Sinusoidal Function Transformations — AP® Precalculus Notes

Every sinusoidal function can be written with one formula,

$f(\theta)$ $= a\sin(b(\theta + c)) + d$ or $f(\theta)$ $= a\cos(b(\theta + c)) + d$ ,

where $a$ and $b$ are not zero. Each constant is one of the transformations from Unit 1 applied to the sine graph. This page shows what each one does, then how to run the process backward: read a graph and write its formula.

What each constant does

Constant	Transformation	Feature it sets
$a$	vertical dilation by $\|a\|$ ; reflection across the midline if $a < 0$	amplitude $\|a\|$
$b$	horizontal dilation by $\frac{1}{\|b\|}$	period $\frac{2\pi}{\|b\|}$
$c$	horizontal translation by $-c$	phase shift: left $c$ if $c > 0$
$d$	vertical translation by $d$	midline $y = d$

Why the period is $\frac{2\pi}{|b|}$ . $\sin(b\theta)$ completes one cycle while $b\theta$ runs from $0$ to $2\pi$ , which happens while $\theta$ runs from $0$ to $\frac{2\pi}{b}$ . A bigger $b$ squeezes more cycles into the same space, so the period shrinks. Turned around, this is the rule the exam asks for again and again:

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

In context the period is often hidden in a frequency. A spring that bounces 2 times per second has period $\frac{1}{2}$ second, so $b = \frac{2\pi}{1/2} = 4\pi$ , not $\frac{2\pi}{2}$ .

Why the shift is $-c$ . The sine graph starts a cycle (crosses its midline going up) where its input is $0$ . For $\sin(b(\theta + c))$ that is where $\theta + c = 0$ , at $\theta = -c$ . So $c > 0$ moves the graph left. This is the same rule as $f(x + 3)$ moving a graph left 3.

Figure 1. A phase shift of

\frac{\pi}{4}

to the right, an amplitude of 2 and a midline of

y = 1

. LevlPrep original diagram.

Worked example 1. Describe $f(\theta)$ $= -3\sin(2\theta + \pi) + 4$ .

Factor $b$ out of the input first: $2\theta + \pi = 2\left(\theta + \frac{\pi}{2}\right)$ , so $c = \frac{\pi}{2}$ .

Amplitude $|-3| = 3$ , reflected across the midline. Period $\frac{2\pi}{2} = \pi$ . Phase shift $\frac{\pi}{2}$ to the left. Midline $y = 4$ , so the values run from $1$ to $7$ .

Sine and cosine are shifts of each other

Cosine is sine moved $\frac{\pi}{2}$ to the left: $\cos\theta = \sin\left(\theta + \frac{\pi}{2}\right)$ . So every sinusoid has a sine formula and a cosine formula. A reflection gives more: $-\cos\theta = \sin\left(\theta - \frac{\pi}{2}\right)$ . On the exam, any correct formula earns the points; choose the one that makes $c$ easiest to read.

From a graph to a formula

Midline: $d = \frac{\max + \min}{2}$ .
Amplitude: $|a| = \frac{\max - \min}{2}$ .
Period: measure one cycle, then $b = \frac{2\pi}{\text{period}}$ .
Shift: pick a landmark. For sine with $a > 0$ , use a midline crossing where the graph is rising: $c = -(\text{its input})$ . For cosine with $a > 0$ , use a highest point. For cosine with $a < 0$ , use a lowest point.

Worked example 2. A sinusoid has a highest point at $(\pi, 4)$ and lowest points at $(0, -2)$ and $(2\pi, -2)$ . Write a formula.

Midline $d = \frac{4 + (-2)}{2} = 1$ . Amplitude $\frac{4 - (-2)}{2} = 3$ . Period $2\pi$ (lowest to lowest), so $b = 1$ .

With cosine and $a = 3$ , the highest point should sit where the input is $0$ ; here it is at $\theta = \pi$ , a shift right $\pi$ : $f(\theta) = 3\cos(\theta - \pi) + 1$ .

With cosine and $a = -3$ (a reflected cosine starts at a lowest point), the lowest point at $\theta = 0$ needs no shift: $f(\theta) = -3\cos\theta + 1$ . With sine and $a = 3$ : the graph crosses the midline going up halfway between $(0, -2)$ and $(\pi, 4)$ , at $\theta = \frac{\pi}{2}$ , so $f(\theta)$ $= 3\sin\left(\theta - \frac{\pi}{2}\right) + 1$ . All three are the same function.

Worked example 3. Write a sinusoid with amplitude 2, period $\pi$ and midline $y = -1$ whose graph crosses its midline going up at $\theta = 0$ .

$b = \frac{2\pi}{\pi} = 2$ . A sine with $a > 0$ already rises through its midline at $0$ , so $c = 0$ : $f(\theta) = 2\sin(2\theta) - 1$ .

Common slips

Reading the shift before factoring: $\sin(3\theta - \pi)$ shifts right $\frac{\pi}{3}$ , not $\pi$ .
Using the period as $b$ , or the frequency as the period: $b$ is $2\pi$ divided by the period.
Using $\max - \min$ as the amplitude: it is half of that.
Forgetting the sign of the shift: $\theta - 2$ moves right, $\theta + 2$ moves left.