Phase shift: why minus inside moves right

Explain it like I am five

Picture your watch, x, and a movie's own counter, the inside of the parentheses. In sin(x − 2) the counter runs 2 behind your watch. The movie starts when the counter reads 0, which happens when your watch reads 2. So the whole wave starts 2 later, 2 units to the right on a graph. That sideways slide is the phase shift.

With a number in front of x, solve for the start. Take y = sin(2x − $π over 4$ ), where ω (omega) = 2 and φ (phi) = $π over 4$ . Set the inside to 0: 2x − $π over 4$ = 0. Add $π over 4$ : 2x = $π over 4$ . Divide by 2, which doubles the bottom: x = $π over 8$ . The wave starts $π over 8$ to the right, and the phase shift formula h = $\frac{φ}{ω}$ = $π over 4$ ÷ 2 = $π over 8$ agrees.

Check: 2 × $π over 8$ − $π over 4$ = $π over 4$ − $π over 4$ = 0. A plus sign inside, as in sin(x + 2), makes the counter run 2 ahead, so the start comes earlier: 2 units to the left.

In plain words

Picture a music player whose song starts when its counter reads 0. The outside number x is your clock, but sine or cosine listens to the counter inside the parentheses. In sin(x − 1), the counter is one behind your clock. When your clock reads 0, the counter reads −1, so the song has not reached its original start. Your clock must reach 1 before x − 1 becomes 0. That puts the original start one unit to the right. The inside offset is φ (phi). A Phase shift is this horizontal slide of the whole pattern. A plus inside makes the counter run ahead, so the start happens earlier, on the left.

Read the cycle's starting input and its distance to the next matching point.

Reminder

Solving a linear equation. 2x − π = 0 becomes 2x = π and x = $π over 2$ ; substituting gives π − π = 0.
Subtracting a negative. x + 1 = x − (−1), so φ = −1.
Factoring. 2(x − 1) = 2x − 2; the coefficient multiplies both terms.
Fraction division. $π over 4$ ÷ $1 over 3$ = $3 π over 4$ .

Why it works. Match the old inside input u with the new one: ωx − φ = u. Adding φ and dividing by positive ω gives x = + . The unshifted wave sin(ωx) receives u at x = . Every matching point therefore moves by the same distance . Positive distance adds to every x-coordinate and moves right; negative distance subtracts and moves left. This applies to the entire curve, not only its start.

RuleRule: h = , with ω > 0. Factor ωx − φ = ω(x − h). h > 0 moves right; h < 0 moves left; h = 0 gives no horizontal shift. We name the signed horizontal slide h.

The same idea, five ways

Say it

Phase shift is how far the repeating picture slides left or right.

Write it

For positive ω, the horizontal shift equals the signed inside offset divided by the inside coefficient.

In math

h = $\frac{φ}{ω}$
ωx − φ = ω(x − h)
h > 0: right; h < 0: left
Graph words: matching stages move by h horizontally.

A delayed music counter reaches the same part of the song later.

See it

Read the cycle's starting input and its distance to the next matching point.

The same idea, other ways

As a delayed counter

A counter x − 1 is one behind. To show 0 it needs x = 1. A counter x + 1 is one ahead and shows 0 at x = −1.

The input 1 produces the old starting input 0.

As matching points

The old point (u, sin u) becomes (u + 1, sin u) for sin(x − 1). Its height stays the same, and its horizontal coordinate grows by 1.

Read the cycle's starting input and its distance to the next matching point.

As factoring

2x − π = 2(x − $π over 2$ ). The factor 2 is input speed; the subtraction $π over 2$ from x is the horizontal slide.

With a peak

The basic sine peak receives $π over 2$ . For sin(x − 1), x = $π over 2$ + 1 gives that same inside input, so even the peak moves right 1.

.1Minus inside

Subtracting a positive number makes the inside counter lag behind. The graph reaches every original stage later.

Rule: sin(x − 1) is sin x moved right 1.

Read the cycle's starting input and its distance to the next matching point.

The same idea, five ways

Say it

Minus inside

Write it

Subtracting a positive number makes the inside counter lag behind. The graph reaches every original stage later.

In math

Rule: sin(x − 1) is sin x moved right 1.

