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Precalculus · F.TF.B.5

Finding the period from the equation

A hummingbird’s wings sweep back and forth dozens of times a second. When a wave like that is written as y = sin(bx), where in the formula is its speed, and how long does one cycle last?

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Writing an equation from a wavePrecalculus
Where you are 4 in Amplitude, period and phase shift
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Trigonometric functions and their graphs8
Amplitude, period and phase shift4
Amplitude and reflection
Finding the period from the equation
Midline and vertical shiftPhase shift

When you reach for it

Reach for this move whenever a sine or cosine has a number multiplying its input, as in \displaystyle y = \cos\left(\frac{2}{3}x\right) or y = \sin(4\pi x), and you need to know how long one cycle lasts: to sketch the graph, to place its key points, to say how often something repeats, or to choose the number that makes a model repeat at the right rate.

The graphs of sine and cosine have period 2\pi. Multiplying the input by a constant b keeps the shape of the wave and changes only how wide a cycle is.

The procedure

One cycle of the inside. Take b > 0. The graph of y = \sin u makes one full cycle while u runs from 0 to 2\pi. In y = \sin(bx) the sine is applied to u = bx, so one cycle runs while bx goes from 0 to 2\pi, that is, while x goes from 0 to \displaystyle \frac{2\pi}{b}. The same is true of cosine.

Why that is exactly the period. Let b be any nonzero number. A shift p is a repeat shift of \sin(bx) when \sin\big(b(x + p)\big) = \sin(bx) for every x. As x runs over all real numbers so does u = bx, so this says \sin(u + bp) = \sin u for every u: the number bp is a repeat shift of sine. By Periodic functions, the repeat shifts of sine are exactly 2\pi k for nonzero integers k. So \displaystyle p = \frac{2\pi k}{b}, and the smallest positive choice gives the period of both \sin(bx) and \cos(bx):

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

What happens to the graph. If (x_0, y_0) is on y = \sin x, then \displaystyle \left(\frac{x_0}{b}, y_0\right) is on y = \sin(bx), because \displaystyle \sin\left(b \cdot \frac{x_0}{b}\right) = \sin x_0 = y_0. Every point keeps its height and has its x-coordinate divided by b. For b > 1 the graph is squeezed toward the y-axis, a horizontal compression by the factor \displaystyle \frac{1}{b}. For 0 < b < 1 it is pulled away from the y-axis, a horizontal stretch by the factor \displaystyle \frac{1}{b}. The number inside acts on x in reverse, unlike the vertical stretches made by multiplying the output: to feed the sine the same input x_0, the new x only has to be \displaystyle \frac{x_0}{b}.

y = \sin x
y = \sin 3x
\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
-1
1
x
y
Figure 1Multiplying the input by 3 squeezes the wave toward the y-axis. The graph of y = \sin(3x) fits three cycles into the width of one cycle of y = \sin x.

Frequency. The frequency of the wave is the number of cycles per unit of input:

f = \frac{1}{P} = \frac{|b|}{2\pi}

When x is time in seconds, f counts cycles per second, a unit called the hertz. Turned around, |b| = 2\pi f: the coefficient counts how many radians the inside angle turns per unit of x, and is sometimes called the angular frequency. A larger |b| means a faster wave and a shorter period.

Key points. One cycle of \sin u has its key points at the quarter turns \displaystyle u = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi. Dividing by b > 0 puts the key points of y = \sin(bx) at \displaystyle x = 0, \frac{P}{4}, \frac{P}{2}, \frac{3P}{4}, P, with heights 0, 1, 0, -1, 0. For y = \cos(bx) the heights are 1, 0, -1, 0, 1.

A negative b. By Unit circle symmetry and even-odd identities, sine is odd and cosine is even: \sin(-u) = -\sin u and \cos(-u) = \cos u. So for b < 0,

\begin{aligned} \sin(bx) &= -\sin(|b|x) \\ \cos(bx) &= \cos(|b|x) \end{aligned}

The period is unchanged. The cosine graph is the same graph; the sine graph is reflected across the x-axis. (For an odd function, reflecting across the x-axis and across the y-axis give the same picture.)

From a period back to b. Solving \displaystyle P = \frac{2\pi}{|b|} for |b| gives \displaystyle |b| = \frac{2\pi}{P}. A positive b is the usual choice.

Worked examples

Example 1

A fraction for b

Find the period and the frequency of \displaystyle y = \cos\left(\frac{2}{3}x\right), say whether its graph is a stretch or a compression of y = \cos x, and give the key points of the cycle starting at x = 0.

