Precalculus · F.TF.B.5

Writing an equation from a wave

A wave on a screen rises to a peak, falls to a trough and repeats. Which formula draws exactly that wave, and how much of the picture do you need to find it?

You need

Amplitude and reflectionPrecalculus Finding the period from the equationPrecalculus Midline and vertical shiftPrecalculus Phase shiftPrecalculus +1 more

Unlocks

Modeling things that repeatGrades 10 to 12

Where you are 8 in Trigonometric functions and their graphs

Trigonometry7

Trigonometric functions and their graphs8

▴ 3 above Periodic functions Inverse trig functions

Writing an equation from a wave

Composing trig and inverse trig functions Modeling things that repeat3

When you reach for it

The four earlier lessons read a wave off its formula: the Amplitude and reflection from A, the period from b, the phase shift from c and the midline from k in

\begin{aligned} y &= A\sin\bigl(b(x - c)\bigr) + k \quad\text{or} \\ y &= A\cos\bigl(b(x - c)\bigr) + k \end{aligned}

This lesson runs the reading backward. You have a picture of a wave, or a sentence about one, or a table of its values, and you want a formula that produces it. You need one whenever a wave has to be computed rather than looked at: its height at a time the picture does not show, or a formula to hand to a calculator. The lesson runs over two days. Part 1 (Examples 1 to 3) works from a graph. Part 2 (Examples 4 to 7) works from a description or a table and settles which of sine or cosine to use.

The procedure

Part 1: from a graph

Midline and amplitude. Find the highest value M and the lowest value m. The wave swings the same distance above and below its midline, so the midline sits halfway between and the amplitude is half the gap:

k = \frac{M + m}{2}, \qquad |A| = \frac{M - m}{2}

Period. The period P is the horizontal distance from one peak to the next. When only a peak and the next trough are visible, that distance is half a period, and a peak to the next midline crossing is a quarter period. Then, taking b positive,

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

because a cycle of sine or cosine is 2\pi wide and b squeezes it to P.

Where a cycle starts. Each of the four basic shapes, a sine or a cosine with A positive or negative, starts its cycle at a point you can recognize on any graph:

Shape	Sign of A	At x = c the wave is
cosine	A > 0	at a peak
cosine	A < 0	at a trough
sine	A > 0	on the midline, rising
sine	A < 0	on the midline, falling

Choose one such point on the graph. Its x-coordinate is c.

Why the formula matches the whole graph. Take the first row. At x = c the cosine’s input is 0 and \cos 0 = 1, so y = k + |A|, the top of the wave. The formula runs between k - |A| = m and k + |A| = M and repeats every \displaystyle \frac{2\pi}{b} = P. So formula and graph have the same top, bottom and period, and a peak in the same place. Those four facts leave no freedom in a wave y = A\cos\bigl(b(x - c)\bigr) + k with A > 0 and b > 0: the top and bottom give k and A, the period gives b, and a peak gives c up to a multiple of P, which (as shown below) does not change the wave. The graph is such a wave too, so it has these same letters, and the two agree at every x. The other rows work the same way, because -\cos starts at its lowest value and \pm\sin start on the midline.

More than one right answer. Any peak will do, and consecutive peaks are P apart, so c may be changed by any integer multiple of P. Any of the four rows will do as well. All of these formulas draw the same graph.

Part 2: from a description or a table, and choosing sine or cosine

A description such as “highest value 11 at time 1, lowest value 3 at time 4” carries the same numbers as the graph, so the same three readings apply. A frequency f, in cycles per unit of input, gives the period as \displaystyle P = \frac{1}{f}.

A table gives sampled points. Its largest and smallest entries are M and m only if the table actually lands on a peak and a trough. That is sure to happen when the steps divide a quarter period evenly and one entry falls on a peak, a trough or a midline crossing, since the features come a quarter period apart. The period is the smallest shift that carries every entry to an equal entry.

Choosing sine or cosine. Look for the feature with the simplest x-coordinate, often x = 0. A peak or trough there calls for \pm\cos; a midline crossing calls for \pm\sin. Either way c comes out 0 and the formula is shortest. The choice is only about convenience, because a cosine wave is a sine wave slid a quarter period, as the graphs of sine and cosine showed: \displaystyle \cos\theta = \sin\left(\theta + \frac{\pi}{2}\right) for every \theta.

Worked examples

Example 1

The lead wave

Find two equations for the wave in the figure, which has a highest point at (1, 11) and the next lowest point at (4, 3): one using cosine and one using sine.

