Quarry School

One cycle: solve the inside-input inequality

Explain it like I am five

Picture a recipe with its own step counter, the inside of the parentheses, running from 0 to 2π while your kitchen clock reads x. One cycle is the stretch of clock time that walks the counter from 0 all the way to 2π. The three-part inequality 0 ≤ inside ≤ 2π, read "the inside is at least 0 and at most 2π", holds that whole trip in one line.

Take y = sin(2x − π4). Write 0 ≤ 2x − π4 ≤ 2π. Add π4 to all three parts: π4 ≤ 2x ≤ 2π + π4. Match bottoms to add: 2π = 8π4, so the right end is 9π4. Divide all three parts by 2, a positive number, so the ≤ signs stay: π8 ≤ x ≤ 9π8.

The left end, π8, is the phase shift. The width, 9π8 − π8 = 8π8 = π, is the period, matching 2πω = 2π2 = π. Check both ends in the inside: 2 × π8 − π4 = 0, and 2 × 9π8 − π4 = 9π4 − π4 = 2π.

In plain words

Think of a recipe that runs from its first instruction to its last. Your clock and the recipe's progress need not use the same numbers. You must find when your clock makes the progress counter read the beginning, and when it makes the counter read the end. A sine or cosine cycle runs through inside inputs from 0 to 2π. A three-part inequality keeps that whole inside trip between its two endpoints. Undoing the operations finds the corresponding x-interval. Its left endpoint is the selected cycle start. Its length is the period. This method finds both without relying on a memorized sign trick.

π/2π3π/22π−8−7−6−5−4−3−2−1one periodcycle starts
Read the cycle's starting input and its distance to the next matching point.
Reminder
  • Three-part inequality. 0 ≤ 2x − π ≤ 2π becomes π ≤ 2x ≤ 3π after adding π everywhere.
  • Common denominators. 2π + π4 = 8π4 + π4 = 9π4.
  • Interval notation. [0, π] means 0 ≤ x ≤ π; square brackets include both endpoints.
  • Horizontal distance. From 2 to 5 the distance is 5 − 2 = 3.
Why it works. The inequality 0 ≤ ωx − φ ≤ 2π covers one full inside turn. Adding φ everywhere gives φ ≤ ωx ≤ φ + 2π. Dividing by positive ω gives h ≤ x ≤ h + T. Subtracting the endpoints cancels h, leaving T = 2πω. Sine and cosine both repeat after that full turn. Dividing T into four equal steps gives the Quarter period T4; each step advances the inside by π2 and connects the five familiar cycle stages.
RuleRule: 0 ≤ ωx − φ ≤ 2π gives h ≤ x ≤ h + T, where h = φω and T = 2πω, provided ω > 0.
The same idea, five ways
Say it

Find the horizontal interval that makes the inside do one full turn.

Write it

A selected cycle begins at h and ends at h + T.

In math
  • 0 ≤ ωx − φ ≤ 2π
  • h ≤ x ≤ h + T
  • [h, h + T]
  • T = (h + T) − h
Like

Find when the recipe's counter reaches its first and last instruction.

See it
π/2π3π/22π−8−6−4−2one periodcycle starts
Read the cycle's starting input and its distance to the next matching point.
The same idea, other ways
As one complete lap

The inside travels one lap from 0 to 2π. The interval asks which horizontal inputs make that happen.

π/2π3π/22π−8−6−4−2one periodcycle starts
Read the cycle's starting input and its distance to the next matching point.
As a window

The inequality is a window containing one cycle. Sliding the window changes its position; stretching its width changes the period.

