Math Core

Real waves rarely have height $1$ and period $2\pi$ . A tide might swing $3$ meters around an average level of $5$ meters and repeat every $12.4$ hours. To model waves like that, you stretch and shift the basic sine and cosine graphs, using the same transformation ideas you've applied to parabolas and absolute value graphs.

The general form

Every transformed sine or cosine graph can be written as

y = a\sin\big(b(x - h)\big) + k \qquad\text{or}\qquad y = a\cos\big(b(x - h)\big) + k.

Each parameter controls one feature of the wave.

What each parameter does

Amplitude $= |a|$ . The wave rises $|a|$ above the midline and falls $|a|$ below it. If $a < 0$ , the graph is also reflected across the midline.
Period $= \dfrac{2\pi}{|b|}$ . A larger $b$ squeezes the cycles closer together.
Phase shift $= h$ . The graph moves $h$ units right (left if $h < 0$ ).
Midline $y = k$ . The graph moves $k$ units up (down if $k < 0$ ).

The maximum value is $k + |a|$ and the minimum value is $k - |a|$ .

Amplitude and midline

Multiplying by $a$ stretches the wave vertically. Adding $k$ lifts the whole wave, midline and all.

y = sin x and y = 3 sin x + 1. The second wave has amplitude 3 and midline y = 1, so it ranges from -2 to 4.Open in grapher →

For $y = 3\sin x + 1$ , the midline is $y = 1$ , the maximum is $1 + 3 = 4$ and the minimum is $1 - 3 = -2$ . The range is $-2 \le y \le 4$ .

Period

Why is the period $\dfrac{2\pi}{b}$ ? In $y = \sin(bx)$ , the input to sine is $bx$ . One full cycle happens as $bx$ goes from $0$ to $2\pi$ , that is, as $x$ goes from $0$ to $\dfrac{2\pi}{b}$ .

y = sin x and y = sin(2x). With b = 2, the period shrinks from 2π to π, so two cycles fit where one used to.Open in grapher →

Worked example: Identifying the features

State the amplitude, period, midline and range of $y = -2\cos(4x) + 5$ .

Amplitude: $|a| = |-2| = 2$ . The negative sign flips the graph, so it starts at a minimum instead of a maximum.
Period: $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$ .
Midline: $y = 5$ .
Range: from $5 - 2 = 3$ to $5 + 2 = 7$ , so $3 \le y \le 7$ .

Phase shift

The phase shift is the horizontal translation. In $y = \sin\left(x - \dfrac{\pi}{3}\right)$ , every point of the sine graph moves $\dfrac{\pi}{3}$ to the right. The "minus means right" rule is the same one you used for $y = (x - h)^2$ .

Common mistake

When $b \ne 1$ , factor $b$ out before reading the phase shift. In $y = \sin(2x - \pi)$ , the shift is not $\pi$ . Rewrite it as

y = \sin\left(2\left(x - \frac{\pi}{2}\right)\right),

so the shift is $\dfrac{\pi}{2}$ to the right. You can check: the cycle starts where the input $2x - \pi$ equals $0$ , which is $x = \dfrac{\pi}{2}$ .

Graphing with key points

To sketch one cycle of $y = a\sin\big(b(x - h)\big) + k$ :

Draw the midline $y = k$ and mark the max $k + |a|$ and min $k - |a|$ .
The cycle starts at $x = h$ and ends at $x = h + \text{period}$ .
Split the cycle into four equal steps of $\dfrac{\text{period}}{4}$ . These five $x$ -values carry the key points.
Sine follows the pattern midline, max, midline, min, midline. Cosine follows max, midline, min, midline, max. If $a < 0$ , swap max and min.

Worked example: Sketching a transformed cosine

Find the five key points of one cycle of $y = 2\cos\left(x - \dfrac{\pi}{4}\right) + 1$ .

Here $a = 2$ , $b = 1$ , $h = \dfrac{\pi}{4}$ , $k = 1$ . The period is $2\pi$ , so each step is $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$ . The max is $3$ and the min is $-1$ .

$x$	$\dfrac{\pi}{4}$	$\dfrac{3\pi}{4}$	$\dfrac{5\pi}{4}$	$\dfrac{7\pi}{4}$	$\dfrac{9\pi}{4}$
$y$	$3$ (max)	$1$	$-1$ (min)	$1$	$3$ (max)

y = 2cos(x - π/4) + 1 with its midline y = 1 dashed.Open in grapher →

Writing an equation from a graph

Work backward: read the max and min, then the period, then decide where a cycle starts.

Worked example: From graph to equation

A sinusoid has a maximum of $4$ at $x = 0$ , and its next maximum is at $x = \pi$ . Its minimum value is $-2$ . Write an equation for it.

Midline: halfway between $4$ and $-2$ , so $k = \dfrac{4 + (-2)}{2} = 1$ .
Amplitude: $a = \dfrac{4 - (-2)}{2} = 3$ .
Period: from one max to the next is $\pi$ , so $\dfrac{2\pi}{b} = \pi$ and $b = 2$ .
The graph has a maximum at $x = 0$ , which is where cosine starts, so use cosine with no phase shift.

y = 3\cos(2x) + 1

y = 3cos(2x) + 1: maximum 4, minimum -2, period π.Open in grapher →

Tip

Many equations describe the same graph. The wave above is also $y = 3\sin\left(2\left(x + \dfrac{\pi}{4}\right)\right) + 1$ , since a sine cycle starts a quarter period before the cosine peak. Choose whichever makes the phase shift simplest.