Phase Shift of Sine and Cosine Graphs

A phase shift moves a sine or cosine graph left or right. In $y = a \sin(b(x - h)) + k$ , the graph shifts right $h$ units (left if $h$ is negative), and $k$ shifts it up or down so the middle line is $y = k$ . For $y = a \sin(bx - c) + d$ , the phase shift is $\frac{c}{b}$ .

Before this lesson: You should know Amplitude and Period, Graphs of Sine and Cosine and Transformations of Functions.

Sliding the Wave

Changing $x$ to $x - h$ moves any graph right $h$ units, as you saw in transformations of functions. For a sine wave, this horizontal slide is called the phase shift. The wave keeps its shape; it just starts its cycle at $x = h$ instead of $x = 0$ .

Adding $k$ outside moves the graph up $k$ units. The wave then rises and falls around the line $y = k$ , called the midline, instead of around the x-axis.

The blue wave is y = sin x moved right π/2 (the phase shift) and up 1 (the vertical shift). Its midline is y = 1.

The General Sine and Cosine Equations

y = a \sin(b(x - h)) + k \qquad y = a \cos(b(x - h)) + k

Number	Effect	Formula
$a$	Amplitude (and a flip if negative)	Amplitude $= \|a\|$
$b$	Period	Period $= \frac{2\pi}{\|b\|}$
$h$	Phase shift (right if positive, left if negative)	Phase shift $= h$
$k$	Vertical shift and midline	Midline: $y = k$

If the equation is written as $y = a \sin(bx - c) + d$ , factor out $b$ first: $bx - c = b(x - \frac{c}{b})$ . So the phase shift is $\frac{c}{b}$ , not $c$ .

Factor $b$ out of the inside so it looks like $b(x - h)$ .
Find the amplitude $|a|$ , the period $\frac{2\pi}{|b|}$ , the phase shift $h$ and the midline $y = k$ .
The cycle starts at $x = h$ . Add a quarter period four times to get five key x-values.
Use the parent pattern for y, multiply by $a$ , then add $k$ .
Plot the points and draw the wave.

Worked Examples

Example 1: A phase shift only

Find the phase shift and key points of $y = \sin(x - \frac{\pi}{4})$ .

Match the form: $h = \frac{\pi}{4}$ , so shift right $\frac{\pi}{4}$

The period is $2\pi$ , so the spacing is $\frac{\pi}{2}$

\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}, \frac{9\pi}{4}

Start at $\frac{\pi}{4}$ and add $\frac{\pi}{2}$ each time.

0, 1, 0, -1, 0

Use the sine pattern.

Answer: right $\frac{\pi}{4}$ ; key points $(\frac{\pi}{4}, 0)$ , $(\frac{3\pi}{4}, 1)$ , $(\frac{5\pi}{4}, 0)$ , $(\frac{7\pi}{4}, -1)$ , $(\frac{9\pi}{4}, 0)$

Example 2: A left shift and a vertical shift

Describe the graph of $y = \cos(x + \frac{\pi}{3}) + 2$ .

x + \frac{\pi}{3} = x - (-\frac{\pi}{3})

Rewrite the inside as $x - h$ .

h = -\frac{\pi}{3}

Read h from the inside.

The phase shift is left $\frac{\pi}{3}$

The vertical shift is up 2, so the midline is $y = 2$

The amplitude is 1, so the maximum is $2 + 1 = 3$ and the minimum is $2 - 1 = 1$

Answer: the cosine wave moved left $\frac{\pi}{3}$ and up 2. It has midline $y = 2$ and range $[1, 3]$ , and its first maximum is at $x = -\frac{\pi}{3}$ .

Example 3: Factor out b first

Find the amplitude, period, phase shift and key points of $y = 3 \sin(2x - \pi)$ .

2x - \pi = 2(x - \frac{\pi}{2})

Factor 2 out of the inside.

|3| = 3

Find the amplitude.

\frac{2\pi}{2} = \pi

Find the period: divide 2π by b.

Phase shift: right $\frac{\pi}{2}$

Spacing: $\frac{\pi}{4}$ ; x-values: $\frac{\pi}{2}, \frac{3\pi}{4}, \pi, \frac{5\pi}{4}, \frac{3\pi}{2}$

0, 3, 0, -3, 0

Multiply the sine pattern by 3.