Strategy: step by step
  1. Write 0 ≤ the entire inside input ≤ 2π.
  2. Undo addition or subtraction in all three parts.
  3. Undo multiplication by dividing all three parts by positive ω.
  4. Plug both endpoints back into the inside; get 0 and 2π.
  5. Subtract start from end to get the period.
Strategy
Strategy: find cycle endpoints
1
Is the inside coefficient positive?
YesDivide all three parts by it.
NoRewrite the negative coefficient with Lesson 7 first, then return.
  1. Choose inside endpoints 0 and 2π.
  2. Treat the three-part inequality as two inequalities joined together.
  3. Perform the same undoing step on all three parts.
  4. Check both boundaries by substitution and subtract them.
Worked exampleRung 1: the basic cycle

Find one cycle interval of y = sin x using the inside input from 0 to 2π. This asks for the horizontal beginning and end.

π/2π3π/22π−2−112
The marked stages from x = 0 to x = 2π cover one complete cycle of width 2π.
  1. The inside is x, so 0 ≤ x ≤ 2π.The base sine completes one turn as its inside moves through that interval.
  2. The selected start is x = 0 and end is x = 2π.These inputs send the inside to its required endpoints.
  3. Plug back: at x = 0 the inside is 0; at x = 2π the inside is 2π.Both endpoints match the desired stages.
  4. Length = 2π − 0 = 2π.A horizontal distance is right endpoint minus left endpoint.
Answer
  • One cycle: [0, 2π]
  • Phase shift = 0
  • Period = 2π
Check The graph begins at height 0 going up and returns to that same stage at 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: the basic cycle

Find one cycle interval of y = sin x using the inside input from 0 to 2π. This asks for the horizontal beginning and end.

π/2π3π/22π−2−112
The marked stages from x = 0 to x = 2π cover one complete cycle of width 2π.
  1. The inside is x, so 0 ≤ x ≤ 2π.The base sine completes one turn as its inside moves through that interval.
  2. The selected start is x = 0 and end is x = 2π.These inputs send the inside to its required endpoints.
  3. Plug back: at x = 0 the inside is 0; at x = 2π the inside is 2π.Both endpoints match the desired stages.
  4. Length = 2π − 0 = 2π.A horizontal distance is right endpoint minus left endpoint.
Answer
  • One cycle: [0, 2π]
  • Phase shift = 0
  • Period = 2π
Check The graph begins at height 0 going up and returns to that same stage at 2π.
Rung 2Rung 2: speed up the inside

Find one cycle interval of sin(2x). Find the two x-values that make the inside run from 0 to 2π.

π/2π3π/22π−2−112one period
Read the cycle's starting input and its distance to the next matching point.
  1. Write 0 ≤ 2x ≤ 2π.This restricts the inside input to one full turn.
  2. Divide all three parts by positive 2: 0 ≤ x ≤ π.Positive division preserves their order.
  3. Plug back: 2 × 0 = 0 and 2 × π = 2π.The endpoints produce a full inside cycle.
  4. Length = π − 0 = π.Subtracting endpoints measures the width.
Answer
  • One cycle: [0, π]
  • Period = π
Check T = 2π2 = π independently.
Rung 3Rung 3: Speed and shift together: one cycle of y = sin(2x − 2π3)

Solve the inside-input inequality to find the x-interval that holds exactly one cycle of y = sin(2x − 2π3). Check both endpoints in the inside input, then find the period. The figure graphs the inside input u = 2x − 2π3 against x. It is a line with slope 2 that crosses the vertical axis at u = −2π3.

