Topic 3.6A

Sinusoidal Function Transformations

In 3.5 we measured sinusoids — amplitude, midline, period, and frequency — and saw how $a$ and $d$ in $y = a \sin \theta + d$ control the height of a wave. But every function in 3.5 still had period $2\pi$ . Real periodic data almost never cycles every $2\pi$ units: the pedal from 3.1 takes 1.2 seconds per turn. This note adds the constant that controls the period and then puts all the pieces together to write equations. Sliding a graph left or right — the last transformation — is the topic of 3.6B.

9 MIN READ6 IDEAS32 PROBLEMS14 flashcards

Practice 32 problems Next: Chapter 3.6B Sinusoidal Function Transformations

One Constant for Each Transformation

Remember from 1.12: multiplying a function's outputs stretches its graph vertically, adding to the outputs shifts it up or down, and multiplying the input stretches or squeezes it horizontally. A sinusoidal function uses exactly these moves.

f(\theta)=a\sin(b\theta)+d\qquad g(\theta)=a\cos(b\theta)+d

Constant	Transformation	What it controls
a	vertical dilation (plus a reflection if a < 0)	amplitude = \|a\|
d	vertical translation	midline: y = d
b	horizontal dilation by a factor of 1/\|b\|	period = 2π/\|b\|

How b Changes the Period

Why is the period $2\pi/|b|$ ? The sine function completes one cycle while its input runs from 0 to $2\pi$ . In $\sin (b\theta )$ the input is $b\theta$ , so one cycle is finished when $b\theta$ reaches $2\pi$ — that is, when $\theta = 2\pi /b$ (for $b > 0$ ).

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

Top: $b = 3$ packs three cycles into $2\pi$ , so each cycle has length $2\pi /3$ . Bottom: $b = 1/2$ stretches one cycle to length $4\pi$ .

So a larger $|b|$ means a SHORTER period — the wave is squeezed together — while $a |b|$ between 0 and 1 means a longer period. The frequency, $|b|/(2\pi )$ , grows as $|b|$ grows: more cycles fit into each unit of input.

COMMON MISTAKE

"In $y = \sin (3\theta )$ , the period is 3." or "…the period is $3 \cdot 2\pi = 6\pi$ ."

$b$ is not the period, and you do not multiply by it. $b$ tells you how many times faster the input runs. With $b = 3$ , the input $3\theta$ reaches $2\pi$ when $\theta$ is only $2\pi /3$ , so each cycle is three times SHORTER: period = $2\pi \div 3 = 2\pi /3$ .

Sanity check with the picture: the graph of $\sin (3\theta )$ fits three full waves between 0 and $2\pi$ .

Periods that are not multiples of π

In real situations the period is usually a plain number of seconds, days, or hours. Solving period = $2\pi /b$ for $b$ gives

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

For example, a period of 10 needs $b = 2\pi /10 = \pi /5$ . That is why $\pi$ so often appears inside the parentheses of a real-world model: it is there to cancel the $\pi$ in $2\pi$ .

Quick check

What is the period of $y = 3\sin\left(\frac{\pi}{4}\theta\right)$ ?

Reading an Equation

Worked example

Example 1. For $f(\theta ) = -2 \sin (\theta /3) + 5$ , find the amplitude, midline, period, frequency, maximum value, and minimum value. Then describe how the graph starts at $\theta = 0$ .

01
Match the form $a \sin (b\theta ) + d$ : $a = -2$ , $b = 1/3$ , and $d = 5$ .
02
Amplitude: $|a| = 2$ . Midline: $y = 5$ . Maximum value: $5 + 2 = 7$ . Minimum value: $5 - 2 = 3$ .
03
Period: $2\pi \div (1/3) = 2\pi \times 3 = 6\pi$ . Frequency: 1/( $6\pi$ ) ≈ 0.053.
04
At $\theta = 0$ the graph is on its midline: $f(0) = 5$ . Because a is negative, the graph is flipped, so it heads DOWN from the midline first and reaches its minimum, 3, a quarter period later, at $\theta = 6\pi /4 = 3\pi /2$ .