Solution

Here \displaystyle b = \frac{2}{3}, so

P = \frac{2\pi}{\frac{2}{3}} = 2\pi \cdot \frac{3}{2} = 3\pi

and the frequency is \displaystyle \frac{1}{3\pi} cycle per unit. Since 0 < b < 1, the graph is a horizontal stretch of y = \cos x by the factor \displaystyle \frac{1}{b} = \frac{3}{2}: each cycle is 3\pi wide instead of 2\pi.

The key points are a quarter period, \displaystyle \frac{3\pi}{4}, apart:

\begin{aligned} &(0, 1),\ \left(\tfrac{3\pi}{4}, 0\right),\ \left(\tfrac{3\pi}{2}, -1\right), \\ &\left(\tfrac{9\pi}{4}, 0\right),\ (3\pi, 1) \end{aligned}

Check one: at \displaystyle x = \frac{3\pi}{2} the input is \displaystyle \frac{2}{3} \cdot \frac{3\pi}{2} = \pi, and \cos \pi = -1.

y = \cos\left(\tfrac{2}{3}x\right)
y = \cos x
\displaystyle \frac{3\pi}{4}
\displaystyle \frac{3\pi}{2}
\displaystyle \frac{9\pi}{4}
3\pi
-1
1
x
y
Figure 2Example 1. One cycle of \displaystyle y = \cos\left(\frac{2}{3}x\right) is 3\pi wide, half again as wide as a cycle of y = \cos x. Its key points sit a quarter period, \displaystyle \frac{3\pi}{4}, apart.
Example 2

A pi inside b

Find the period and frequency of y = \sin(4\pi x) and the key points of its first cycle.

Solution

With b = 4\pi,

P = \frac{2\pi}{4\pi} = \frac{1}{2}

The \pi cancels, so the period is a plain number, and the frequency is \displaystyle \frac{1}{P} = 2 cycles per unit. Writing \pi into b is how a model gets a period that is a whole number of seconds or meters. The quarter period is \displaystyle \frac{1}{8}, so the key points are

\begin{aligned} &(0, 0),\ \left(\tfrac{1}{8}, 1\right),\ \left(\tfrac{1}{4}, 0\right), \\ &\left(\tfrac{3}{8}, -1\right),\ \left(\tfrac{1}{2}, 0\right) \end{aligned}

At \displaystyle x = \frac{1}{8} the input is \displaystyle 4\pi \cdot \frac{1}{8} = \frac{\pi}{2}, where sine is 1.

y = \sin(4\pi x)
\displaystyle \frac{1}{8}
\displaystyle \frac{1}{4}
\displaystyle \frac{3}{8}
\displaystyle \frac{1}{2}
\displaystyle \frac{5}{8}
\displaystyle \frac{3}{4}
\displaystyle \frac{7}{8}
1
-1
1
x
y
Figure 3Example 2. The graph of y = \sin(4\pi x) completes a cycle every \displaystyle \frac{1}{2} unit, so two cycles fit between 0 and 1.
Example 3

A negative b

Compare y = \sin(-3x) and y = \cos(-3x) with y = \sin(3x) and y = \cos(3x). Give the period of each and the key points of one cycle of y = \sin(-3x).

Solution

All four have period \displaystyle \frac{2\pi}{|{-3}|} = \frac{2\pi}{3}, since only |b| enters the period. Because cosine is even, \cos(-3x) = \cos(3x) for every x: the two cosine graphs are the same graph. Because sine is odd, \sin(-3x) = -\sin(3x): every height of y = \sin(3x) changes sign, so the graph is reflected across the x-axis.

The key points of y = \sin(3x) from 0 to \displaystyle \frac{2\pi}{3} have heights 0, 1, 0, -1, 0; for y = \sin(-3x) the signs flip:

\begin{aligned} &(0, 0),\ \left(\tfrac{\pi}{6}, -1\right),\ \left(\tfrac{\pi}{3}, 0\right), \\ &\left(\tfrac{\pi}{2}, 1\right),\ \left(\tfrac{2\pi}{3}, 0\right) \end{aligned}

Check: \displaystyle \sin\left(-3 \cdot \frac{\pi}{6}\right) = \sin\left(-\frac{\pi}{2}\right) = -1.

