawing one hill and one valley on a strip of wallpaper. You do not need a dot at every possible place. You need a few reliable landmarks, then a smooth line between them. A wave works the same way. Its key points are five landmarks that divide one complete repeat, called a cycle, into four equal steps. The period is the width of that repeat. The amplitude is the distance from the middle to the top. You find the five horizontal positions from the period, then choose the heights from the sine or cosine pattern. After that, you can copy the finished cycle across the page.

The five points divide the reflected cycle into four equal horizontal steps.

Reminder

Input, output, and ordered pair. A point (x, y) places x horizontally and y vertically: for y = 6 sin x, x = $π over 2$ gives y = 6, so plot ( $π over 2$ , 6).
Radian and unit circle. One quarter-turn is $π over 2$ radians; at it, sin( $π over 2$ ) = 1 and cos( $π over 2$ ) = 0.
Absolute value. Amplitude is a distance: |−4| = 4.
Signed multiplication. A negative times a negative is positive: −4 × (−1) = 4.
Divide fractions. Divide by a fraction by multiplying by its reciprocal: 2π ÷ $2 over 3$ = 2π × $3 over 2$ = 3π.
Divide by a whole number. $π over 2$ ÷ 4 = $π over 2$ × $1 over 4$ = $π over 8$ .
Solve and substitute. To find a full cycle for ω = 3, solve 3T = 2π to get T = $2 π over 3$ ; plugging back gives 3 × $2 π over 3$ = 2π.
Closed intervals. [0, 3π] means 0 ≤ x ≤ 3π, including both endpoints.
Sine and cosine signs. sin( $3 π over 2$ ) = −1, while cos π = −1. A negative A reverses those outputs.
Negative inputs. sin(−u) = −sin u and cos(−u) = cos u: sin(− $π over 2$ ) = −1 and cos(−π) = −1.
Exact value. Keep π exact in calculations: $π over 8$ is exact, while its decimal would be rounded.

Why it works. One turn around the unit circle contains four quarter-turns. At those quarter-turns, sine and cosine have outputs 0, 1, or −1. Multiplying x by ω changes where these angles occur but preserves their order when ω > 0. Multiplying the outputs by A changes their heights. Therefore the five transformed points remain equally spaced by , and direct substitution produces their heights. The connecting curve stays smooth because the circle coordinates change continuously.

RuleRule: For A ≠ 0 and ω > 0, y = A sin(ωx) or y = A cos(ωx) has T = . Use the input columns 0, , , , T in the key-point picture. Multiply the parent heights by A, connect smoothly, and repeat every T.

The same idea, five ways

Say it

Say: find one repeat, split its width into quarters, and draw the five landmarks.

Write it

The five key points divide a complete sine or cosine cycle into four equal horizontal steps.

In math

T = $2 π over ω$ , A ≠ 0, ω > 0
Quarter period = $T over 4$
y = A sin(ωx) or y = A cos(ωx)
One cycle: 0 ≤ x ≤ T, or [0, T]

Place evenly spaced fence posts, then run a smooth ribbon through their chosen heights.

See it

Cosine uses the same horizontal posts and different heights.

The same idea, other ways

As a picture

The graph is a road with five landmarks: a starting position, a quarter-cycle position, a halfway position, a three-quarter position, and the matching endpoint.

Four equal gaps separate the five landmarks.

With numbers

If T = 12, the quarter period is 12 ÷ 4 = 3. For y = sin( $π over 6$ x), each step of 3 makes the inside angle advance by $π over 2$ . The picture holds the input/output pairs.

The step of 3 is one quarter of a period of 12.

.1Sine key-point pattern

For positive sine, you begin at the middle, climb to the top, return to the middle, descend to the bottom, and return to the middle. Think of one rise and one fall on a ribbon. Multiplying all heights by a negative A reflects the ribbon across y = 0. It then begins by going down. Read heights from the table instead of guessing the curve.

Rule: For y = A sin(ωx), A ≠ 0, ω > 0, use the sine heights in the five-column picture.
Rule: Positive A starts at the middle going up; negative A starts at the middle going down.

Positive sine starts at zero and first rises.

The same idea, five ways

Say it

Say: middle, high, middle, low, middle for positive sine.

Write it

Sine begins on the middle line and its sign chooses whether it initially rises or falls.