24−4−22468start (u = 0)end (u = 2π)
The inside input u = 2x − 2π3 plotted against x; the vertical axis shows u, not y. It is a line of slope 2 that crosses the vertical axis at −2π3 ≈ −2.09. The line is at u = 0 when x = π3 ≈ 1.05 and at u = 2π ≈ 6.28 when x = 4π3 ≈ 4.19, so one cycle of y = sin(2x − 2π3) runs between those two x-values.
  1. Write the entire inside input between 0 and 2π: 0 ≤ 2x − 2π3 ≤ 2π.A basic sine cycle runs while its input climbs from 0 to 2π. Here the input of sine is the whole expression 2x − 2π3, so that whole expression, not just x, must run from 0 to 2π.
  2. Add 2π3 to all three parts: 2π3 ≤ 2x ≤ 2π + 2π3. Since 2π = 6π3, the right side is 8π3. So 2π3 ≤ 2x ≤ 8π3.Subtracting 2π3 was the last thing done to x, so it is undone first by adding 2π3. Adding the same number to every part keeps the double inequality true.
  3. Divide all three parts by 2. 2π3 divided by 2 is 2π6 = π3, and 8π3 divided by 2 is 8π6 = 4π3. So π3 ≤ x ≤ 4π3.x is multiplied by ω = 2, and dividing by a positive number keeps both inequality signs pointing the same way. The speed also divides the shift, so the cycle starts at π3, not at 2π3.
  4. Plug both endpoints back into the inside: 2·π3 − 2π3 = 2π3 − 2π3 = 0, and 2·4π3 − 2π3 = 8π3 − 2π3 = 6π3 = 2π.The start of the cycle must make the inside input 0 and the end must make it 2π. Getting exactly 0 and 2π confirms that the interval is one full cycle.
  5. Subtract start from end: 4π3 − π3 = 3π3 = π. The period is π.The period is the horizontal length of one cycle, and this cycle runs from x = π3 to x = 4π3.
Answer
  • One cycle: π3 ≤ x ≤ 4π3
  • period π.
Check Use the rule with ω = 2 and φ = 2π3. h = φω = 2π3·2 = π3 and T = 2πω = 2π2 = π. Then h ≤ x ≤ h + T gives π3 ≤ x ≤ π3 + π = 4π3, which matches. Also, the midpoint x = 5π6 gives 2·5π6 − 2π3 = 5π3 − 2π3 = π. That is exactly halfway between 0 and 2π, as the middle of a cycle should be.

Work to write

  1. 0 ≤ 2x − 2π3 ≤ 2π
  2. 2π3 ≤ 2x ≤ 8π3
  3. π3 ≤ x ≤ 4π3
  4. 2·π3 − 2π3 = 0 and 2·4π3 − 2π3 = 2π
  5. Period = 4π3 − π3 = π

One cycle: π3 ≤ x ≤ 4π3; period π.

Rung 4Rung 4: One cycle with a fractional coefficient: y = cos(23x − π6)

For y = cos(23x − π6), solve the inside-input inequality to find the x-interval that holds exactly one cycle. Plug both endpoints back into the inside input, then find the period. The figure graphs the inside input u = 23x − π6 against x.

24681012−22468u = 0 (start)u = 2π (end)
The inside input u = 23x − π6 graphed against x, a line of slope 23. It reaches u = 0 at x = π4 ≈ 0.79 and u = 2π ≈ 6.28 at x = 13π4 ≈ 10.21. One cycle of the cosine fits on that stretch of x, which has width 3π ≈ 9.42.
  1. Write 0 ≤ 23x − π6 ≤ 2π.Cosine completes one cycle as its input runs from 0 to 2π. The input here is the entire expression 23x − π6, not x alone.
  2. Add π6 to all three parts: π6 ≤ 23x ≤ 2π + π6. Since 2π = 12π6, the right side is 13π6, so π6 ≤ 23x ≤ 13π6.Subtracting π6 was the last thing done to x, so it is undone first. Adding the same amount to all three parts keeps the inequality true.
  3. Divide all three parts by 23 by multiplying by its reciprocal 32: π6 · 32 = 3π12 = π4 and 13π6 · 32 = 39π12 = 13π4. So π4 ≤ x ≤ 13π4.Dividing by a fraction is the same as multiplying by its reciprocal. Because ω = 23 is positive, the inequality signs keep their direction. The usual slip is to multiply by 23 instead, and the endpoint check in the next step would catch it.
  4. Plug both endpoints into the inside. At x = π4: 23 · π4 − π6 = 2π12 − π6 = π6 − π6 = 0. At x = 13π4: 23 · 13π4 − π6 = 26π12 − π6 = 13π6 − π6 = 12π6 = 2π.The start of the cycle must make the inside 0 and the end must make it 2π. Getting exactly those two values confirms the solving.
  5. Subtract start from end: 13π4 − π4 = 12π4 = 3π.The width of the interval that holds exactly one cycle is the period. It agrees with T = 2πω = 2π · 32 = 3π.
Answer
One cycle lies on π4 ≤ x ≤ 13π4, and the period is 3π.
Check By the rule, h = φω = π6 ÷ 23 = π6 · 32 = π4 and T = 2πω = 2π · 32 = 3π. So h + T = π4 + 12π4 = 13π4, which gives the same interval. The cosine values also show one full cycle. At x = π4 the inside is 0, so y = cos 0 = 1. At the midpoint x = 7π4 the inside is 23 · 7π4 − π6 = 7π6 − π6 = π, so y = cos π = −1. At x = 13π4 the inside is 2π, so y = cos 2π = 1. The values go max, min, max, which is exactly one cycle.