y = \sin(-3x)
y = \sin 3x
\displaystyle \frac{\pi}{6}
\displaystyle \frac{\pi}{3}
\displaystyle \frac{\pi}{2}
\displaystyle \frac{2\pi}{3}
\displaystyle \frac{5\pi}{6}
\pi
\displaystyle \frac{7\pi}{6}
\displaystyle \frac{4\pi}{3}
-1
1
x
y
Figure 4Example 3. The graph of y = \sin(-3x) (solid) is the graph of y = \sin(3x) (dashed) reflected across the x-axis. The period, \displaystyle \frac{2\pi}{3}, is the same.
Example 4

From a period to an equation

Write an equation y = \cos(bx) with period 6, and an equation y = \sin(bx) with period \displaystyle \frac{5\pi}{2}, each with b > 0.

Solution

From \displaystyle |b| = \frac{2\pi}{P}: for period 6, \displaystyle b = \frac{2\pi}{6} = \frac{\pi}{3}, so \displaystyle y = \cos\left(\frac{\pi}{3}x\right). Check: one cycle ends when \displaystyle \frac{\pi}{3}x = 2\pi, at x = 6.

For period \displaystyle \frac{5\pi}{2},

b = \frac{2\pi}{\frac{5\pi}{2}} = 2\pi \cdot \frac{2}{5\pi} = \frac{4}{5}

so \displaystyle y = \sin\left(\frac{4}{5}x\right). The choice \displaystyle b = -\frac{\pi}{3} would give the same cosine graph, but \displaystyle b = -\frac{4}{5} would give the sine graph reflected, a different function with the same period.

Example 5

Tsubasa’s hummingbird

Tsubasa films a hummingbird hovering at a feeder. Its wings sweep forward and back 50 times a second. She models the position of one wingtip, in centimeters forward of the shoulder (negative means behind it), by w(t) = 3\sin(bt), with t in seconds from a moment the wingtip passes the shoulder moving forward. The 3 sets how far the wingtip swings (the Amplitude and reflection) and has no effect on timing. Find b, the first time the wingtip is farthest forward, and the number of full beats in 3 seconds.

Solution

Fifty beats a second is a frequency of 50 hertz, so the period is \displaystyle P = \frac{1}{50} second. Then

b = \frac{2\pi}{P} = 2\pi \cdot 50 = 100\pi

and the model is w(t) = 3\sin(100\pi t). The wingtip is farthest forward at the first peak of the sine, a quarter period in: \displaystyle t = \frac{1}{4} \cdot \frac{1}{50} = \frac{1}{200} = 0.005 second. Check: \displaystyle 100\pi \cdot \frac{1}{200} = \frac{\pi}{2}. In 3 seconds the wings make 3 \cdot 50 = 150 full beats.

Example 6

The period of a sum

Find the period of f(x) = \sin(2x) + \cos(3x).

Solution

The parts have periods \displaystyle \frac{2\pi}{2} = \pi and \displaystyle \frac{2\pi}{3}. A shift of 2\pi is two periods of the first and three of the second, so f(x + 2\pi) = f(x). To see that nothing smaller works, suppose f(x + p) = f(x) for every x:

\begin{aligned} &\sin(2x + 2p) + \cos(3x + 3p) \\ &\quad = \sin(2x) + \cos(3x) \end{aligned}

Put -x - p in place of x, so that f(-x) = f(-x - p). Since sine is odd and cosine even, this reads

\begin{aligned} &-\sin(2x + 2p) + \cos(3x + 3p) \\ &\quad = -\sin(2x) + \cos(3x) \end{aligned}

Subtracting the two lines gives \sin(2x + 2p) = \sin(2x) for every x, so p is a repeat shift of \sin(2x), a multiple of \pi. Adding them gives \cos(3x + 3p) = \cos(3x), so p is a multiple of \displaystyle \frac{2\pi}{3}. The smallest positive number that is both is 2\pi (\displaystyle \pi m = \frac{2\pi}{3}n means 3m = 2n, first met at m = 2). The period is 2\pi, the least common multiple from Periodic functions.

Boundaries

b must not be 0. Then \sin(0x) = 0 and \cos(0x) = 1 are constant: periodic, with no period.

The input must be linear in x. In y = \sin(bx + c) the period is still \displaystyle \frac{2\pi}{|b|}, but reading the sideways slide needs the input factored first, which is Phase shift. Numbers outside the sine, as in A\sin(bx) + D, leave the period alone; they set the Amplitude and reflection and the midline. If the input is not linear in x the formula says nothing: \sin(x^2) is not periodic at all, since its zeros \displaystyle x = \sqrt{k\pi} crowd closer together, so two intervals of the same length can hold different numbers of zeros.