In math

y = A sin(ωx), A ≠ 0, ω > 0
f(0) = 0
f( $T over 4$ ) = A
f( $3 T over 4$ ) = −A

Walk away from the middle floor, return, walk below it, then return again.

See it

Multiplying sine's five parent heights by A produces these columns.

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

✗ Not this: Counterexample: y = −4 sin x starts at the middle going up.

The first positive sine height is multiplied by −4, so it becomes negative.

✓ Instead: It begins at the middle going down; the first trough is ( , −4 ).

Tips and tricks

Tip: Memory cue: sine starts at the center. Keep A's sign beside your height row.

.2Cosine key-point pattern

For positive cosine, you begin at the top, pass through the middle, reach the bottom, pass through the middle again, and return to the top. This is like starting a ride at its highest seat instead of at the side. The same quarter-period spacing still works. If A is negative, every height reverses and you start at the bottom.

Rule: For y = A cos(ωx), A ≠ 0, ω > 0, multiply the cosine heights in the picture by A.
Rule: Positive A begins at the top; negative A begins at the bottom.

The maximum is 6, so the positive cosine cycle starts there.

The same idea, five ways

Say it

Say: high, middle, low, middle, high for positive cosine.

Write it

Cosine begins at an extreme, and A's sign selects the top or bottom.

In math

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

Begin a rocking movement at its farthest position instead of at the center.

See it

Cosine's start and end have the same extreme height.

Worked exampleThe cosine pattern with an output multiplier

Sketch y = 6 cos x on 0 ≤ x ≤ 2π. This asks you to draw one cycle that has the cosine shape but reaches six units from its middle.

Match the five plotted points to the five columns of the key-point table.

Each input column tells you where to place its output height.

Read A = 6 and ω = 1. The amplitude is 6 and T = 2π.The outside multiplier changes heights, while the inside coefficient leaves the cycle width unchanged.
Compute the quarter period $2 π over 4$ = $π over 2$ .Four steps of this size complete the cycle.
Multiply every parent cosine output by 6 and use the output row in the table.Cosine's circle coordinate starts at 1, passes through 0, reaches −1, passes through 0, and returns to 1.
Start at the top, plot the table columns, and connect smoothly.Positive cosine begins at its maximum, so it does not begin at the origin.

Answer

Amplitude = 6
Period = 2π
Quarter period = $π over 2$
The curve starts and ends at y = 6.

Check Direct substitution gives 6 cos 0 = 6, 6 cos() = 0, 6 cos π = −6, 6 cos() = 0, and 6 cos(2π) = 6. The span from −6 to 6 is 12, so its half is 6. The inside angle advances by 1 × 2π = 2π.

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

✗ Not this: Counterexample: put the first point of y = 6 cos x at (0, 0).

Cosine at 0 is 1, and 6 × 1 = 6.

✓ Instead: The first point is (0, 6).

Tips and tricks

Tip: Memory cue: cosine starts at the crest when A is positive. Substitute x = 0 to check.

.3Repeating the cycle

A finished cycle is a pattern you can stamp again. To move a point one cycle to the right, add T to its input while keeping its height. To move it left, subtract T. This is like copying a wallpaper strip next to itself. At each shared endpoint, the copied curve continues in the same direction. The period tells you the stamp's width.

Rule: For every integer k, f(x + kT) = f(x). An integer is a whole-number count such as −2, −1, 0, 1, 2.
Rule: Copy (x, y) to (x + kT, y). The vertical coordinate stays unchanged.

One sine cycle copied left and right fills the fresh interval with three cycles.

The same idea, five ways

Say it

Say: add one period to the input and keep the output.

Write it

A periodic function repeats its entire pattern after every full period.

In math

f(x + kT) = f(x), k an integer
(x, y) becomes (x + kT, y)
For these waves, T = $2 π over ω$

Copy one tile beside itself without stretching or turning it.

See it

Three matching cycles share the same height pattern.

Worked exampleMove a cycle left and right

Sketch y = 2 sin(3x) on − $2 π over 3$ ≤ x ≤ $4 π over 3$ . This asks you to show three complete cycles by copying one correct cycle across the requested interval.

One sine cycle copied left and right fills the fresh interval with three cycles.

Columns one period apart have equal heights.