A clock running one minute behind reaches the start one minute later.

See it

Read the cycle's starting input and its distance to the next matching point.

Worked exampleRung 1: subtract 1 inside

Compare sin(x − 1) with sin x. Find where the original inside input 0 now happens. This asks where the original starting point moves.

Read the cycle's starting input and its distance to the next matching point.

The basic sine has inside input 0 at x = 0.Its starting height is sin 0 = 0.
Set x − 1 = 0 to find where the shifted formula receives that same inside input.Matching the inside input matches the stage of the sine motion.
Add 1 to both sides: x = 1.Adding undoes subtraction and locates the shifted starting point.
Plug back: 1 − 1 = 0.The new input 1 does feed the original sine input 0.
For any old input u, choose x = u + 1; then x − 1 = u.This shows every point moves right 1, not only the starting point.

Answer

Phase shift = 1 unit right

Check The old peak at x = now occurs at x = 1 + , because (1 + ) − 1 = .

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

✗ Not this: Counterexample: Minus means move left.

The inside value is smaller, so x must be larger to compensate.

✓ Instead: Make x − 1 equal to 0: x = 1, which is right.

Tips and tricks

Tip: Make x − 1 equal to 0: x = 1, which is right.

.2Plus inside

Adding a positive number makes the inside counter run ahead. It reaches the original start when x is negative.

Rule: sin(x + 1) is sin x moved left 1.

Read the cycle's starting input and its distance to the next matching point.

The same idea, five ways

Say it

Plus inside

Write it

Adding a positive number makes the inside counter run ahead. It reaches the original start when x is negative.

In math

Rule: sin(x + 1) is sin x moved left 1.

A clock running one minute ahead reaches the start one minute earlier.

See it

Read the cycle's starting input and its distance to the next matching point.

Worked exampleRung 2: add 1 inside

For sin(x + 1), locate the same original inside input 0. This asks whether its starting point moves left or right.

Read the cycle's starting input and its distance to the next matching point.

Set x + 1 = 0.You are finding the x that makes sine receive its original start input.
Subtract 1: x = −1.The x must be negative to cancel the added 1.
Plug back: −1 + 1 = 0.The shifted formula is at the original sine start there.
Rewrite x + 1 = x − (−1).This identifies φ = −1 and h = −1.

Answer

Phase shift = 1 unit left

Check At the old peak input u = , use x = − 1, and x + 1 = . Every point is 1 unit left.

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

✗ Not this: Counterexample: The plus in sin(x + 1) means right 1.

At x = 1, its inside is 2, not the original start 0.

✓ Instead: Set x + 1 = 0; x = −1, and −1 + 1 = 0.

Tips and tricks

Tip: Set x + 1 = 0; x = −1, and −1 + 1 = 0.

.3φ versus horizontal distance

φ measures the offset in the inside input. Dividing by ω translates that offset into units along the x-axis.

Rule: cos(πx + 2) has φ = −2 and h = − $2 over π$ .

Read the cycle's starting input and its distance to the next matching point.

The same idea, five ways

Say it

φ versus horizontal distance

Write it

φ measures the offset in the inside input. Dividing by ω translates that offset into units along the x-axis.

In math

Rule: cos(πx + 2) has φ = −2 and h = − $2 over π$ .

Convert an offset on a fast clock into elapsed minutes on your own clock.

See it

Read the cycle's starting input and its distance to the next matching point.

Worked exampleKeep φ and h separate

Find the signed offset and shift in cos(πx + 2). You are reading an inside number and converting it to a horizontal distance.

Read the cycle's starting input and its distance to the next matching point.

πx + 2 = πx − (−2), so φ = −2 and ω = π.This matches subtraction of a signed constant.
Set πx + 2 = 0, giving πx = −2 and x = − $2 over π$ .Solving locates the original cosine start.
Plug back: π × (− $2 over π$ ) + 2 = −2 + 2 = 0.The proposed horizontal shift feeds cosine 0.

Answer

φ = −2
Shift = $2 over π$ left

Check Factoring gives π(x + ), confirming the leftward distance.

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

✗ Not this: Counterexample: The shift is −2 because the offset is −2.

π multiplies x, so 2 inside units correspond to horizontal units.

