Phase Shift

Learn how a phase shift moves sine and cosine graphs left or right, why a minus sign means right, and how to find the shift in sin(Bx - c).

Advanced25 min18 problems
Practice Problems

Definition

A phase shift (or horizontal shift) moves a sine or cosine graph to the left or to the right without changing its shape. In the form
y=sin⁡(B(x−C))ory=cos⁡(B(x−C))y = \sin(B(x - C)) \quad \text{or} \quad y = \cos(B(x - C))
the **phase shift is CC**:
  • C>0C > 0: the graph moves CC units to the right
  • C<0C < 0: the graph moves ∣C∣|C| units to the left
Why does a minus sign mean right? The plain sine starts a cycle where its input is 00. In sin⁡(x−C)\sin(x - C) the input is x−Cx - C, which is 00 at x=Cx = C. So every point of the curve now arrives CC units later, that is, further to the right. For example y=sin⁡(x−π3)y = \sin\left(x - \frac{\pi}{3}\right) is y=sin⁡(x)y = \sin(x) moved π3\frac{\pi}{3} to the right, and y=sin⁡(x+π4)=sin⁡(x−(−π4))y = \sin\left(x + \frac{\pi}{4}\right) = \sin\left(x - \left(-\frac{\pi}{4}\right)\right) is moved π4\frac{\pi}{4} to the left.

Watch out: factor out BB first

When the equation is written as y=sin⁡(Bx−c)y = \sin(Bx - c), the shift is not cc. Factor out BB:
sin⁡(Bx−c)=sin⁡(B(x−cB))⇒phase shift=cB\sin(Bx - c) = \sin\left(B\left(x - \frac{c}{B}\right)\right) \quad \Rightarrow \quad \text{phase shift} = \frac{c}{B}
Example: sin⁡(2x−π2)=sin⁡(2(x−π4))\sin\left(2x - \frac{\pi}{2}\right) = \sin\left(2\left(x - \frac{\pi}{4}\right)\right) is shifted π4\frac{\pi}{4} to the right, not π2\frac{\pi}{2}.

Shift and period are different things

What it changesFormula
BBthe width of one cycleperiod $= \frac{2\pi}{B}$
CCwhere the cycle startsshift =C= C
One cycle of y=sin⁡(B(x−C))y = \sin(B(x - C)) starts at x=Cx = C and ends at x=C+2πBx = C + \frac{2\pi}{B}. Shifting by a whole period gives the same graph again: sin⁡(x−2π)=sin⁡(x)\sin(x - 2\pi) = \sin(x).

Sine and cosine are shifts of each other

cos⁡(x)=sin⁡(x+π2)sin⁡(x)=cos⁡(x−π2)\cos(x) = \sin\left(x + \frac{\pi}{2}\right) \qquad \sin(x) = \cos\left(x - \frac{\pi}{2}\right)
The cosine is a sine moved π2\frac{\pi}{2} to the left. In the complete form y=Asin⁡(B(x−C))+Dy = A\sin(B(x - C)) + D, AA and BB come from the lesson on amplitude and period, CC is this lesson, and DD (the vertical shift) comes next.

Try it now

Which equation is y=cos⁡(x)y = \cos(x) shifted π4\frac{\pi}{4} to the left?

Worked Examples

Describe the graph of y=sin⁡(x−π4)y = \sin\left(x - \frac{\pi}{4}\right) compared with y=sin⁡(x)y = \sin(x). Where does the first cycle start and where is its maximum?
1

Compare with y=sin⁡(B(x−C))y = \sin(B(x - C))

x−π4x - \frac{\pi}{4} already has the form 1⋅(x−C)1 \cdot (x - C) → B=1B = 1, C=π4C = \frac{\pi}{4}
2

Read the direction

CC is positive → Shift π4\frac{\pi}{4} to the right
3

Find the start of the cycle

The input is 00 when x−π4=0x - \frac{\pi}{4} = 0 → Cycle starts at x=π4x = \frac{\pi}{4}
4

Move the maximum

sin⁡(x)\sin(x) has its maximum at π2\frac{\pi}{2}, so add π4\frac{\pi}{4}: π2+π4\frac{\pi}{2} + \frac{\pi}{4} → Maximum at x=3π4x = \frac{3\pi}{4}

Common Mistakes

Reading y=sin⁡(x−π3)y = \sin\left(x - \frac{\pi}{3}\right) as a shift to the left

Why it's wrong: The minus sign looks like it should point left, as on a number line. But the curve reaches each value later, when x−π3x - \frac{\pi}{3} equals what xx used to be.

