3.6 - Sinusoidal Function Transformations

Overview

Every sinusoidal function is the graph of

y = \sin\theta

y = \cos\theta

after some shifting, stretching, or flipping. In this topic you'll learn exactly what each of the four constants in

f(\theta) = a\sin\big(b(\theta + c)\big) + d

does to the graph. Then you can read the amplitude, period, phase shift, and midline straight off an equation, and write an equation from a graph. You'll use this in every periodic model for the rest of the unit, including Free-Response Question 3 on the AP exam.

The general form

A sinusoidal function is any function that can be written as

f(\theta) = a\sin\big(b(\theta + c)\big) + d \quad \text{or} \quad g(\theta) = a\cos\big(b(\theta + c)\big) + d

where

a

b

c

, and

d

are real numbers,

a \neq 0

, and

b \neq 0

. Each constant controls one feature of the graph. The constants

a

and

b

are multiplicative transformations (they multiply), and

c

and

d

are additive transformations (they add).

Why do sine and cosine follow the exact same rules? Because cosine is just sine shifted. From Topic 3.5,

\cos\theta = \sin\left(\theta + \frac{\pi}{2}\right)

, so the cosine graph is the sine graph slid

\frac{\pi}{2}

units to the left. Any transformation moves both graphs the same way.

Here is each transformation on its own. In every panel, the dashed red curve is the parent function

y = \sin\theta

and the black curve is the transformed function.

3.6 - Sinusoidal Function Transformations

Vertical shift: $d$

Adding a constant to the output moves the whole graph up or down:

g(\theta) = \sin\theta + d

If $d > 0$ , the graph moves up $d$ units. If $d < 0$ , it moves down $|d|$ units.

The midline moves with it. The midline of $y = \sin\theta$ is $y = 0$ , so the midline of $y = \sin\theta + d$ is $y = d$ .

The maximum and minimum move too. $y = \sin\theta + 2$ runs from $1$ to $3$ instead of from $-1$ to $1$ .

Cosine works the same way:

y = \cos\theta - 4

is the cosine graph moved down 4 units, with midline

y = -4

Phase shift: $c$

Adding a constant to the input moves the graph left or right. This horizontal translation is called a phase shift.

g(\theta) = \sin(\theta + c)

The graph of

g

is the graph of

\sin\theta

shifted

-c

units horizontally. The direction is the opposite of the sign you see:

$y = \sin\left(\theta + \frac{\pi}{3}\right)$ shifts left $\frac{\pi}{3}$ . Here $c = \frac{\pi}{3}$ , so $-c = -\frac{\pi}{3}$ .

$y = \sin\left(\theta - \frac{\pi}{3}\right)$ shifts right $\frac{\pi}{3}$ . Here $c = -\frac{\pi}{3}$ , so $-c = \frac{\pi}{3}$ .

Why backwards? Think about where a cycle starts. The parent

\sin\theta

starts a cycle (on the midline, going up) when its input is

0

. For

\sin(\theta + c)

, the input is

\theta + c

, which equals

0

when

\theta = -c

. So the whole cycle now starts at

\theta = -c

The phase shift panel above shows

y = \sin\left(\theta + \frac{\pi}{2}\right)

: the sine graph moved left

\frac{\pi}{2}

, which lands exactly on the cosine graph.

Vertical dilation: $a$

Multiplying the whole function by

a

stretches or compresses the graph vertically:

g(\theta) = a\sin\theta

Every output is multiplied by

a

, so the distance from the midline to a maximum changes from

1

|a|

. That distance is the amplitude, so the amplitude of

g

|a|

If $|a| > 1$ , the graph stretches taller. $y = 2\sin\theta$ has amplitude 2.

If $0 < |a| < 1$ , the graph compresses. $y = 0.5\sin\theta$ has amplitude 0.5.

If $a < 0$ , the graph also reflects over its midline. $y = -3\sin\theta$ has amplitude 3 (not $-3$ ), and it goes down first from $\theta = 0$ instead of up.

Amplitude is a distance, so it's never negative. Always use the absolute value.

Horizontal dilation: $b$

Multiplying the input by

b

squeezes or stretches the graph horizontally:

g(\theta) = \sin(b\theta)

The parent sine finishes one cycle as its input goes from

0

2\pi

. For

\sin(b\theta)

with

b > 0

, the input

b\theta

reaches

2\pi

when

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

. So the graph is a horizontal dilation by a factor of

\frac{1}{|b|}

, and the period changes by that same factor:

\text{period} = \frac{2\pi}{|b|}

If $|b| > 1$ , the graph is squeezed and the period is shorter. $y = \sin(2\theta)$ has period $\frac{2\pi}{2} = \pi$ , so it fits two full cycles where $\sin\theta$ fits one.

