Is the phase shift of $\sin(2x - \pi)$ equal to $\pi$ or $\frac{\pi}{2}$?

It is $\frac{\pi}{2}$ to the right. Factor out $2$: $\sin(2x - \pi) = \sin\left(2\left(x - \frac{\pi}{2}\right)\right)$. Some books call $\pi$ the phase angle, but the distance the graph moves on the $x$-axis is $\frac{\pi}{2}$.

Can a graph have more than one correct phase shift?

Yes. Adding or subtracting a whole period gives the same graph, so $\sin\left(x - \frac{\pi}{3}\right)$ and $\sin\left(x + \frac{5\pi}{3}\right)$ are identical. Usually you give the shift with the smallest size.

Does the order of stretching and shifting matter?

Yes. $\sin(2(x - 1))$ and $\sin(2x - 1)$ are different: the first is shifted $1$ to the right, the second only $\frac{1}{2}$. Writing the function as $\sin(B(x - C))$ avoids the confusion.

Teacher Resources

Phase Shift

Q: Why is it called a phase shift?

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.

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).

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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)) \quad \text{or} \quad y = \cos(B(x - C))

the **phase shift is

C

**:

$C > 0$ : the graph moves $C$ units to the right
$C < 0$ : the graph moves $|C|$ units to the left

Why does a minus sign mean right? The plain sine starts a cycle where its input is

0

. In

\sin(x - C)

the input is

x - C

, which is

0

x = C

. So every point of the curve now arrives

C

units later, that is, further to the right. For example

y = \sin\left(x - \frac{\pi}{3}\right)

y = \sin(x)

moved

\frac{\pi}{3}

to the right, and

y = \sin\left(x + \frac{\pi}{4}\right) = \sin\left(x - \left(-\frac{\pi}{4}\right)\right)

is moved

\frac{\pi}{4}

to the left.

Watch out: factor out $B$ first

When the equation is written as

y = \sin(Bx - c)

, the shift is not

c

. Factor out

B

\sin(Bx - c) = \sin\left(B\left(x - \frac{c}{B}\right)\right) \quad \Rightarrow \quad \text{phase shift} = \frac{c}{B}

Example:

\sin\left(2x - \frac{\pi}{2}\right) = \sin\left(2\left(x - \frac{\pi}{4}\right)\right)

is shifted

\frac{\pi}{4}

to the right, not

\frac{\pi}{2}

Shift and period are different things

	What it changes	Formula
$B$	the width of one cycle	period $= \frac{2\pi}{	B	}$
$C$	where the cycle starts	shift $= C$

One cycle of

y = \sin(B(x - C))

starts at

x = C

and ends at

x = C + \frac{2\pi}{B}

. Shifting by a whole period gives the same graph again:

\sin(x - 2\pi) = \sin(x)

Sine and cosine are shifts of each other

\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

\frac{\pi}{2}

to the left. In the complete form

y = A\sin(B(x - C)) + D

A

and

B

come from the lesson on amplitude and period,

C

is this lesson, and

D

(the vertical shift) comes next.

Try it now

Which equation is

y = \cos(x)

shifted

\frac{\pi}{4}

to the left?

Worked Examples

Describe the graph of

y = \sin\left(x - \frac{\pi}{4}\right)

compared with

y = \sin(x)

. Where does the first cycle start and where is its maximum?

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

x - \frac{\pi}{4}

already has the form

1 \cdot (x - C)

→

B = 1

C = \frac{\pi}{4}

Read the direction

C

is positive → Shift

\frac{\pi}{4}

to the right

Find the start of the cycle

The input is

0

when

x - \frac{\pi}{4} = 0

→ Cycle starts at

x = \frac{\pi}{4}

Move the maximum

\sin(x)

has its maximum at

\frac{\pi}{2}

, so add

\frac{\pi}{4}

\frac{\pi}{2} + \frac{\pi}{4}

→ Maximum at

x = \frac{3\pi}{4}

Answer: The graph is $y = \sin(x)$ moved $\frac{\pi}{4}$ to the right. The first cycle starts at $x = \frac{\pi}{4}$ and its maximum $1$ is at $x = \frac{3\pi}{4}$ .

Common Mistakes

Reading $y = \sin\left(x - \frac{\pi}{3}\right)$ 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 $x - \frac{\pi}{3}$ equals what $x$ used to be.