Answer: amplitude 3, period $\pi$ , phase shift right $\frac{\pi}{2}$ ; key points $(\frac{\pi}{2}, 0)$ , $(\frac{3\pi}{4}, 3)$ , $(\pi, 0)$ , $(\frac{5\pi}{4}, -3)$ , $(\frac{3\pi}{2}, 0)$

Example 4: Write the equation

A sine wave has amplitude 2, period $2\pi$ and midline $y = -1$ . It rises through its midline at $x = \frac{\pi}{6}$ . Write its equation.

h = \frac{\pi}{6}

A plain sine wave rises through its midline at the start of its cycle.

a = 2

Amplitude 2, and the wave rises.

b = \frac{2\pi}{2\pi} = 1

Divide 2π by the period, $2\pi$ .

k = -1

The midline is $y = -1$ .

Answer: $y = 2 \sin(x - \frac{\pi}{6}) - 1$

Example 5: Word problem with daily temperature

The temperature in a city, in degrees Celsius, is modeled by $T(t) = 10 \sin(\frac{\pi}{12}(t - 9)) + 15$ , where $t$ is hours after midnight. What is the highest temperature, and at what time does it happen?

Read the parts: amplitude 10, midline $T = 15$

15 + 10 = 25

Find the highest temperature: midline plus amplitude.

\frac{\pi}{12}(t - 9) = \frac{\pi}{2}

Sine reaches its maximum when its input is $\frac{\pi}{2}$ .

t - 9 = 6

Multiply both sides by $\frac{12}{\pi}$ .

t = 15

Add 9 to both sides.

Answer: The high is 25°C at $t = 15$ , which is 3:00 p.m. The period is $\frac{2\pi}{\frac{\pi}{12}} = 24$ hours, so the pattern repeats each day.

Common Mistakes

Mistake	Example of the error	Fix
Shifting the wrong direction	$y = \sin(x - \frac{\pi}{4})$ moves left	A minus inside moves the graph right: right $\frac{\pi}{4}$ .
Not factoring out b	Phase shift of $y = 3 \sin(2x - \pi)$ is $\pi$	Factor: $2(x - \frac{\pi}{2})$ , so the shift is $\frac{\pi}{2}$ .
Mixing up the vertical shift and amplitude	Saying $y = \sin x + 4$ has amplitude 4	The 4 outside moves the midline to $y = 4$ . The amplitude is still 1.
Using the old range	Range of $y = \cos x + 2$ is $[-1, 1]$	Add the shift: the range is $[1, 3]$ .
Starting key points at 0	Key points of $y = \sin(x - \frac{\pi}{4})$ start at $x = 0$	The cycle starts at the phase shift, $x = \frac{\pi}{4}$ .

Related Methods

Find the shift from the start of the cycle

Set the inside of the function equal to 0 and solve. For $y = 3 \sin(2x - \pi)$ , $2x - \pi = 0$ gives $x = \frac{\pi}{2}$ , which is where the cycle starts.

Sine and cosine are shifts of each other

A cosine wave is a sine wave shifted left $\frac{\pi}{2}$ : $\cos x = \sin(x + \frac{\pi}{2})$ . So any sinusoid can be written with either function, using a different phase shift.

Midline from the maximum and minimum

The midline is halfway between the highest and lowest values: $k = \frac{\text{max} + \text{min}}{2}$ . For a range of $[1, 3]$ , $k = \frac{1 + 3}{2} = 2$ .

Practice Problems

Find the phase shift of $y = \sin(x - \frac{\pi}{2})$ .
Find the phase shift of $y = \cos(x + \pi)$ .
Find the midline and range of $y = \sin x + 4$ .
Find the amplitude, phase shift, midline and range of $y = 2 \cos(x - \frac{\pi}{6}) - 3$ .
Find the period and phase shift of $y = \sin(3x - \pi)$ .
Find the period and phase shift of $y = -\cos(2x + \frac{\pi}{2})$ .
Find the amplitude, period, phase shift and midline of $y = 4 \sin(\frac{1}{2}(x - \pi)) + 1$ .
List the key points of $y = \sin(x + \frac{\pi}{2})$ for one cycle. What do you notice?
Write a cosine function with amplitude 3, period $2\pi$ , midline $y = 0$ and a maximum at $x = \frac{\pi}{4}$ .
A Ferris wheel seat’s height in meters after $t$ minutes is $h(t) = 20 \sin(\frac{\pi}{5}(t - 2.5)) + 22$ . Find the greatest and least heights and the first time the seat is at the top.