Worked example

Example 2. A quantity is modeled by $g(t) = 7 \cos (\pi t/5) + 1$ , where $t$ is measured in hours. Find the amplitude, midline, period, and frequency.

01
$a = 7$ , $b = \pi /5$ , and $d = 1$ . Amplitude: 7. Midline: $y = 1$ , so the quantity swings between $1 - 7 = -6$ and $1 + 7 = 8$ .
02
Period: $2\pi \div (\pi /5) = 2\pi \times 5/\pi = 10$ hours. The π's cancel, leaving a plain number.
03
Frequency: $1/10$ cycle per hour.

Graphing with Quarter-Period Steps

Remember from 3.5: neighboring key points of a sinusoid are a quarter period apart. So once you know the period, divide it by 4 and step along the horizontal axis.

Worked example

Example 3. Sketch one period of $y = 1.5 \cos (3\theta ) + 2$ , starting at $\theta = 0$ .

01
Amplitude 1.5 and midline $y = 2$ , so the maximum is 3.5 and the minimum is 0.5.
02
Period: $2\pi /3$ . Quarter period: $(2\pi /3) \div 4 = \pi /6$ . So the key points are at $\theta = 0$ , $\pi /6$ , $\pi /3$ , $\pi /2$ , and $2\pi /3$ .
03
A cosine graph with $a > 0$ starts at its maximum, then follows the pattern max → midline → min → midline → max.
θ 0 π/6 π/3 π/2 2π/3
y 3.5 2 0.5 2 3.5
04
Plot the five points and connect them with a smooth wave that bends toward the midline (3.4). Repeat the cycle to extend the graph.

$y = 1.5 \cos (3\theta ) + 2$ : key points every $\pi /6$ , one full period every $2\pi /3$ .

θ	0	π/6	π/3	π/2	2π/3
y	3.5	2	0.5	2	3.5

COMMON MISTAKE

"The period is $2\pi /3$ , so I plot points at 0, $2\pi /3$ , $4\pi /3$ , …"

Those are only the maximums — one per period. They all have the same height, so they show nothing about the shape in between. Always step by a QUARTER period ( $\pi /6$ here) to catch the midline crossings and the minimum.

Writing an Equation from the Features

Going the other way, each feature gives one constant: the midline gives $d$ , the amplitude gives $|a|$ , and the period gives $b$ . Where the graph starts decides whether to use sine or cosine and what sign a should have.

CONCEPT

Choosing the starting function (no horizontal shift)

Starts at a maximum → $a \cos (b\theta ) + d$ with $a > 0$ .

Starts at a minimum → $a \cos (b\theta ) + d$ with $a < 0$ .

Starts on the midline, going up → $a \sin (b\theta ) + d$ with $a > 0$ .

Starts on the midline, going down → $a \sin (b\theta ) + d$ with $a < 0$ .

Worked example

Example 4. A sinusoidal function $f$ has maximum value 11, minimum value 3, and period 8, and it has a maximum at $x = 0$ . Write an equation for $f$ .

01
Midline: $d = (11 + 3) \div 2 = 7$ . Amplitude: $(11 - 3) \div 2 = 4$ .
02
$b = 2\pi$ ÷ period = $2\pi \div 8 = \pi /4$ .
03
The graph starts at a maximum, so use cosine with a positive a: $f(x) = 4 \cos (\pi x/4) + 7$ .
04
Check: $f(0) = 4 \cos 0 + 7 = 11$ , the maximum ✓. $f(4) = 4 \cos \pi + 7 = 3$ , the minimum, half a period later ✓.

COMMON MISTAKE

"The period is 8, so $f(x) = 4 \cos (8x) + 7$ ."