Figure 1The wave of Example 1. The dashed line y = 7 is the midline, halfway between the highest and lowest values.

Solution

The highest value is M = 11 and the lowest is m = 3, so

k = \frac{11 + 3}{2} = 7, \qquad |A| = \frac{11 - 3}{2} = 4

The midline is y = 7, the dashed line. From the peak at x = 1 to the trough at x = 4 is half a period, so P = 2 \cdot 3 = 6 and \displaystyle b = \frac{2\pi}{6} = \frac{\pi}{3}.

Cosine starts at a peak, and there is one at x = 1, so c = 1:

y = 4\cos\left(\frac{\pi}{3}(x - 1)\right) + 7

Sine starts on the midline, rising. The wave rises through the midline a quarter period, \displaystyle \frac{6}{4} = \frac{3}{2}, before the peak, at \displaystyle x = 1 - \frac{3}{2} = -\frac{1}{2}. So \displaystyle c = -\frac{1}{2}:

y = 4\sin\left(\frac{\pi}{3}\left(x + \frac{1}{2}\right)\right) + 7

Check at the trough x = 4: the first input is \displaystyle \frac{\pi}{3} \cdot 3 = \pi and \cos\pi = -1; the second is \displaystyle \frac{\pi}{3} \cdot \frac{9}{2} = \frac{3\pi}{2} and \displaystyle \sin\frac{3\pi}{2} = -1. Both give y = -4 + 7 = 3.

Example 2

Starting at a trough

Find an equation for the wave in the figure, with lowest point \displaystyle \left(\frac{\pi}{6}, -4\right) and next highest point \displaystyle \left(\frac{2\pi}{3}, 2\right).

Figure 2The wave of Example 2, with its midline y = -1 dashed.

Solution

Here M = 2 and m = -4, so \displaystyle k = \frac{2 + (-4)}{2} = -1 and \displaystyle |A| = \frac{2 - (-4)}{2} = 3. Trough to peak is \displaystyle \frac{2\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}, half a period, so P = \pi and \displaystyle b = \frac{2\pi}{\pi} = 2.

The first marked point is a trough, and -\cos starts at a trough, so take A = -3 and \displaystyle c = \frac{\pi}{6}:

y = -3\cos\left(2\left(x - \frac{\pi}{6}\right)\right) - 1

Starting at the peak instead gives a positive cosine with \displaystyle c = \frac{2\pi}{3}:

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

Check the second at the trough: \displaystyle 2\left(\frac{\pi}{6} - \frac{2\pi}{3}\right) = -\pi, and 3\cos(-\pi) - 1 = -4.

Example 3

Otabek’s piston

Otabek graphs the height h, in centimeters, of an engine piston’s top above a fixed mark against time t in milliseconds. The graph crosses h = 11 going up at (2, 11), peaks at (7, 16) and bottoms out at (17, 6). Find an equation, the number of cycles each second, and the height at t = 30.

Figure 3Otabek’s graph of the piston’s height in Example 3.

Solution

With M = 16 and m = 6, the midline is k = 11 and the amplitude is 5, so the point (2, 11) is on the midline. From that rising crossing to the peak is a quarter period, 7 - 2 = 5, so P = 20 milliseconds and \displaystyle b = \frac{2\pi}{20} = \frac{\pi}{10}. A rising midline crossing is where a positive sine starts, so c = 2:

h = 5\sin\left(\frac{\pi}{10}(t - 2)\right) + 11

One cycle takes 20 milliseconds, so there are \displaystyle \frac{1000}{20} = 50 cycles each second.

At t = 30 the input is \displaystyle \frac{\pi}{10} \cdot 28 = \frac{14\pi}{5}. Removing one full turn, \displaystyle \sin\frac{14\pi}{5} = \sin\frac{4\pi}{5} \approx 0.5878, so h \approx 5(0.5878) + 11 \approx 13.9 centimeters.

Example 4

Donncha’s trace: four ways to start

Donncha watches a voltage trace on an oscilloscope. It reaches its lowest value, -1.5 volts, at t = 1 millisecond and its next highest value, 2.5 volts, at t = 3. Write the trace as each of the four basic shapes, then pick the simplest.