If $0 < |b| < 1$ , the graph is stretched and the period is longer. $y = \sin\left(\frac{1}{3}\theta\right)$ has period $\frac{2\pi}{1/3} = 6\pi$ .

If $b < 0$ , the graph also reflects over the vertical axis. You can rewrite it with a positive $b$ : $\sin(-b\theta) = -\sin(b\theta)$ because sine is odd, and $\cos(-b\theta) = \cos(b\theta)$ because cosine is even.

Frequency is the reciprocal of the period (Topic 3.5), so

g

has frequency

\frac{|b|}{2\pi}

: that many cycles per unit of input.

Putting it all together

f(\theta) = a\sin\big(b(\theta + c)\big) + d

Constant	What it does to the graph	Key feature
$a$	Vertical dilation by a factor of $\|a\|$ (reflects over the midline if $a < 0$ )	Amplitude $= \|a\|$
$b$	Horizontal dilation by a factor of $\frac{1}{\|b\|}$	Period $= \frac{2\pi}{\|b\|}$
$c$	Horizontal translation	Phase shift of $-c$
$d$	Vertical translation	Midline $y = d$

The same table works for

g(\theta) = a\cos\big(b(\theta + c)\big) + d

. Two more facts follow from it:

The maximum value is $d + |a|$ and the minimum value is $d - |a|$ , so the range is $\left[d - |a|,\ d + |a|\right]$ .

When $a > 0$ and $b > 0$ , one full sine cycle starts on the midline at $\theta = -c$ and ends one period later, at $\theta = -c + \frac{2\pi}{b}$ .

Here is

f(\theta) = 2\sin\left(2\left(\theta - \frac{\pi}{4}\right)\right) + 1

with every feature labeled. Its constants are

a = 2

b = 2

c = -\frac{\pi}{4}

, and

d = 1

The amplitude is

2

, the period is

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

, the phase shift is

\frac{\pi}{4}

to the right, and the midline is

y = 1

. A sine cycle starts on the midline going up, so this cycle starts at

\left(\frac{\pi}{4}, 1\right)

, reaches its maximum at

\left(\frac{\pi}{2}, 3\right)

, and the next maximum comes one period later, at

\left(\frac{3\pi}{2}, 3\right)

Factor out $b$ first

This is the most common mistake in the topic. In

f(\theta) = \sin(2\theta + \pi)

, the phase shift is not

-\pi

. You have to factor

b

out of the input first:

\sin(2\theta + \pi) = \sin\left(2\left(\theta + \frac{\pi}{2}\right)\right)

Now you can read

c = \frac{\pi}{2}

, so the phase shift is

-\frac{\pi}{2}

(left

\frac{\pi}{2}

). Quick check: a cycle starts where the input is

0

, and

2\theta + \pi = 0

gives

\theta = -\frac{\pi}{2}

. That matches.

Sketching a sinusoid

To sketch

f(\theta) = a\sin\big(b(\theta + c)\big) + d

by hand (with

b > 0

), plot five key points per cycle:

Draw the midline $y = d$ and the lines $y = d + |a|$ and $y = d - |a|$ .

Find where a cycle starts: $\theta = -c$ .

Split one period into four equal parts, each $\frac{1}{4} \cdot \frac{2\pi}{b}$ long.

For sine with $a > 0$ , the five points go midline, max, midline, min, midline. For cosine with $a > 0$ , they go max, midline, min, midline, max.

If $a < 0$ , swap every max and min.

Example 1: Reading an equation

Find the amplitude, period, phase shift, midline, and range of

f(\theta) = -3\sin\left(2\theta - \frac{\pi}{2}\right) + 1

. Then find where

f

reaches its maximum on

0 \le \theta < \pi

Step 1: Factor

b

out of the input.

f(\theta) = -3\sin\left(2\left(\theta - \frac{\pi}{4}\right)\right) + 1

a = -3

b = 2

c = -\frac{\pi}{4}

, and

d = 1

Step 2: Read the features.

Amplitude: $|-3| = 3$ . Since $a < 0$ , the graph is reflected over its midline.

Period: $\frac{2\pi}{2} = \pi$ .

Phase shift: $-c = \frac{\pi}{4}$ , so $\frac{\pi}{4}$ to the right.

Midline: $y = 1$ .