✓ Instead: Divide φ by ω to get h = −.

Tips and tricks

Tip: Divide φ by ω to get h = − $2 over π$ .

Strategy: step by step

Match the entire inside input to ωx − φ, including the sign of φ.
Set the inside input equal to 0 to locate the chosen cycle start.
Undo the subtraction, then divide by the positive coefficient of x.
Plug the answer back into the inside and confirm 0.
State the distance and direction separately.

Strategy

Strategy: find a phase shift without guessing its direction

Is ω negative?

YesRewrite with the even or odd rule in Lesson 7 first.

NoContinue.

↓

Is there a plus constant inside?

YesRecord φ as the negative of that constant.

NoRecord the signed subtracted constant as φ.

↓

Is h positive?

YesSay h units to the right.

NoIf h is negative, say |h| units to the left; if 0, no shift.

Rewrite a plus inside as subtraction of a negative.
Find the x that makes the entire inside input 0.
Substitute that x back.
Read positive x-distance as right and negative as left.

Worked exampleRung 1: subtract 1 inside

Compare sin(x − 1) with sin x. Find where the original inside input 0 now happens. This asks where the original starting point moves.

Read the cycle's starting input and its distance to the next matching point.

The basic sine has inside input 0 at x = 0.Its starting height is sin 0 = 0.
Set x − 1 = 0 to find where the shifted formula receives that same inside input.Matching the inside input matches the stage of the sine motion.
Add 1 to both sides: x = 1.Adding undoes subtraction and locates the shifted starting point.
Plug back: 1 − 1 = 0.The new input 1 does feed the original sine input 0.
For any old input u, choose x = u + 1; then x − 1 = u.This shows every point moves right 1, not only the starting point.

Answer

Phase shift = 1 unit right

Check The old peak at x = now occurs at x = 1 + , because (1 + ) − 1 = .

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: subtract 1 inside

Compare sin(x − 1) with sin x. Find where the original inside input 0 now happens. This asks where the original starting point moves.

Read the cycle's starting input and its distance to the next matching point.

The basic sine has inside input 0 at x = 0.Its starting height is sin 0 = 0.
Set x − 1 = 0 to find where the shifted formula receives that same inside input.Matching the inside input matches the stage of the sine motion.
Add 1 to both sides: x = 1.Adding undoes subtraction and locates the shifted starting point.
Plug back: 1 − 1 = 0.The new input 1 does feed the original sine input 0.
For any old input u, choose x = u + 1; then x − 1 = u.This shows every point moves right 1, not only the starting point.

Answer

Phase shift = 1 unit right

Check The old peak at x = now occurs at x = 1 + , because (1 + ) − 1 = .

Rung 2Rung 2: add 1 inside

For sin(x + 1), locate the same original inside input 0. This asks whether its starting point moves left or right.

Read the cycle's starting input and its distance to the next matching point.

Set x + 1 = 0.You are finding the x that makes sine receive its original start input.
Subtract 1: x = −1.The x must be negative to cancel the added 1.
Plug back: −1 + 1 = 0.The shifted formula is at the original sine start there.
Rewrite x + 1 = x − (−1).This identifies φ = −1 and h = −1.

Answer

Phase shift = 1 unit left

Check At the old peak input u = , use x = − 1, and x + 1 = . Every point is 1 unit left.

Rung 3Rung 3: divide the inside subtraction by its speed

Find the phase shift of sin(2x − π). This asks how much the graph slides relative to sin(2x).

Read the cycle's starting input and its distance to the next matching point.

Match 2x − π to ωx − φ: ω = 2 and φ = π.The subtracted number is not yet the x-distance.
Set 2x − π = 0 to locate the selected cycle start.The sine receives inside input 0 at that horizontal position.
Add π: 2x = π. Divide by 2: x = $π over 2$ .You undo the subtraction and then the input multiplier.
Plug back: 2 × $π over 2$ − π = 0.The proposed x gives the intended inside input.
Factor: 2x − π = 2(x − $π over 2$ ).The form x − h displays the horizontal distance h.

Answer

Phase shift = right

Check The unshifted peak needs inside input . Solve 2x = to find its horizontal position: x = . Plug back: 2 × = . Moving it right by h = gives + = . Plug into the shifted inside: 2 × − π = − π = , so sine is 1 at the shifted peak.