The 2\pi belongs to sine and cosine. It is their period (and that of secant and cosecant). Tangent and cotangent have period \pi, so y = \tan(bx) has period \displaystyle \frac{\pi}{|b|}.

A sum is not read from one coefficient. As Example 6 shows, the period of a sum must be worked out.

Units follow x. The period is measured in the units of x, and b in radians per unit of x.

Where this goes next

The period is one of four numbers that describe a sinusoid. The others are the Amplitude and reflection, the phase shift and the midline. Together they let you write an equation from a graph, where the period is measured on the picture and turned into \displaystyle b = \frac{2\pi}{P}. The same step sets the rate in Simple harmonic motion and in models of things that repeat, from tides measured in hours to sound measured in thousandths of a second.

Exercises

Period and frequency from the equation

Give the period and the frequency.

  1. y = \sin(6x)
  2. \displaystyle y = \cos\left(\frac{x}{4}\right)
  3. \displaystyle y = \cos\left(\frac{\pi}{5}x\right)
  4. y = \sin(-5x); also say how its graph is related to that of y = \sin(5x).
  5. \displaystyle y = \cos\left(-\frac{3}{4}x\right); also say how its graph is related to that of \displaystyle y = \cos\left(\frac{3}{4}x\right).
  6. \displaystyle y = \sin\left(\frac{3\pi}{2}x\right)
  7. y = \cos(0.4x)
  8. \displaystyle y = \sin(\sqrt{2}x)

Stretches, compressions and key points

  1. Is \displaystyle y = \cos\left(\frac{5}{3}x\right) a horizontal stretch or compression of y = \cos x, by what factor, and what is its period?
  2. List the five key points of the cycle of \displaystyle y = \sin\left(\frac{x}{3}\right) that starts at x = 0.
  3. List the five key points of the cycle of y = \cos(2\pi x) that starts at x = 0.
  4. The point (\pi, -1) is on y = \cos x. Which point of y = \cos(4x) does it move to?
  5. How many full cycles of y = \sin(5x) lie in [0, 2\pi]? How much of a cycle of \displaystyle y = \cos\left(\frac{x}{2}\right)?

From a period to an equation

  1. Find b > 0 so that y = \cos(bx) has period \displaystyle \frac{4\pi}{3}.
  2. Find b > 0 so that y = \sin(bx) has period 8.
  3. Find b > 0 so that y = \sin(bx) has frequency 3 cycles per unit.

Wingbeats

  1. Hayato models a wingtip of another hummingbird by w(t) = 3\sin(140\pi t), t in seconds. Find the period, the number of beats per second, and the number of beats in 2 seconds.
  2. Yui films a bird whose wings beat 40 times a second. Find b > 0 for the model w(t) = 3\sin(bt), and the first time after t = 0 that the wingtip is farthest forward.
  3. Hayato’s bird beats its wings 60 times a second, and he plays the video at one eighth of real speed. Find b for the real motion and for the motion as seen on the video, with time in seconds of playback.
  4. Find the period of g(x) = \cos(4x) + \sin(6x).