Step 3: Range: from

1 - 3

1 + 3

, which is

[-2, 4]

Step 4: Because

a

is negative,

f

is at its maximum when the sine part equals

-1

2\left(\theta - \frac{\pi}{4}\right) = -\frac{\pi}{2} \quad \Rightarrow \quad \theta = 0

f(0) = -3\sin\left(-\frac{\pi}{2}\right) + 1 = -3(-1) + 1 = 4

Answer: amplitude 3, period

\pi

, phase shift

\frac{\pi}{4}

right, midline

y = 1

, range

[-2, 4]

. On

0 \le \theta < \pi

, the maximum value 4 happens at

\theta = 0

Example 2: Writing an equation from a graph

The graph of a sinusoidal function

f

is shown. It has a maximum at

(1, 5)

, and the next minimum is at

(4, -1)

. Write an equation for

f

Step 1: Midline and amplitude from the max and min.

d = \frac{5 + (-1)}{2} = 2 \qquad a = \frac{5 - (-1)}{2} = 3

Step 2: Period. A maximum to the next minimum is half a period, so half the period is

4 - 1 = 3

and the period is

6

b = \frac{2\pi}{6} = \frac{\pi}{3}

Step 3: Phase shift. Cosine starts a cycle at a maximum, and this graph has a maximum at

x = 1

, so shift cosine right 1.

f(x) = 3\cos\left(\frac{\pi}{3}(x - 1)\right) + 2

Step 4: Check the minimum:

f(4) = 3\cos(\pi) + 2 = -3 + 2 = -1

. That matches.

Sine works too. Sine starts on the midline going up, halfway between a minimum and the next maximum. The minimum before

(1, 5)

is half a period earlier, at

x = -2

, so the graph crosses the midline going up at

x = \frac{-2 + 1}{2} = -0.5

f(x) = 3\sin\left(\frac{\pi}{3}(x + 0.5)\right) + 2

Both equations describe the same graph. In fact there are infinitely many correct answers, because you can shift by any whole number of periods.

Example 3: Transformations in words (AP-style)

The graph of

g

is made from the graph of

y = \cos\theta

by these transformations, in this order: a horizontal dilation by a factor of

\frac{1}{2}

, a horizontal translation

\frac{\pi}{3}

units to the right, a vertical dilation by a factor of 4, and a vertical translation 1 unit down.

(a) Write an expression for

g(\theta)

. (b) Find the period and range of

g

. (c) Find the smallest positive

\theta

where

g

has a maximum.

(a) A horizontal dilation by a factor of

\frac{1}{2}

means

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

, so

b = 2

y = \cos(2\theta)

. Translating right

\frac{\pi}{3}

means replacing

\theta

with

\theta - \frac{\pi}{3}

y = \cos\left(2\left(\theta - \frac{\pi}{3}\right)\right)

Then stretch vertically by 4 and move down 1:

g(\theta) = 4\cos\left(2\left(\theta - \frac{\pi}{3}\right)\right) - 1

Watch the translation step.

\cos\left(2\left(\theta - \frac{\pi}{3}\right)\right) = \cos\left(2\theta - \frac{2\pi}{3}\right)

. Writing

\cos\left(2\theta - \frac{\pi}{3}\right)

instead would only shift the graph right

\frac{\pi}{6}

(b) Period:

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

. Range: from

-1 - 4

-1 + 4

, so

[-5, 3]

(c) Cosine is at its maximum when its input is

0

2\left(\theta - \frac{\pi}{3}\right) = 0 \quad \Rightarrow \quad \theta = \frac{\pi}{3}, \qquad g\left(\frac{\pi}{3}\right) = 4(1) - 1 = 3

The maximum before that is one period earlier, at

\frac{\pi}{3} - \pi = -\frac{2\pi}{3}

, which is negative. Answer:

g(\theta) = 4\cos\left(2\left(\theta - \frac{\pi}{3}\right)\right) - 1

, with period

\pi

, range

[-5, 3]

, and its first positive maximum at

\theta = \frac{\pi}{3}

Common mistakes

Reading the phase shift before factoring out $b$ . Always rewrite $\sin(b\theta + k)$ as $\sin\left(b\left(\theta + \frac{k}{b}\right)\right)$ first.

Getting the shift direction backwards. $\theta + c$ moves the graph left and $\theta - c$ moves it right (for $c > 0$ ).

Writing a negative amplitude. The amplitude is $|a|$ .

Thinking a bigger $b$ means a longer period. It's the opposite: the period is $\frac{2\pi}{|b|}$ , so a bigger $|b|$ packs in more cycles.

Mixing up the midline and the horizontal axis. The midline is $y = d$ , the average of the max and min.

Back Next

3.6 - Sinusoidal Function Transformations

Overview

The general form

Vertical shift: ddd

Phase shift: ccc

Vertical dilation: aaa

Horizontal dilation: bbb

Putting it all together

Factor out bbb first

Sketching a sinusoid

Example 1: Reading an equation

Example 2: Writing an equation from a graph

Example 3: Transformations in words (AP-style)

Common mistakes

Vertical shift: $d$

Phase shift: $c$

Vertical dilation: $a$

Horizontal dilation: $b$

Factor out $b$ first