Read A = 2 and ω = 3. Find amplitude = 2 and T = $2 π over 3$ .A controls height and ω determines how far x travels while the inside angle makes one turn.
Compute $T over 4$ = $2 π over 3$ ÷ 4 = $π over 6$ . Use the five columns from x = 0 through x = $2 π over 3$ for the middle cycle.Five quarter-spaced points give one complete sine cycle.
Copy each point once to the left by subtracting $2 π over 3$ from its input, and once to the right by adding $2 π over 3$ . Keep its height unchanged. For example, copy ( $π over 6$ , 2) left: $π over 6$ − $2 π over 3$ = $π over 6$ − $4 π over 6$ = − $π over 2$ , giving (− $π over 2$ , 2). Copy it right: $π over 6$ + $4 π over 6$ = $5 π over 6$ , giving ( $5 π over 6$ , 2). Only the horizontal coordinate changes.Adding a full period changes the inside angle by 2π, which returns to the same place on the circle.
Use the extended table and connect all the points smoothly.The interval length is $4 π over 3$ − (− $2 π over 3$ ) = 2π = 3 × $2 π over 3$ , so exactly three cycles fit.

Answer

Amplitude = 2
Period = $2 π over 3$
Quarter period = $π over 6$
Three full cycles fill [− $2 π over 3$ , $4 π over 3$ ].

Check Substitute the solved period: 3 × = 2π. For any copied input x + , 3(x + ) = 3x + 2π, so its sine is unchanged. Directly, at the thirteen input columns the inside angles are −2π, −, −π, −, 0, , π, , 2π, , 3π, , 4π. Multiplying their sine values by 2 gives the displayed heights. The highest and lowest outputs are 2 and −2, so = 2.

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

✗ Not this: Counterexample: for y = sin x, copying every point by π gives a full repeat.

sin() = 1, but sin( + π) = sin() = −1. The heights differ.

✓ Instead: Copy by 2π: sin( + 2π) = sin() = 1.

Tips and tricks

Tip: A matching zero alone does not prove a full repeat. Match its direction as well.

Strategy: step by step

1. Check A and ω for zero. If the graph is nonconstant and ω is negative, rewrite using sin(−u) = −sin u or cos(−u) = cos u to obtain positive ω. Then copy the resulting signed A and positive ω. Find amplitude |A| and T = $2 π over ω$ .
2. Divide T by 4 to find the quarter period. Put its successive multiples in an input/output table.
3. Use the sine or cosine parent heights and multiply each by A. The table is the point-plotting plan.
4. Draw each column as an ordered pair. Join the points smoothly, with rounded tops and bottoms.
5. If the requested interval spans several cycles, copy the points by adding or subtracting T from each input. Do not change their heights.

Strategy

Strategy: sketch a wave from its equation

Is A = 0 or ω = 0?

YesThe graph is constant. If A = 0, y = 0. If ω = 0, sine gives y = 0 and cosine gives y = A. Draw that horizontal line and stop. A constant has no smallest positive period.

NoContinue with a changing sine or cosine wave.

↓

Is ω positive?

YesUse the forward quarter-cycle patterns.

NoRewrite using sin(−u) = −sin u or cos(−u) = cos u, then use positive ω. For example, 2 sin(−3x) = −2 sin(3x).

↓

Is the function sine?

YesStart at the middle and use the sine height row.

NoFor cosine, start at height A and use the cosine row.

↓

Is A negative?

YesMultiply the parent heights by the negative A once. That multiplication both scales and reflects them; do not reverse their signs again afterward.

NoMultiply the parent heights by the positive A once to scale them.

↓

Does the requested interval cover more than one cycle?

YesAdd or subtract whole periods from the input columns to repeat the pattern.

NoDraw the single requested portion of the cycle.

Write A, ω, |A|, T, and $T over 4$ .
Draw the five-column table using sine or cosine heights. Multiply each parent height by the signed A exactly once; a negative A already performs the reflection.
Plot the points and connect smoothly. Copy by T to cover the requested interval.

Worked exampleA complete reflected sine cycle

Sketch y = −4 sin( $2 over 3$ x) for 0 ≤ x ≤ 3π. This asks you to draw one whole wave, with its five key points, and state its amplitude and period.

Match the five plotted points to the five columns of the key-point table.

Each input column tells you where to place its output height.

