Math Core

Real waves are rarely exactly $y = \sin x$ . A tide might swing $4$ feet, repeat every $12$ hours and sit $6$ feet above the sea floor on average. Four numbers in the equation stretch, squeeze and slide the basic wave to match any of these situations.

Amplitude: how tall the wave is

Multiplying the output by a number $A$ stretches the graph vertically. In $y = 3\sin x$ , every height of the sine curve is tripled, so the peaks rise to $3$ and the valleys drop to $-3$ . The zeros stay where they were, because $3 \cdot 0 = 0$ .

y = sin x and y = 3 sin x. Same zeros, three times the height.Open in grapher →

The amplitude is half the distance from a peak to a valley. For $y = A\sin x$ or $y = A\cos x$ it equals $|A|$ . If $A$ is negative, the graph is also reflected across the midline: $y = -2\cos x$ starts at a valley $(0, -2)$ instead of a peak.

Period: how long one cycle takes

Multiplying the input by a number $B$ squeezes or stretches the graph horizontally. In $y = \cos(2x)$ , the inside $2x$ runs from $0$ to $2\pi$ while $x$ only runs from $0$ to $\pi$ . So one full cycle fits in a length of $\pi$ : the wave repeats twice as fast.

y = cos x completes one cycle on [0, 2π]; y = cos 2x completes two.Open in grapher →

In general, one cycle happens as $Bx$ goes from $0$ to $2\pi$ , that is, as $x$ goes from $0$ to $\dfrac{2\pi}{B}$ . So the period is $\dfrac{2\pi}{|B|}$ . A large $B$ gives a short period (a fast wave); a $B$ between $0$ and $1$ gives a long period. For example, $y = \sin\left(\dfrac{x}{4}\right)$ has period $\dfrac{2\pi}{1/4} = 8\pi$ .

Vertical shift: where the middle is

Adding a constant $D$ to the output slides the whole graph up or down. The midline moves from $y = 0$ to $y = D$ . The wave now oscillates between

\text{maximum} = D + |A| \qquad \text{and} \qquad \text{minimum} = D - |A|.

For $y = 3\cos x - 2$ , the midline is $y = -2$ , the maximum is $-2 + 3 = 1$ and the minimum is $-2 - 3 = -5$ .

Phase shift: sliding left or right

Replacing $x$ with $x - C$ slides the graph $C$ units to the right (to the left if $C$ is negative), just as with any function. This horizontal shift is called the phase shift. For example, $y = \sin\left(x - \dfrac{\pi}{4}\right)$ is the sine wave moved $\dfrac{\pi}{4}$ to the right: its cycle starts at $x = \dfrac{\pi}{4}$ instead of $x = 0$ .

The general sinusoid

y = A\sin\big(B(x - C)\big) + D \qquad\qquad y = A\cos\big(B(x - C)\big) + D

Feature	Formula
Amplitude	$\lvert A \rvert$ (a negative $A$ also reflects the graph)
Period	$\dfrac{2\pi}{\lvert B \rvert}$
Phase shift	$C$ (right if positive, left if negative)
Midline	$y = D$
Range	$[D - \lvert A \rvert,\ D + \lvert A \rvert]$

Common mistake

Factor out $B$ before reading the phase shift. In $y = \sin(2x - \pi)$ the shift is not $\pi$ . Rewrite the inside as $2x - \pi = 2\left(x - \dfrac{\pi}{2}\right)$ ; the phase shift is $\dfrac{\pi}{2}$ to the right. Likewise, $y = \cos(3x + \pi) = \cos\left(3\left(x + \dfrac{\pi}{3}\right)\right)$ is shifted $\dfrac{\pi}{3}$ to the left.

Graphing with the five key points

Find the start of a cycle, $x = C$ , and the end, $x = C + \text{period}$ .
Split that interval into four equal steps of $\dfrac{\text{period}}{4}$ . These are the $x$ -values of the five key points.
Take the basic $y$ -values ( $0, 1, 0, -1, 0$ for sine or $1, 0, -1, 0, 1$ for cosine), multiply them by $A$ , then add $D$ .
Plot, connect smoothly, and repeat.

Worked example: A stretch and a squeeze

Graph one cycle of $y = 3\sin(2x)$ .

Solution. Amplitude $3$ , period $\dfrac{2\pi}{2} = \pi$ , no phase shift, midline $y = 0$ . The quarter step is $\dfrac{\pi}{4}$ , so the key $x$ -values are $0, \dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4}, \pi$ . Multiply the sine pattern by $3$ :

(0, 0),\ \left(\tfrac{\pi}{4}, 3\right),\ \left(\tfrac{\pi}{2}, 0\right),\ \left(\tfrac{3\pi}{4}, -3\right),\ (\pi, 0).

Worked example: All four changes at once

Graph one cycle of $y = -2\cos\left(x - \dfrac{\pi}{3}\right) + 1$ .

Solution. $A = -2$ (amplitude $2$ , reflected), $B = 1$ (period $2\pi$ ), $C = \dfrac{\pi}{3}$ , $D = 1$ . The cycle runs from $\dfrac{\pi}{3}$ to $\dfrac{\pi}{3} + 2\pi = \dfrac{7\pi}{3}$ in steps of $\dfrac{\pi}{2}$ :

$x$	$\frac{\pi}{3}$	$\frac{5\pi}{6}$	$\frac{4\pi}{3}$	$\frac{11\pi}{6}$	$\frac{7\pi}{3}$
basic $\cos$	$1$	$0$	$-1$	$0$	$1$
$\times(-2)$	$-2$	$0$	$2$	$0$	$-2$
$+1$	$-1$	$1$	$3$	$1$	$-1$

The graph starts at a minimum of $-1$ , crosses the midline $y = 1$ , peaks at $3$ and returns.

y = -2cos(x - π/3) + 1 with its midline y = 1 dashed.Open in grapher →

Worked example: Factoring first

State the amplitude, period, phase shift and range of $y = 4\sin(3x + \pi) - 2$ .

Solution. Factor the inside: $3x + \pi = 3\left(x + \dfrac{\pi}{3}\right)$ , so $C = -\dfrac{\pi}{3}$ .

Amplitude: $4$
Period: $\dfrac{2\pi}{3}$
Phase shift: $\dfrac{\pi}{3}$ to the left
Midline $y = -2$ , so the range is $[-2 - 4, -2 + 4] = [-6, 2]$ .

Worked example: Writing the equation from a graph

A sinusoid with a peak at (0, 5) and a valley at (π/2, 1).Open in grapher →

Write an equation for this graph.

Solution. Read off the maximum $5$ and minimum $1$ .

Midline: $D = \dfrac{5 + 1}{2} = 3$ . Amplitude: $A = \dfrac{5 - 1}{2} = 2$ .
A peak to the next valley is half a cycle: from $0$ to $\dfrac{\pi}{2}$ . So the period is $\pi$ and $B = \dfrac{2\pi}{\pi} = 2$ .
The graph starts at a peak when $x = 0$ , which is how cosine starts, so use cosine with $C = 0$ .

The equation is $y = 2\cos(2x) + 3$ . (A sine equation with a phase shift would also work, such as $y = 2\sin\left(2\left(x + \dfrac{\pi}{4}\right)\right) + 3$ .)

Tip

Check an equation by plugging in one key point. For $y = 2\cos(2x) + 3$ at $x = \dfrac{\pi}{2}$ : $2\cos(\pi) + 3 = -2 + 3 = 1$ , the valley. It matches.