Reasoning

  1. Explain why multiplying the input of \sin x by 2 halves the period instead of doubling it.
  2. Show that if \displaystyle 0 < p < \frac{2\pi}{|b|}, then p is not a repeat shift of \cos(bx).
  3. Explain why the graph of y = \sin(-bx) is both the reflection of y = \sin(bx) across the x-axis and its reflection across the y-axis.
  4. Show that the frequency of y = \sin(bx) times its period is 1, and explain what that says in words.
  5. Suppose b > 0, c > 0, and y = \cos(bx) and y = \cos(cx) have the same graph. Show that b = c.
Answers
  1. Period \displaystyle \frac{2\pi}{6} = \frac{\pi}{3}; frequency \displaystyle \frac{3}{\pi} cycles per unit.
  2. Period \displaystyle \frac{2\pi}{1/4} = 8\pi; frequency \displaystyle \frac{1}{8\pi}.
  3. Period \displaystyle \frac{2\pi}{\pi/5} = 10; frequency \displaystyle \frac{1}{10}.
  4. Period \displaystyle \frac{2\pi}{5}; frequency \displaystyle \frac{5}{2\pi}. Since \sin(-5x) = -\sin(5x), it is the graph of y = \sin(5x) reflected across the x-axis.
  5. Period \displaystyle \frac{2\pi}{3/4} = \frac{8\pi}{3}; frequency \displaystyle \frac{3}{8\pi}. Since cosine is even, it is the same graph.
  6. Period \displaystyle \frac{2\pi}{3\pi/2} = \frac{4}{3}; frequency \displaystyle \frac{3}{4}.
  7. Period \displaystyle \frac{2\pi}{0.4} = 5\pi; frequency \displaystyle \frac{1}{5\pi}.
  8. Period \displaystyle \frac{2\pi}{\sqrt{2}} = \sqrt{2}\pi; frequency \displaystyle \frac{1}{\sqrt{2}\pi}.
  9. A compression by the factor \displaystyle \frac{3}{5}, since \displaystyle b = \frac{5}{3} > 1; period \displaystyle \frac{2\pi}{5/3} = \frac{6\pi}{5}.
  10. Period 6\pi, quarter period \displaystyle \frac{3\pi}{2}: (0, 0), \displaystyle \left(\frac{3\pi}{2}, 1\right), (3\pi, 0), \displaystyle \left(\frac{9\pi}{2}, -1\right), (6\pi, 0).
  11. Period 1, quarter period \displaystyle \frac{1}{4}: (0, 1), \displaystyle \left(\frac{1}{4}, 0\right), \displaystyle \left(\frac{1}{2}, -1\right), \displaystyle \left(\frac{3}{4}, 0\right), (1, 1).
  12. \displaystyle \left(\frac{\pi}{4}, -1\right): the x-coordinate is divided by 4, and \displaystyle \cos\left(4 \cdot \frac{\pi}{4}\right) = \cos\pi = -1.
  13. 5 full cycles, since the period \displaystyle \frac{2\pi}{5} fits 5 times into 2\pi. For \displaystyle \cos\left(\frac{x}{2}\right) the period is 4\pi, so [0, 2\pi] holds half a cycle.
  14. \displaystyle b = \frac{2\pi}{4\pi/3} = \frac{3}{2}.
  15. \displaystyle b = \frac{2\pi}{8} = \frac{\pi}{4}.
  16. Frequency 3 means period \displaystyle \frac{1}{3}, so b = 2\pi \cdot 3 = 6\pi.
  17. Period \displaystyle \frac{2\pi}{140\pi} = \frac{1}{70} second; 70 beats per second; 140 beats in 2 seconds.
  18. b = 2\pi \cdot 40 = 80\pi; farthest forward a quarter period in, at \displaystyle t = \frac{1}{4} \cdot \frac{1}{40} = \frac{1}{160} second.
  19. Real: b = 2\pi \cdot 60 = 120\pi. On the video each beat lasts \displaystyle 8 \cdot \frac{1}{60} = \frac{2}{15} second, so \displaystyle b = \frac{2\pi}{2/15} = 15\pi. Equivalently, playback time T = 8t turns \sin(120\pi t) into \sin(15\pi T).
  20. Period \pi. Both parts repeat after \pi (two periods of \displaystyle \frac{\pi}{2}, three of \displaystyle \frac{\pi}{3}). If g(x + p) = g(x) for all x, putting -x - p for x gives that \cos(4x + 4p) - \sin(6x + 6p) equals \cos(4x) - \sin(6x). Adding and subtracting this and the original shows p is a multiple of \displaystyle \frac{\pi}{2} and of \displaystyle \frac{\pi}{3}, hence of \pi.
  21. The sine needs its input to run through 2\pi for one cycle. With input 2x, that happens while x runs through only \pi, so the cycle is half as wide.
  22. If \cos\big(b(x + p)\big) = \cos(bx) for every x, then with u = bx, bp is a repeat shift of cosine, so |bp| \ge 2\pi and \displaystyle p \ge \frac{2\pi}{|b|}.
  23. \sin(-bx) = -\sin(bx) changes the sign of every height: reflection across the x-axis. Replacing x by -x in y = \sin(bx) also gives y = \sin(-bx): reflection across the y-axis.
  24. \displaystyle f \cdot P = \frac{|b|}{2\pi} \cdot \frac{2\pi}{|b|} = 1. One period holds exactly one cycle, so cycles per unit and units per cycle are reciprocals.
  25. Equal graphs have equal periods, so \displaystyle \frac{2\pi}{b} = \frac{2\pi}{c}, which gives b = c.
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