Solution

The midline is \displaystyle k = \frac{2.5 + (-1.5)}{2} = 0.5 and \displaystyle |A| = \frac{2.5 - (-1.5)}{2} = 2. Lowest to highest is half a period, so P = 4 milliseconds and \displaystyle b = \frac{2\pi}{4} = \frac{\pi}{2}. There are \displaystyle \frac{1000}{4} = 250 cycles each second.

Each shape needs its own starting point. The trough at t = 1 gives -\cos with c = 1. The peak at t = 3 gives \cos with c = 3. The rising midline crossing is halfway from trough to peak, at t = 2, giving \sin with c = 2. The falling crossing comes a quarter period before the trough, at t = 0, giving -\sin with c = 0:

\begin{aligned} V &= -2\cos\left(\frac{\pi}{2}(t - 1)\right) + 0.5 \\ V &= 2\cos\left(\frac{\pi}{2}(t - 3)\right) + 0.5 \\ V &= 2\sin\left(\frac{\pi}{2}(t - 2)\right) + 0.5 \\ V &= -2\sin\left(\frac{\pi}{2}t\right) + 0.5 \end{aligned}

All four draw the same trace. The last has no shift at all, so it is the one to use.

Example 5

From a table

Zvonimir records these values of a sinusoid. Find an equation with c = 0.

x	y
0	-1
1	2
2	5
3	2
4	-1
5	2
6	5
7	2
8	-1

Solution

The values repeat every 4 steps, so P = 4 and \displaystyle b = \frac{\pi}{2}. Half a period apart, a sinusoid sits the same distance above and below its midline, because \cos(\theta + \pi) = -\cos\theta. The pairs x = 0, 2 and x = 1, 3 give \displaystyle \frac{-1 + 5}{2} = \frac{2 + 2}{2} = 2, so k = 2. Then x = 1 is a midline crossing, and x = 0 and x = 2, a quarter period on either side, are a trough and a peak: M = 5, m = -1 and |A| = 3. At x = 0 the wave is at a trough, where -\cos starts:

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

Check at x = 2: -3\cos\pi + 2 = 5. A sine form needs the rising crossing at x = 1, giving \displaystyle y = 3\sin\left(\frac{\pi}{2}(x - 1)\right) + 2, the same wave with a shift to carry.

Example 6

Zvonimir’s signal generator

Zvonimir sets a signal generator to swing between -3 and 5 volts at 200 cycles each second. At t = 0 the voltage is 1 volt and rising. Find an equation with t in milliseconds, and the voltage at t = 2.

Solution

The midline is \displaystyle k = \frac{5 + (-3)}{2} = 1 and the amplitude is \displaystyle \frac{5 - (-3)}{2} = 4. One cycle takes \displaystyle \frac{1}{200} of a second, which is 5 milliseconds, so \displaystyle b = \frac{2\pi}{5}. At t = 0 the voltage is 1, on the midline, and rising, which is where a positive sine starts:

V = 4\sin\left(\frac{2\pi}{5}t\right) + 1

At t = 2: \displaystyle V = 4\sin\frac{4\pi}{5} + 1 \approx 4(0.5878) + 1 \approx 3.35 volts.

Example 7

Why the two answers agree

Show that the two equations of Example 1 are the same function.

Solution

Write \displaystyle \theta = \frac{\pi}{3}(x - 1). By the distributive property,

\begin{aligned} \frac{\pi}{3}\left(x + \frac{1}{2}\right) &= \frac{\pi}{3}(x - 1) + \frac{\pi}{3} \cdot \frac{3}{2} \\ &= \theta + \frac{\pi}{2} \end{aligned}

so the sine equation is \displaystyle y = 4\sin\left(\theta + \frac{\pi}{2}\right) + 7. Since \displaystyle \sin\left(\theta + \frac{\pi}{2}\right) = \cos\theta for every \theta, this is y = 4\cos\theta + 7, the cosine equation. The shift by a quarter period, \displaystyle \frac{3}{2}, became the \displaystyle \frac{\pi}{2} that turns sine into cosine.

Boundaries

The procedure needs a true sinusoid: every peak at the same height, every trough at the same depth, peaks equally spaced and each trough exactly halfway between its two peaks. A wave whose peaks shrink, or that rises faster than it falls, has no equation of this form.
The formula is only as exact as the readings. Read points that are labeled or fall on grid lines.
A table gives M and m only if it contains a peak and a trough. A table that steps past them shows values that are too small, and the midline and amplitude come out wrong (Exercise 21).
With x in degrees, a cycle is 360 wide and \displaystyle b = \frac{360}{P}.
The answer is never unique: c may change by any integer multiple of P, and \pm\sin and \pm\cos all work.
A physical model is an approximation. A real piston’s height is close to a sine wave but not exactly one, because the rod linking it to the crankshaft tilts as it turns.