Rung 4Rung 4: cosine with a plus inside

Find the phase shift of 6 cos(2x + $π over 2$ ). This asks where the cosine's inside input 0 now occurs, regardless of its height 6.

Read the cycle's starting input and its distance to the next matching point.

2x + $π over 2$ = 2x − (− $π over 2$ ), so φ = − $π over 2$ .Standard form subtracts the signed number φ.
h = (− $π over 2$ ) ÷ 2 = − $π over 4$ .The horizontal distance is φ divided by ω = 2.
Plug back: 2 × (− $π over 4$ ) + $π over 2$ = − $π over 2$ + $π over 2$ = 0.The proposed start is where cosine receives 0 and gives its peak.
Factor the input as 2(x + $π over 4$ ) = 2(x − (− $π over 4$ )).The factored expression confirms the negative horizontal shift.

Answer

Phase shift = left

Check At x = −, cosine gives 1, so the wave reaches 6 there. Moving right by period π reaches the next peak.

Rung 5Rung 5: fractional input speed

Find the phase shift of sin( $x over 3$ − $π over 4$ ). You want the x-distance needed to make the inside input 0.

Read the cycle's starting input and its distance to the next matching point.

ω = $1 over 3$ and φ = $π over 4$ .The term $x over 3$ is one third times x.
Set $x over 3$ − $π over 4$ = 0, so $x over 3$ = $π over 4$ .This finds the selected cycle start by undoing the inside subtraction.
Multiply by 3: x = $3 π over 4$ .The inside travels only a third as far as x, so x must travel three times the inside offset.
Plug back: $3 π over 4$ ÷ 3 − $π over 4$ = $π over 4$ − $π over 4$ = 0.The start really does make the inside input 0.
Factor $1 over 3$ (x − $3 π over 4$ ).Distributing one third gives $x over 3$ − $π over 4$ , so the x-distance is visible.

Answer

Phase shift = right

Check Using h = φ ÷ ω gives () ÷ () = , the same answer by a second route.

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

✗ Not this: Counterexample: sin(x − 1) moves left because of the minus.

At x = −1 the inside is −2, not 0. At x = 1 the inside is 0.

✓ Instead: It moves right 1.

✗ Not this: Counterexample: sin(2x − π) shifts right π.

The inside travels 2 units for each x-unit. Moving x by π makes the inside advance 2π, twice the offset.

✓ Instead: The shift is right.

Tips and tricks

Tip: Memory device: inside works backward. Explain it by making the inside 0.
Tip: Draw a small arrow from 0 to h on the x-axis after solving.
Tip: A cycle start is a chosen reference stage of a repeating wave. The graph also exists to its left. In sin(x − 1), h = 1 starts the chosen cycle; x = 0 still gives the valid output sin(−1).

Trap. Do not read the phase shift as φ alone. First translate from inside units to x-units by dividing by ω.

Keep in mind

Minus inside moves right and plus inside moves left: sin(x − 2) starts at x = 2, and sin(x + 2) starts at x = −2.
Divide by ω before you read the shift: in sin(2x − $π over 4$ ) the shift is $π over 8$ , not $π over 4$ .
Plug the start back into the inside to check that it gives 0: 2 × $π over 8$ − $π over 4$ = 0.
A phase shift slides the whole wave, so the period and the amplitude stay put: sin(2x − $π over 4$ ) still has period π and amplitude 1.

Memory hookInside is opposite: minus moves right, plus moves left. Set the inside equal to 0 and solve: that x is where the wave starts.

Flash cards: say the answer out loud, then flip

What is a phase shift?

The horizontal slide of the whole wave, h = .

Which way and how far is y = cos(x − 6) shifted?

6 units to the right

What is the phase shift of y = cos(6x − 3)?

6x − 3 = 0 gives x = = , so unit to the right.

What is the phase shift of y = sin(2x + π)?

2x + π = 0 gives x = −, so to the left.

Why is the shift and not φ?

The start is where ωx − φ = 0, and solving that gives x = .

Is y = sin(x + 4) shifted 4 to the right?

No. Plus inside moves left: x + 4 = 0 at x = −4, so 4 to the left.