Work to write

  1. 0 ≤ 23x − π6 ≤ 2π
  2. π6 ≤ 23x ≤ 13π6
  3. Multiply all three parts by 32: π4 ≤ x ≤ 13π4
  4. 23 · π4 − π6 = 0 and 23 · 13π4 − π6 = 2π
  5. Period = 13π4 − π4 = 3π

One cycle lies on π4 ≤ x ≤ 13π4, and the period is 3π.

Counterexamples: what it is not, and the tempting wrong moves
✗ Not this: Counterexample: in 0 ≤ 2x − π ≤ 2π, add π only to the middle.
Changing only the middle changes which x-values satisfy the inequality. At x = 0 the original middle is −π, outside the required interval.
✓ Instead: Add π everywhere: π ≤ 2x ≤ 3π.
Tips and tricks
  • Tip: Write three copies of each addition or division mentally, one for each part.
  • Tip: A and B never enter the horizontal cycle inequality.
  • Tip: We select this one cycle to organize the calculation. The wave keeps repeating on both sides of the interval.
Trap. Matching endpoint heights alone does not prove a full cycle. A sine wave also has equal heights half a cycle apart, but it then moves in the opposite direction.
Keep in mind
  • Do the same move to all three parts: adding π4 to 0, to the middle and to 2π keeps the inequality true.
  • Match bottoms before adding fractions with π: 2π + π4 = 8π4 + π4 = 9π4.
  • Dividing every part by a positive number keeps the ≤ signs, but dividing by a negative flips them: −2x ≤ 6 becomes x ≥ −3.
  • Equal heights do not prove a full cycle: sin x is 0 at both x = 0 and x = π, yet π is only halfway, where the wave heads down instead of up.
Memory hookSandwich the inside: 0 ≤ inside ≤ 2π. Undo the minus, then the multiply, in all three parts. Left end = phase shift, width = period.
Flash cards: say the answer out loud, then flip
What does 1 ≤ x ≤ 4 mean?
x is at least 1 and at most 4, endpoints included.
What must the inside do during one cycle of sine or cosine?
Run from 0 to 2π.
Find one cycle of y = sin(x − π6).
  • 0 ≤ x − π6 ≤ 2π
  • π6 ≤ x ≤ 13π6
Find one cycle of y = cos(2x + π), then its period and phase shift.
  • 0 ≤ 2x + π ≤ 2π
  • −π ≤ 2x ≤ π
  • −π2 ≤ x ≤ π2
  • Period: π
  • Phase shift: π2 to the left
When do the ≤ signs flip?
Only when you multiply or divide every part by a negative number.
One cycle of y = sin(2x − π4) runs from π8 to 9π8. Is the period 9π8?
No. The period is the width, 9π8 − π8 = π. 9π8 is only the end point.