Where this goes next

Writing the equation is the step that turns a wave into a model. Height on a turning wheel builds the formula from a wheel’s size and speed, Tides, daylight and sound fits waves to repeating data, and Modeling things that repeat uses such a formula to predict and to answer questions about times the data never showed.

Exercises

Part 1: from a graph

Exercises 1 and 2 use the first figure, Exercises 3 and 4 the second.

Figure 4The wave for Exercises 1 and 2.

Figure 5The wave for Exercises 3 and 4.

Find an equation using cosine for the wave in the first figure.
Find an equation using sine, with c = 0, for the wave in Exercise 1.
Find an equation with c = 0 for the wave in the second figure.
Find an equation using cosine with A > 0 for the wave in Exercise 3.
A graph has a highest point at (-1, 6) and the next lowest point at (4, 0). Find an equation using cosine.
A graph has a lowest point at \displaystyle \left(\frac{\pi}{4}, 1\right) and the next highest point at \displaystyle \left(\frac{3\pi}{4}, 7\right). Find an equation with c = 0.
A graph crosses its midline y = -2 going up at x = 3 and next reaches its highest value, 1, at x = 5. Find an equation using sine.
A graph has a highest point at (17, 4) and the next lowest point at (24, -2). Find a cosine equation with 0 \le c < P.
Otabek graphs a larger engine. The piston’s height is highest, 18 centimeters, at t = 5 milliseconds and next lowest, 8 centimeters, at t = 17. Find a cosine equation, the number of cycles each second, and the height at t = 0.

Part 2: from a description or a table

A sinusoid has highest value 10 at x = 0, lowest value -4, and period 12. Find an equation.
A sinusoid crosses its midline y = -1 going down at x = 0, has amplitude 5 and period 4\pi. Find an equation.
Donncha’s trace reaches 1.2 volts, its highest, at t = 0.5 milliseconds and -0.4 volts, its next lowest, at t = 1.5. Find an equation with c = 0 and the number of cycles each second.
Zvonimir’s generator runs at 400 cycles each second between -2 and 6 volts and is at its lowest at t = 0. Find an equation with t in milliseconds.
Find the voltage at t = 1 millisecond for the generator in Exercise 13.
A table of a sinusoid lists x = 0, 1, 2, 3, 4, 5, 6 with y = 3, 6, 3, 0, 3, 6, 3. Find an equation.
A table of a sinusoid lists x = 0, 2, 4, 6, 8, 10, 12 with y = -2, -5, -8, -5, -2, -5, -8. Find an equation using cosine.
Find an equation using a positive sine for the table in Exercise 16.
A table of a sinusoid lists \displaystyle x = 0, \frac{\pi}{8}, \frac{\pi}{4}, \frac{3\pi}{8}, \frac{\pi}{2} with y = 1, 4, 1, -2, 1. Find an equation.
Zvonimir writes \displaystyle y = 2\cos\left(\frac{\pi}{4}(x - 1)\right) and Donncha writes \displaystyle y = 2\sin\left(\frac{\pi}{4}(x + 1)\right) for the same graph. Show that both are right.

Reasoning

Using y = A\cos\bigl(b(x - c)\bigr) + k with A > 0 and b > 0, explain why a peak and the next trough are exactly half a period apart.
Zvonimir samples y = 2\sin x + 1 at x = 0, 1, 2, 3, 4, 5, 6 and takes the largest and smallest entries as M and m. What midline and amplitude does he get, and why are they wrong?
Explain why c may be changed by any integer multiple of the period without changing the graph.
Prove that for b > 0 and \displaystyle P = \frac{2\pi}{b}, the wave y = A\cos\bigl(b(x - c)\bigr) + k is the same as \displaystyle y = A\sin\left(b\left(x - c + \frac{P}{4}\right)\right) + k.
A sketch shows peaks at x = 1 and x = 5 and a trough between them at x = 4. Explain why no equation y = A\sin\bigl(b(x - c)\bigr) + k fits it.

Answers

Other correct equations exist for most of these: c may change by a multiple of P, or another of the four shapes may be used.