Read A = −4 and ω = $2 over 3$ . The amplitude is |−4| = 4.A multiplies the output, so the negative sign reflects the wave while the distance from its middle remains positive.
Find T = 2π ÷ $2 over 3$ = 2π × $3 over 2$ = 3π.A full cycle makes the inside angle advance by 2π.
Find the quarter period: $T over 4$ = $3 π over 4$ . Use successive multiples of this spacing in the table. Start with 0. One gap gives $3 π over 4$ . Two gaps give $6 π over 4$ = $3 π over 2$ . Three gaps give $9 π over 4$ . Four gaps give $12 π over 4$ = 3π. These are the input columns in the picture.Four equal steps span one whole period.
Multiply the sine heights by −4. The table shows the reflected heights in their matching columns.The sine quarter-turn heights are shown in the pattern table. Multiplying each height by −4 reflects it across the middle line.
Plot the table columns as ordered pairs and join them with a smooth curve. Begin at the middle and travel downward.The sine function changes continuously between its key points; straight segments would change the shape.

Answer

Amplitude = 4
Period T = 3π
Quarter period = $3 π over 4$
One cycle begins at the middle going down and returns to the middle going down at x = 3π.

Check Check by direct substitution, using the circle values rather than the sketch. The inside products are × 0 = 0, × = , × = π, × = , and × 3π = 2π. Thus the outputs are −4 × 0 = 0, −4 × 1 = −4, −4 × 0 = 0, −4 × (−1) = 4, and −4 × 0 = 0. The extremes give = 4, and ωT = × 3π = 2π.

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: make a sine cycle taller

Sketch y = 6 sin x on 0 ≤ x ≤ 2π. You are keeping the ordinary sine width and making its heights six times as large.

Match the five plotted points to the five columns of the key-point table.

Each input column tells you where to place its output height.

Read A = 6 and ω = 1. Amplitude = 6 and T = 2π.A scales output heights, and an inside coefficient of 1 leaves the period unchanged.
Compute $T over 4$ = $π over 2$ . Use the table's x-row.Four quarter-steps span 2π.
Multiply the parent sine outputs by 6 and plot the columns.Every y-value is multiplied by the outside coefficient.
Draw a smooth curve through the points, starting at the middle going up.Positive sine keeps its starting direction.

Answer

Amplitude = 6
Period = 2π
Quarter period = $π over 2$
The maximum is 6 and the minimum is −6.

Check Direct substitution gives 6 sin 0 = 0, 6 sin() = 6, 6 sin π = 0, 6 sin() = −6, and 6 sin(2π) = 0. The height span is 12, so half is 6. Also ωT = 1 × 2π = 2π checks the period.

Rung 2Rung 2: a wider cosine cycle

Sketch y = cos( $1 over 5$ x) on 0 ≤ x ≤ 10π. You are stretching the cosine cycle horizontally while keeping its height unchanged.

Match the five plotted points to the five columns of the key-point table.

Each input column tells you where to place its output height.

Read A = 1 and ω = $1 over 5$ . Amplitude = 1.Only the inside coefficient changed.
Find T = 2π ÷ $1 over 5$ = 2π × 5 = 10π.Dividing by one fifth multiplies by five.
Compute $T over 4$ = $10 π over 4$ = $5 π over 2$ . Use the table's input columns.Each horizontal step is a quarter of the new width.
Use the ordinary cosine outputs and join the five points smoothly.An inside multiplier moves the x-locations but does not multiply the output heights.

Answer

Amplitude = 1
Period = 10π
Quarter period = $5 π over 2$
The cycle starts at the top.

Check Substitute the table inputs: × 0 = 0, × = , × 5π = π, × = , and × 10π = 2π. Their cosine outputs match the table. The extremes give = 1, and ωT = × 10π = 2π.

Rung 3Rung 3: reflect and compress a sine wave

Sketch y = − $3 over 2$ sin(4x) on 0 ≤ x ≤ $π over 2$ . You need one short cycle with a reflection and a fractional amplitude.

Match the five plotted points to the five columns of the key-point table.

Each input column tells you where to place its output height.