Show answers

h = \frac{\pi}{2}

Match the form.

Answer: right $\frac{\pi}{2}$

x + \pi = x - (-\pi)

Rewrite.

h = -\pi

Read h from the inside.

Answer: left $\pi$

y = 4

The 4 outside moves the midline up.

[4 - 1, 4 + 1] = [3, 5]

Find the range: midline minus and plus the amplitude.

Answer: midline $y = 4$ , range $[3, 5]$

|2| = 2

Find the amplitude.

Phase shift: right $\frac{\pi}{6}$

y = -3

Find the midline.

[-3 - 2, -3 + 2] = [-5, -1]

Find the range: midline minus and plus the amplitude.

Answer: amplitude 2, right $\frac{\pi}{6}$ , midline $y = -3$ , range $[-5, -1]$

3x - \pi = 3(x - \frac{\pi}{3})

Factor.

\frac{2\pi}{3}

Find the period: divide 2π by b.

Answer: period $\frac{2\pi}{3}$ , right $\frac{\pi}{3}$

2x + \frac{\pi}{2} = 2(x + \frac{\pi}{4})

Factor.

\frac{2\pi}{2} = \pi

Find the period: divide 2π by b.

Answer: period $\pi$ , left $\frac{\pi}{4}$ , and the negative sign flips the graph

|4| = 4

Find the amplitude.

\frac{2\pi}{\frac{1}{2}} = 4\pi

Find the period: divide 2π by b.

y = 1

Phase shift: right $\pi$ ; midline.

Answer: amplitude 4, period $4\pi$ , right $\pi$ , midline $y = 1$

Phase shift: left $\frac{\pi}{2}$ , so start at $-\frac{\pi}{2}$

-\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}

List the x-values of the key points.

0, 1, 0, -1, 0

List the y-values of the key points.

Answer: these are the points of $y = \cos x$ , so $\sin(x + \frac{\pi}{2}) = \cos x$

Cosine starts its cycle at a maximum, so $h = \frac{\pi}{4}$
Amplitude 3 means $a = 3$ ; period $2\pi$ means $b = 1$ ; midline 0 means $k = 0$
Answer: $y = 3 \cos(x - \frac{\pi}{4})$

10.

Greatest height: $22 + 20 = 42$ ; least height: $22 - 20 = 2$

\frac{\pi}{5}(t - 2.5) = \frac{\pi}{2}

The top is where the sine input is $\frac{\pi}{2}$ .

t - 2.5 = 2.5

Multiply both sides by $\frac{5}{\pi}$ .

t = 5

Add 2.5 to both sides.

Answer: 42 meters and 2 meters; the seat first reaches the top at 5 minutes

FAQ

What is a phase shift?

It is a horizontal shift of a sine or cosine graph. In $y = a \sin(b(x - h)) + k$ , the phase shift is $h$ : right if positive, left if negative.

How do you find the phase shift of y = a sin(bx − c)?

Factor out $b$ to get $a \sin(b(x - \frac{c}{b}))$ . The phase shift is $\frac{c}{b}$ . You can also set $bx - c = 0$ and solve for $x$ .

What is the midline of a trig graph?

It is the horizontal line halfway between the maximum and minimum. For $y = a \sin(b(x - h)) + k$ , the midline is $y = k$ .

Does a phase shift change the period?

No. A phase shift slides the graph sideways but does not stretch it. Only $b$ changes the period.

Summary

The general form is $y = a \sin(b(x - h)) + k$ (or cosine).
$h$ is the phase shift: right for positive $h$ , left for negative $h$ .
$k$ is the vertical shift; the midline is $y = k$ .
Factor out $b$ first: for $bx - c$ , the phase shift is $\frac{c}{b}$ .
Key points start at $x = h$ and are a quarter period apart.

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