Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Section 2.3 Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift

Let’s consider the function

g (x) = \sin (2 x - \frac{2 π}{3}) .

Using what we study in MTH 111 about graph transformations, it should be apparent that the graph of

g (x) = \sin (2 x - \frac{2 π}{3})

can be obtained by transforming the graph of

g (x) = \sin (x) .

(To confirm this, notice that

g (x)

can be expressed in terms of

f (x) = \sin (x),

g (x) = f (2 x - \frac{2 π}{3}) .

) Since the constants “

2

” and “

\frac{2 π}{3}

” are multiplied by and subtracted from the input variable,

x,

what we study in MTH 111 tells us that these constants represent a horizontal stretch/compression and a horizontal shift, respectively.

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It is often recommended in MTH 111 that we factor-out the horizontal stretching/compressing factor before transforming the graph, i.e., it’s often recommended that we first re-write

g (x) = \sin (2 x - \frac{2 π}{3})

g (x) = \sin (2 (x - \frac{π}{3})) .

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After writing

g

in this format, we can draw its graph by performing the following sequence of transformations of the “base function”

f (x) = \sin (x) :

Compress horizontally by a factor of $\frac{1}{2} .$
Shift $\frac{π}{3}$ units to the right.

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The advantage of this method is that the

y

-intercept of

f (x) = \sin (x),

(0, 0),

ends-up exactly where the horizontal shift suggests: when we compress the graph by a factor of

\frac{1}{2},

the the

y

-interceot of the graph doesn’t move since

\frac{1}{2} \cdot 0 = 0;

then, when we shift the graph

\frac{π}{3}

units to the right, the point

(0, 0)

ends up at

(\frac{π}{3}, 0);

so the

y

-intercept ends up moving

\frac{π}{3}

units to the right, exactly how far we shifted.

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Compare this with the alternative method: we can leave

g (x) = \sin (2 x - \frac{2 π}{3})

as-is and skip factoring-out the horizontal stretching/compressing factor, but then we need the following sequence to transform

f (x) = \sin (x)

into the graph of

g :

Shift $\frac{2 π}{3}$ units to the right.
Compress horizontally by a factor of $\frac{1}{2} .$

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The disadvantage of this method is that the

y

-intercept of

f (x) = \sin (x)

doesn’t end-up where the horizontal shift suggests: when we shift the graph of

f (x) = \sin (x)

to the right by

\frac{2 π}{3}

units, the

y

-intercept moves from

(0, 0)

(\frac{2 π}{3}, 0);

then, when we compress the graph by a factor of

\frac{1}{2},

it moves to

(\frac{π}{3}, 0),

so the

y

-intercept doesn’t end up shifted

\frac{2 π}{3}

units to the right.

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Figure 2.3.1 shows the graphs of

y = f (x)

and

y = g (x) .

Notice that the behavior of

y = g (x)

x = \frac{π}{3}

is like the behavior of

y = f (x)

x = 0,

i.e.,

y = g (x)

appears to have been shifted

\frac{π}{3}

units to the right. For this reason,

\frac{π}{3}

is called the horzontal shift of

g (x) = \sin (2 x - \frac{2 π}{3}) = \sin (2 (x - \frac{π}{3})) .

This is a graph showing both f(x) and g(x) where the function g(x) is a horizontal shift of f(x). — 🔗
Figure 2.3.1. $y = g (x)$ with $f (x) = \sin (x)$

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The constant

\frac{2 π}{3}

is given a different name, phase shift, since it can be used to determine how far “out-of-phase” a sinusoidal function is in comparison with

y = \sin (x)

y = \cos (x) .

To determine how far out-of-phase a sinusoidal function is, we can determine the ratio of the phase shift and

2 π .

(We use

2 π

because it’s the period of

y = \sin (x)

and

y = \cos (x) .

) Since

\frac{2 π}{3}

is the phase shift for

g (x) = \sin (2 x - \frac{2 π}{3}),

the graph of

y = g (x)

is out-of-phase

\frac{2 π / 3}{2 π} = \frac{1}{3}

of a period. (Since this number is positive, it represents a horizontal shift to the right

\frac{1}{3}

of a period.)

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Definition 2.3.2.