Correct: Minus inside the bracket means right: sin⁡(x−C)\sin(x - C) moves CC to the right, sin⁡(x+C)\sin(x + C) moves CC to the left.

Saying the shift of y=sin⁡(2x−π2)y = \sin\left(2x - \frac{\pi}{2}\right) is π2\frac{\pi}{2}

Why it's wrong: The number π2\frac{\pi}{2} is subtracted from 2x2x, not from xx. The shift belongs to xx alone.

Correct: Factor out BB: sin⁡(2(x−π4))\sin\left(2\left(x - \frac{\pi}{4}\right)\right). The shift is cB=π4\frac{c}{B} = \frac{\pi}{4} to the right.

Thinking a phase shift changes the period or the height

Why it's wrong: Moving a graph sideways does not stretch it. Only BB changes the period and only AA changes the amplitude.

Correct: After a phase shift the curve has the same shape, period and amplitude. Only the place where each cycle starts changes.

Thinking y=sin⁡(x−2π)y = \sin(x - 2\pi) is a different graph from y=sin⁡(x)y = \sin(x)

Why it's wrong: Sine repeats every 2π2\pi, so moving it by a whole period puts every wave exactly on the next one.

Correct: Shifts that differ by a whole period give the same graph: sin⁡(x−2π)=sin⁡(x)\sin(x - 2\pi) = \sin(x), and sin⁡(x+π2)=sin⁡(x−3π2)\sin\left(x + \frac{\pi}{2}\right) = \sin\left(x - \frac{3\pi}{2}\right).

Interactive Visual

Function graph

-2π-π0π2π-2-1012xy

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  • Phase shift C = π/3

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Function graph

y = sin(x)

Challenge 1 of 5

y = sin(x − π/3)

Minus inside the bracket means the curve moves to the right.

Move the sliders until your curve lies exactly on the dashed target curve.

How close

0 of 5 solved

  • B: right
  • C: bigger
-2π-π0π2π-2-1012xy

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y = sin(B(x − C))y = sin(x − π/3)

Function graph

-2π-π0π2π-2-1012xy

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  • Shift c/B = π/4
  • Period = π

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Interactive Sandbox

Interactive Grapher

-3π-2π-π0π2π3π-4-2024xy

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sin(x)

sin(x - pi/3)

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Practice Problems

18 problems
Problem 1 of 18
Easy
How is the graph of y=sin⁡(x−π6)y = \sin\left(x - \frac{\pi}{6}\right) shifted compared with y=sin⁡(x)y = \sin(x)?

Why It Matters

A phase shift tells you when a repeating pattern happens, not only how big or how fast it is:
  • Tides: two harbors have tides with the same period, but high tide arrives at different times. The difference is a phase shift.
  • Electricity: the three wires of a three-phase power line carry the same sine wave, shifted by one third of a period each.
  • Noise-cancelling headphones: they play the noise again shifted by half a period, so the two waves cancel.
  • Seasons: the warmest day of the year comes about a month after the longest day. The temperature curve is the daylight curve with a phase shift.
  • Music and sound: two speakers playing the same tone out of phase can make the sound weaker in some places.
Reading the phase shift correctly (direction and size) is what lets you predict the next peak.

Real World Applications

Predicting High Tide

Harbor masters model the water level with a shifted sine. The phase shift tells them when the water passes the mean level, and from there they find high tide.

Example:

If the water level is h(t)=1.5sin⁡(π6(t−2))h(t) = 1.5\sin\left(\frac{\pi}{6}(t - 2)\right) meters above the mean level, with tt in hours after midnight, the water rises through the mean level at t=2t = 2, at 2:00.

1Try It Yourself

The water level in a harbor is h(t)=1.5sin⁡(π6(t−2))h(t) = 1.5\sin\left(\frac{\pi}{6}(t - 2)\right) meters above the mean level, where tt is the time in hours after midnight.

At what time tt (in hours after midnight) is the first high tide?

Step 1: Write the mathematical expression

Start of the cycle plus a quarter of the period:

Alternating Current

Power grids in Europe run at 50 Hz. Two voltages with the same frequency can be out of step, and engineers measure that difference as a phase shift in seconds.

Example:

The voltage V1=sin⁡(100πt)V_1 = \sin(100\pi t) has period 2π100π=150\frac{2\pi}{100\pi} = \frac{1}{50} second. A second voltage V2=sin⁡(100πt−π2)V_2 = \sin\left(100\pi t - \frac{\pi}{2}\right) reaches each peak a little later.