$b$ is not the period. $\cos (8x)$ has period $2\pi /8 = \pi /4 \approx 0.785$ — far shorter than 8. Use $b = 2\pi$ ÷ period = $\pi /4$ .

Check by plugging in: with $b = \pi /4$ , $f(8) = 4 \cos (2\pi ) + 7 = 11$ , back at the maximum after exactly one period. ✓

REAL-LIFE EXAMPLE

The pedal equation, at last

In 3.1 we described the right pedal of a bike: crank arm 17 cm, crank center 28 cm above the ground, one turn every 1.2 seconds, starting at the lowest point.

Midline 28 → $d = 28$ . Amplitude 17 → $|a| = 17$ . Period 1.2 → $b = 2\pi \div 1.2 = 5\pi /3$ .

It starts at a minimum, so use cosine with a negative a: $h(t) = -17 \cos (5\pi t/3) + 28$ .

Check against the 3.1 table: $h(0) = 11$ , $h(0.3) = 28$ , $h(0.6) = 45$ , $h(0.9) = 28$ , $h(1.2) = 11$ . Every value matches.

The pedal from 3.1, now described by an equation.

Quick check

A sinusoid has midline $y = 2$ , amplitude $1.5$ , period $\frac{2\pi}{3}$ , and a maximum at $\theta = 0$ . Write an equation.

Common slips

"In $y = \sin (3\theta )$ , the period is 3." or "…the period is $3 \cdot 2\pi = 6\pi$ ."

$b$ is not the period, and you do not multiply by it. $b$ tells you how many times faster the input runs. With $b = 3$ , the input $3\theta$ reaches $2\pi$ when $\theta$ is only $2\pi /3$ , so each cycle is three times shorter: period = $2\pi \div 3 = 2\pi /3$ .

Sanity check with the picture: the graph of $\sin (3\theta )$ fits three full waves between 0 and $2\pi$ .
"The period is $2\pi /3$ , so I plot points at 0, $2\pi /3$ , $4\pi /3$ , …"

Those are only the maximums — one per period. They all have the same height, so they show nothing about the shape in between. Always step by a quarter period ( $\pi /6$ here) to catch the midline crossings and the minimum.
"The period is 8, so $f(x) = 4 \cos (8x) + 7$ ."

$b$ is not the period. $\cos (8x)$ has period $2\pi /8 = \pi /4 \approx 0.785$ — far shorter than 8. Use $b = 2\pi$ ÷ period = $\pi /4$ .

Check by plugging in: with $b = \pi /4$ , $f(8) = 4 \cos (2\pi ) + 7 = 11$ , back at the maximum after exactly one period. ✓

Lock it in

Try the flashcards

14 cards · Sinusoids, Sine or cosine graph?

Start

Recap card

6 lines to re-read the night before.

01
$f(\theta ) = a \sin (b\theta ) + d$ and $g(\theta ) = a \cos (b\theta ) + d$ : a is a vertical dilation, $d$ a vertical translation, $b$ a horizontal dilation.
02
Amplitude $|a|$ , midline $y = d$ , period $2\pi/|b|$ , frequency $|b|/(2\pi )$ .
03
Larger $|b|$ → shorter period. $b$ is not the period: $b = 2\pi$ ÷ period.
04
To graph, step by a quarter period and follow the pattern of key points.
05
To write an equation, find $d$ , $|a|$ , and $b$ from the features; the starting point picks sine or cosine and the sign of a.
06
Coming up in 3.6B: horizontal shifts (phase shifts), and writing equations for graphs that start anywhere.

Practice 32 problems Next: Chapter 3.6B Sinusoidal Function Transformations

Unlock every question

Free

Premium

Sign in to FiveWay

Sinusoidal Function Transformations

One Constant for Each Transformation

How b Changes the Period

Periods that are not multiples of π

Reading an Equation

Graphing with Quarter-Period Steps

Writing an Equation from the Features

Recap card

Unlock Premium