Read A = − $3 over 2$ and ω = 4. Amplitude = $3 over 2$ .Amplitude measures distance, so it uses the size of A rather than its sign.
Find T = $2 π over 4$ = $π over 2$ and $T over 4$ = $π over 2$ ÷ 4 = $π over 8$ .Dividing a fraction by 4 multiplies its denominator by 4.
Put successive quarter-period inputs in the table and multiply the sine outputs by − $3 over 2$ .The negative multiplier reverses the heights, while 4 compresses their horizontal spacing.
Join the five points smoothly, beginning at the middle going down.Reflected sine crosses the middle in the opposite direction from positive sine.

Answer

Amplitude = $3 over 2$
Period = $π over 2$
Quarter period = $π over 8$
The sine cycle is reflected across y = 0.

Check Substitute the inputs into 4x: 4 × 0 = 0, 4 × = , 4 × = π, 4 × = , and 4 × = 2π. Multiplying their sine values by − gives 0, −, 0, , and 0, which match the five table columns. The extrema give × ( − (−)) = × 3 = . Also 4 × = 2π verifies T.

Rung 4Rung 4: reflected cosine with π inside

Sketch y = − $5 over 2$ cos( $π over 3$ x) on 0 ≤ x ≤ 6. This asks you to combine a fractional height, a reflection, and a cycle width measured by an ordinary number.

Match the five plotted points to the five columns of the key-point table.

Each input column tells you where to place its output height.

Read A = − $5 over 2$ and ω = $π over 3$ . The amplitude is $5 over 2$ .The minus sign chooses the reflection, while absolute value gives the height.
Find T = 2π ÷ $π over 3$ = 2π × $3 over π$ = 6.Dividing by a fraction multiplies by its reciprocal, and π cancels because π is nonzero.
Compute $T over 4$ = $6 over 4$ = $3 over 2$ . Use the quarter-spaced inputs in the table.The period is 6, so the graph's input labels are ordinary numbers even though its inside angles are radians.
Multiply the cosine output row by − $5 over 2$ , plot the columns, and connect smoothly.Negative cosine begins at its minimum and returns there after a complete cycle.

Answer

Amplitude = $5 over 2$
Period = 6
Quarter period = $3 over 2$
The cycle starts at the bottom.

Check Direct substitution gives inside angles × 0 = 0, × = , × 3 = π, × = , and × 6 = 2π. Their cosine values multiplied by − match the five heights. The extreme span is 5, whose half is , and × 6 = 2π checks the period.

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

✗ Not this: Counterexample: the period of y = sin(4x) is 4, so its quarter spacing is 1.

Four is the inside multiplier. A complete cycle requires 4T = 2π.

✓ Instead: T = and = . Substitution gives 4 × = .

Tips and tricks

Tip: Memory cue: five points, four gaps. The number of gaps explains why you divide T by 4.
Tip: Rebuild the height rows from the circle instead of memorizing many transformed graphs.
Cheat sheet tip: Write T = $2 π over |ω|$ , quarter period = $T over 4$ , and the signed five-point tables.
Tip: Mark the middle as y = 0. The equation x = 0 names the vertical axis.

Trap. Trap: Drawing straight segments between the key points makes a zigzag. These points locate a smooth sinusoidal function, so round the turns and cross the middle smoothly.

Keep in mind

The five inputs are always 0, $T over 4$ , $T over 2$ , $3 T over 4$ and T: four equal steps across one period.
Copy a finished cycle by adding T to every input and keeping the heights: for y = 6 cos(πx), the next top is at (2 + 2, 6) = (4, 6).
A negative A flips every height: y = −6 cos(πx) starts at the bottom, (0, −6).
Round the tops and bottoms, because straight segments between key points make a zigzag, the wrong shape.

Memory hookFive points, four equal steps of . Sine: middle, top, middle, bottom, middle. Cosine: top, middle, bottom, middle, top.

Flash cards: say the answer out loud, then flip

What are key points?

Five landmarks that split one cycle into four equal steps: start, quarter, half, three quarters, end.

What is a cycle?

One complete repeat of the wave
its width is the period.

How far apart are neighboring key points?

A quarter period,

A wave has period 12. What are its five key-point inputs?

0, 3, 6, 9, 12

List the key points of y = 9 sin(x).

Period 2π ÷ $π over 3$ = 6, so:
(0, 0), ( $3 over 2$ , 9), (3, 0), ( $9 over 2$ , −9), (6, 0)

Should you join the key points with straight lines?

No. Use a smooth, rounded curve
straight lines make a zigzag.