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Given a sinusoidal function of the form

y = A \sin (w x - C) + k

y = A \cos (w x - C) + k,

the phase shift is

C

and

\frac{| C |}{2 π}

represents the fraction of a period that the graph has been shifted (shift to the right if

C

is positive or to the left if

C

is negative).

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Definition 2.3.3.

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If we re-write the function as

y = A \sin (w (x - \frac{C}{w})) + k

y = A \cos (w (x - \frac{C}{w})) + k,

we can see that the horizontal shift is

\frac{C}{w}

units (shift to the right if

\frac{C}{w}

is positive or to the left if

\frac{C}{w}

is negative).

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Example 2.3.4.

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Identify the phase shift and horizontal shift of

g (x) = \cos (3 x - \frac{π}{4}) .

Solution.

The phase shift of

g (x) = \cos (3 x - \frac{π}{4})

\frac{π}{4} .

This tells us that the graph of

y = g (x)

is out of phase

\frac{| π / 4 |}{2 π} = \frac{1}{8}

of a period, i.e., compared with

y = \cos (x),

the graph of

g (x) = \cos (3 x - \frac{π}{4})

has been shifted one-eighth of a period to the right.

To find the horizontal shift, we need to factor-out

3

from

3 x - \frac{π}{4} .

\begin{aligned} g (x) & = \cos (3 x - \frac{π}{4}) \\ = \cos (3 (x - \frac{π}{3 \cdot 4})) \\ = \cos (3 (x - \frac{π}{12})) \end{aligned}

So the horizontal shift is

\frac{π}{12} .

This tells us, that compared with

y = \cos (x),

the graph of

g (x) = \cos (3 x - \frac{π}{4})

has been shifted

\frac{π}{12}

to the right.

Notice that the period of

g (x) = \cos (3 x - \frac{π}{4})

2 π \cdot \frac{1}{3} = \frac{2 π}{3},

and one-eighth of

\frac{2 π}{3}

\frac{2 π}{3} \cdot \frac{1}{8} = \frac{π}{12},

so a shift of one-eighth of a period is the same as a shift of

\frac{π}{12}

units!

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Example 2.3.5.

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Draw a graph

q (t) = 2 \sin (4 t + π) + 1 .

First, find it’s amplitude, period, midline, phase shift, and horiontal shift.

Solution.

Amplitude:

| A | = | 2 | = 2

Period:

P = 2 π \cdot \frac{1}{| w |} = \frac{2 π}{4} = \frac{π}{2}

Midline:

y = 1

Phase shift:

- π

(this tells is that the graph is out-of-phase

\frac{| - π |}{2 π} = \frac{1}{2}

of a period)

Horizontal shift:

\frac{π}{4}

units to the left since:

\begin{aligned} q (t) & = 2 \sin (4 t + π) + 1 \\ = 2 \sin (4 (t + \frac{π}{4})) + 1 \\ = 2 \sin (4 (t - (- \frac{π}{4}))) + 1 \end{aligned}

Now we can draw a graph of

q (t) = 2 \sin (4 t + π) + 1

by drawing a sinusoidal function with the necessary features; see Figure 2.3.6.

Graph of q(t) = 2sin(4t+pi)+1. — Figure 2.3.6. $y = q (t)$

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Exercises Exercises

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Sketch Sinusoidal Graphs.

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In Exercises 1–4, draw a graph of each of the following functions. List the amplitude, midline, period, phase shift, and horizontal shift.

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1.

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f (x) = 3 \sin (3 x - \frac{π}{2})

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2.

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g (t) = \cos (4 t + π) + 3

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3.

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m (θ) = 2 \cos (2 π θ - π) + 4

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4.

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n (x) = - 4 \sin (π x + \frac{π}{4}) - 2

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Find the Formula.

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In Exercises 5–6, find two algebraic rules (one involving sine and one involving cosine) for each of the functions graphed below.

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5.

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y = p (t)

$Graph of y=p(t). The highest points are at 2 and the lowest are at -6. On the y-axis the intercept is (0, -6). The period is pi.$

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6.

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y = q (x)

$Graph of y=q(t). The highest points are at 2 and the lowest are at -4. It has points at (0.25, 2), (0.75, -1), and (1.25, -4). The period is 2.$

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