\displaystyle y = 4\cos\left(\frac{\pi}{4}(x - 2)\right) + 5. Here k = 5, |A| = 4, \displaystyle \frac{P}{2} = 4, so P = 8 and \displaystyle b = \frac{\pi}{4}; the peak is at x = 2.
\displaystyle y = 4\sin\left(\frac{\pi}{4}x\right) + 5. The rising crossing is a quarter period, 2, before the peak at 2, so at x = 0.
y = -4\cos 3x - 2. Here k = -2, |A| = 4, \displaystyle P = \frac{2\pi}{3}, b = 3, and the trough is at x = 0.
\displaystyle y = 4\cos\left(3\left(x - \frac{\pi}{3}\right)\right) - 2, starting at the peak.
\displaystyle y = 3\cos\left(\frac{\pi}{5}(x + 1)\right) + 3. Here k = 3, |A| = 3, P = 10.
y = -3\sin 2x + 4. Here k = 4, |A| = 3, P = \pi, b = 2; the falling crossing is \displaystyle \frac{\pi}{4} before the trough, at x = 0.
\displaystyle y = 3\sin\left(\frac{\pi}{4}(x - 3)\right) - 2. Here |A| = 1 - (-2) = 3 and \displaystyle \frac{P}{4} = 2, so P = 8.
\displaystyle y = 3\cos\left(\frac{\pi}{7}(x - 3)\right) + 1. Here P = 14, and the peak at 17 means peaks also at 17 - 14 = 3.
\displaystyle h = 5\cos\left(\frac{\pi}{12}(t - 5)\right) + 13; P = 24 milliseconds, so about 41.7 cycles each second; \displaystyle h(0) = 5\cos\frac{5\pi}{12} + 13 \approx 14.3 centimeters.
\displaystyle y = 7\cos\left(\frac{\pi}{6}x\right) + 3. Here k = 3, |A| = 7, \displaystyle b = \frac{2\pi}{12}.
\displaystyle y = -5\sin\frac{x}{2} - 1, since \displaystyle b = \frac{2\pi}{4\pi} = \frac{1}{2}.
V = 0.8\sin(\pi t) + 0.4; 500 cycles each second. Here k = 0.4, |A| = 0.8, P = 2, and the rising crossing is a quarter period before the peak, at t = 0.
\displaystyle V = -4\cos\left(\frac{4\pi}{5}t\right) + 2. Here \displaystyle P = \frac{1000}{400} = 2.5 milliseconds and \displaystyle b = \frac{2\pi}{2.5}.
\displaystyle V = -4\cos\frac{4\pi}{5} + 2 \approx 5.24 volts.
\displaystyle y = 3\sin\left(\frac{\pi}{2}x\right) + 3. Period 4, midline 3, rising at x = 0.
\displaystyle y = 3\cos\left(\frac{\pi}{4}x\right) - 5. Period 8, peak at x = 0.
\displaystyle y = 3\sin\left(\frac{\pi}{4}(x - 6)\right) - 5. The wave rises through -5 at x = 6.
y = 3\sin 4x + 1. Period \displaystyle \frac{\pi}{2}, so b = 4; rising through the midline at x = 0.
Both are right: \displaystyle \frac{\pi}{4}(x + 1) = \frac{\pi}{4}(x - 1) + \frac{\pi}{2}, and \displaystyle \sin\left(\theta + \frac{\pi}{2}\right) = \cos\theta.
Peaks are where b(x - c) = 2\pi n and troughs where b(x - c) = \pi + 2\pi n, n any integer. From the peak x = c, the next trough is at \displaystyle x = c + \frac{\pi}{b} = c + \frac{P}{2}.
Largest entry 2.82 (at x = 2), smallest -0.92 (at x = 5), so midline about y = 0.95 and amplitude about 1.87, instead of 1 and 2. No whole-number sample lands on a peak or trough.
Replacing c by c + nP subtracts bnP = 2\pi n from the input, and cosine and sine repeat every 2\pi.
The sine’s input is \displaystyle b\left(x - c + \frac{P}{4}\right), which by the distributive property is \displaystyle b(x - c) + \frac{bP}{4}, that is, \displaystyle b(x - c) + \frac{\pi}{2}. Since \displaystyle \sin\left(\theta + \frac{\pi}{2}\right) = \cos\theta for every \theta, the two agree at every x.
Peaks at 1 and 5 force P = 4, and on a sinusoid the trough lies halfway between, at x = 3. A trough at 4 means the curve falls for 3 units and rises for 1, which no sine or cosine does.