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

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

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

Correct: Factor out $B$ : $\sin\left(2\left(x - \frac{\pi}{4}\right)\right)$ . The shift is $\frac{c}{B} = \frac{\pi}{4}$ 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 $B$ changes the period and only $A$ 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 $y = \sin(x - 2\pi)$ is a different graph from $y = \sin(x)$

Why it's wrong: Sine repeats every $2\pi$ , 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: $\sin(x - 2\pi) = \sin(x)$ , and $\sin\left(x + \frac{\pi}{2}\right) = \sin\left(x - \frac{3\pi}{2}\right)$ .

Interactive Visual

Function graph

C = π/3

Tap or hover over the graph to read off values.

Phase shift C = π/3

Tap a name to show or hide its graph.

Function graph

B = 1C = 0

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

Tap or hover over the graph to read off values.

y = sin(B(x − C))y = sin(x − π/3)

Function graph

B = 2c = π/2

Tap or hover over the graph to read off values.

Shift c/B = π/4
Period = π

Tap a name to show or hide its graph.

Interactive Sandbox

Interactive Grapher

Tap or hover over the graph to read off values.

Try these examples:

sin(x)

sin(x - pi/3)

Expression Calculator

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Practice Problems

18 problems

Problem 1 of 18

Easy

How is the graph of

y = \sin\left(x - \frac{\pi}{6}\right)

shifted compared with

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

1Try It Yourself

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

At what time $t$ (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 $V_1 = \sin(100\pi t)$ has period $\frac{2\pi}{100\pi} = \frac{1}{50}$ second. A second voltage $V_2 = \sin\left(100\pi t - \frac{\pi}{2}\right)$ reaches each peak a little later.

2Try It Yourself

Two voltages are $V_1 = \sin(100\pi t)$ and $V_2 = \sin\left(100\pi t - \frac{\pi}{2}\right)$ , with $t$ in seconds.

By how many seconds is $V_2$ shifted to the right of $V_1$ ?

Step 1: Write the mathematical expression

Divide $c$ by $B$ :

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 $y = \sin(x)$ the copy is $y = \sin(x - \pi)$ , which equals $-\sin(x)$ . Their sum is $0$ everywhere.

3Try It Yourself

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

Which shift $C$ , between $0$ 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 $y = \sin(B(x - C))$ and $y = \cos(B(x - C))$ the phase shift is $C$
2Minus inside the bracket means right, plus means left: $\sin\left(x + \frac{\pi}{4}\right)$ moves $\frac{\pi}{4}$ to the left
3For $\sin(Bx - c)$ factor out $B$ first: the shift is $\frac{c}{B}$ , not $c$
4The period $\frac{2\pi}{|B|}$ and the amplitude do not change when the graph is shifted
5One cycle starts at $x = C$ and ends at $x = C + \frac{2\pi}{B}$ ; shifting by a whole period gives the same graph
6Cosine is a shifted sine: $\cos(x) = \sin\left(x + \frac{\pi}{2}\right)$

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.

Glossary

Phase shift: The horizontal shift of a sine or cosine graph; equals $C$ in $y = \sin(B(x - C))$
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 $\frac{2\pi}{|B|}$
Factored form: Writing the inside of the function as $B(x - C)$ so that the shift $C$ can be read directly

Formula Card

Shifted sine and cosine

y = \sin(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\left(B\left(x - \frac{c}{B}\right)\right)

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

One cycle

C \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\left(x + \frac{\pi}{2}\right)

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

Today's challengeOne problem, three tries

Phase Shift

Definition

Watch out: factor out $B$ first

Shift and period are different things

Sine and cosine are shifts of each other

Worked Examples

Common Mistakes

Interactive Visual

Function graph

Function graph

Function graph

Interactive Sandbox

Interactive Grapher

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

Predicting High Tide

Alternating Current

Noise-Cancelling Headphones

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Shifted sine and cosine

Shift from the expanded form

One cycle

Sine and cosine

More in This Topic

Graphing the Sine Function

Graphing Cotangent

Graphing the Cosine Function

Amplitude and Period

Definition

Watch out: factor out BBB first

Shift and period are different things

Sine and cosine are shifts of each other

Worked Examples

Common Mistakes

Interactive Visual

Function graph

Function graph

Function graph

Interactive Sandbox

Interactive Grapher

Expression Calculator

History

Practice Problems

Why It Matters

Real World Applications

Predicting High Tide

Alternating Current

Noise-Cancelling Headphones

Key Takeaways

Frequently Asked Questions

Glossary

Formula Card

Shifted sine and cosine

Shift from the expanded form

One cycle

Sine and cosine

More in This Topic

Graphing the Sine Function

Graphing Cotangent

Graphing the Cosine Function

Amplitude and Period

Watch out: factor out $B$ first