2Try It Yourself

Two voltages are V1=sin⁡(100πt)V_1 = \sin(100\pi t) and V2=sin⁡(100πt−π2)V_2 = \sin\left(100\pi t - \frac{\pi}{2}\right), with tt in seconds.

By how many seconds is V2V_2 shifted to the right of V1V_1?

Step 1: Write the mathematical expression

Divide cc by BB:

Noise-Cancelling Headphones

A microphone picks up the noise, and the headphones play the same wave shifted by half a period. Where the noise goes up, the copy goes down, and the two cancel.

Example:

For the noise y=sin⁡(x)y = \sin(x) the copy is y=sin⁡(x−π)y = \sin(x - \pi), which equals −sin⁡(x)-\sin(x). Their sum is 00 everywhere.

3Try It Yourself

A noise is modeled by y=sin⁡(2x)y = \sin(2x). The headphones play y=sin⁡(2(x−C))y = \sin(2(x - C)).

Which shift CC, between 00 and one period, makes the copy exactly the opposite of the noise?

Step 1: Write the mathematical expression

Half of the period:

Key Takeaways

  • 1In y=sin⁡(B(x−C))y = \sin(B(x - C)) and y=cos⁡(B(x−C))y = \cos(B(x - C)) the phase shift is CC
  • 2Minus inside the bracket means right, plus means left: sin⁡(x+π4)\sin\left(x + \frac{\pi}{4}\right) moves π4\frac{\pi}{4} to the left
  • 3For sin⁡(Bx−c)\sin(Bx - c) factor out BB first: the shift is cB\frac{c}{B}, not cc
  • 4The period 2π∣B∣\frac{2\pi}{|B|} and the amplitude do not change when the graph is shifted
  • 5One cycle starts at x=Cx = C and ends at x=C+2πBx = C + \frac{2\pi}{B}; shifting by a whole period gives the same graph
  • 6Cosine is a shifted sine: cos⁡(x)=sin⁡(x+π2)\cos(x) = \sin\left(x + \frac{\pi}{2}\right)

Frequently Asked Questions

The phase is the position inside one cycle (start, peak, zero, trough). A phase shift moves every point of the cycle by the same amount, so the whole wave starts earlier or later.
The phase is the position inside one cycle (start, peak, zero, trough). A phase shift moves every point of the cycle by the same amount, so the whole wave starts earlier or later.
It is π2\frac{\pi}{2} to the right. Factor out 22: sin⁡(2x−π)=sin⁡(2(x−π2))\sin(2x - \pi) = \sin\left(2\left(x - \frac{\pi}{2}\right)\right). Some books call π\pi the phase angle, but the distance the graph moves on the xx-axis is π2\frac{\pi}{2}.
Yes. Adding or subtracting a whole period gives the same graph, so sin⁡(x−π3)\sin\left(x - \frac{\pi}{3}\right) and sin⁡(x+5π3)\sin\left(x + \frac{5\pi}{3}\right) are identical. Usually you give the shift with the smallest size.
Yes. sin⁡(2(x−1))\sin(2(x - 1)) and sin⁡(2x−1)\sin(2x - 1) are different: the first is shifted 11 to the right, the second only 12\frac{1}{2}. Writing the function as sin⁡(B(x−C))\sin(B(x - C)) avoids the confusion.

Glossary

Phase shift
The horizontal shift of a sine or cosine graph; equals CC in y=sin⁡(B(x−C))y = \sin(B(x - C))
Horizontal shift
Moving a whole graph left or right without changing its shape
Phase
The position inside one cycle of a wave, such as the start, the peak or the trough
Period
The length of one complete cycle; equals 2π∣B∣\frac{2\pi}{|B|}
Factored form
Writing the inside of the function as B(x−C)B(x - C) so that the shift CC can be read directly

Formula Card

Shifted sine and cosine

y=sin⁡(B(x−C))y = \sin(B(x - C)), y=cos⁡(B(x−C))y = \cos(B(x - C))

Phase shift $C$: right if $C > 0$, left if $C < 0$

Shift from the expanded form

sin⁡(Bx−c)=sin⁡(B(x−cB))\sin(Bx - c) = \sin\left(B\left(x - \frac{c}{B}\right)\right)

The phase shift is $\frac{c}{B}$

One cycle

C≤x≤C+2πBC \le x \le C + \frac{2\pi}{B}

Where one full cycle of the shifted graph starts and ends

Sine and cosine

cos⁡(x)=sin⁡(x+π2)\cos(x) = \sin\left(x + \frac{\pi}{2}\right)

Cosine is sine moved $\frac{\pi}{2}$ to the left

Today's challengeOne problem, three tries

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