Phase shift

When you reach for it

Reach for this move whenever a sine or cosine has a number added to or subtracted from the input after the input has been multiplied, as in

and you want to know where its cycles start: which way the graph slid, and how far. The number you can see, \displaystyle \frac{\pi}{2}, is not the answer. The input has to be rewritten as the coefficient times a shifted variable, \displaystyle 3\left(x - \frac{\pi}{6}\right), and only then can the slide be read. Here it is \displaystyle \frac{\pi}{6}.

The procedure

The sliding rule you already have

From shifting a graph: if g is any function and h is a number, the graph of y = g(x - h) is the graph of y = g(x) moved h to the right (to the left when h is negative). The rule has one condition that is easy to pass over. It is about x itself being replaced by x - h. Nothing else about the formula may change.

Why the subtracted number is not the slide

Start from g(x) = \sin 3x, a sine wave squeezed so that one cycle takes \displaystyle \frac{2\pi}{3} instead of 2\pi (see Finding the period from the equation). Is \displaystyle y = \sin\left(3x - \frac{\pi}{2}\right) the same as \displaystyle g\left(x - \frac{\pi}{2}\right)? Replace x by \displaystyle x - \frac{\pi}{2} and see:

That is a different function. At \displaystyle x = \frac{\pi}{3} the given function is \displaystyle \sin\left(\pi - \frac{\pi}{2}\right) = \sin\frac{\pi}{2} = 1, while \displaystyle \sin\left(3 \cdot \frac{\pi}{3} - \frac{3\pi}{2}\right) = \sin\left(-\frac{\pi}{2}\right) = -1. One is at a peak where the other is at a trough.

The trouble is the order of operations. In \displaystyle 3x - \frac{\pi}{2} the input is first multiplied by 3 and only then has \displaystyle \frac{\pi}{2} taken away. To see a slide, x itself must be shifted first, so factor out the 3:

Multiplying back by the distributive property returns \displaystyle 3x - \frac{\pi}{2}, which confirms the factoring. Now

and the sliding rule applies: the graph of y = \sin 3x moved \displaystyle \frac{\pi}{6} to the right. The \displaystyle \frac{\pi}{2} was measured after the squeeze by 3, so it was three times the distance the graph actually moved along the x-axis. Dividing by 3 undoes that.

A second view: where the cycle starts

The function \sin u runs through one cycle as u goes from 0 to 2\pi. Here \displaystyle u = 3x - \frac{\pi}{2}, so the cycle that sat at [0, 2\pi] now sits where

Adding \displaystyle \frac{\pi}{2} to each part and dividing by 3 gives \displaystyle \frac{\pi}{6} \le x \le \frac{5\pi}{6}. The cycle starts at \displaystyle \frac{\pi}{6}, which is the same number factoring gave, and lasts \displaystyle \frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}. This is no coincidence: \displaystyle 3\left(x - \frac{\pi}{6}\right) is zero exactly when \displaystyle x = \frac{\pi}{6}, so solving “input = 0” and factoring are the same piece of algebra. The five key points of the cycle come from \displaystyle u = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, which give \displaystyle x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3}, \frac{5\pi}{6}, a quarter period \displaystyle \frac{\pi}{6} apart. These are the dots in the figure.

The general statement

For any number b \ne 0, factoring b out of bx - d gives

and the same for cosine. When an equation is written as

the number c is the phase shift: the graph is the graph of y = A\sin bx + k (or the cosine one) moved c to the right when c > 0 and |c| to the left when c < 0. So the phase shift of \sin(bx - d) is \displaystyle \frac{d}{b}, and the phase shift of \sin(bx + d) is \displaystyle -\frac{d}{b}, a slide to the left. The Amplitude and reflection A and the vertical shift k act on the output, so they do not change c.

Why this is right: let g(x) = A\sin bx + k. Then, by the distributive property,

so A\sin(bx - bc) + k is the graph of g slid c. For \sin(bx - d) to match, bc must equal d, that is, \displaystyle c = \frac{d}{b}.

Worked examples

Boundaries

• The coefficient b must not be 0: factoring divides by b, and \sin(0 \cdot x - d) is a constant with no slide to read.
• The input must be a line in x, bx - d. Something like \sin\left(x^2 - 1\right) is not a slid sine wave, and has no phase shift.
• A phase shift is only fixed up to whole periods. If c is a phase shift of a wave with period p, so is c + np for every integer n. The usual choice is the one of smallest size.
• The shift is measured from a named parent. The same graph is a sine slid one amount and a cosine slid another, as in Example 5, so say which parent you mean.
• With a negative b, factor out b with its sign, or first turn the coefficient positive with \sin(-\theta) = -\sin\theta or \cos(-\theta) = \cos\theta.

Where this goes next

With the Amplitude and reflection, the period, the midline and the phase shift, every number in y = A\sin\bigl(b(x - c)\bigr) + k has a meaning on the graph, and Writing an equation from a wave runs the reading backward, from a picture to a formula. The factoring move is not special to waves: \displaystyle y = \sqrt{2x - 6} = \sqrt{2(x - 3)} is the square root graph squeezed and then slid 3 to the right, not 6.

Exercises

Factor and read the shift

For each graph, factor the input, give the phase shift, and say which way the graph slides.

1. \displaystyle y = \sin\left(2x - \frac{\pi}{2}\right)
2. y = \sin\left(3x + \pi\right)
3. \displaystyle y = \cos\left(5x - \frac{\pi}{3}\right)
4. \displaystyle y = \cos\left(\frac{x}{3} + \frac{\pi}{6}\right)
5. \displaystyle y = \sin\left(\pi x - \frac{\pi}{4}\right)
6. y = \sin\left(6x - 2\right)

One cycle

7. Find the cycle of \displaystyle y = \sin\left(4x - \frac{\pi}{2}\right) that starts at the phase shift, and its five key points.
8. Find the cycle of \displaystyle y = \cos\left(2x + \frac{\pi}{3}\right) that starts at the phase shift, and its five key points.
9. Find the smallest positive x at which \displaystyle y = \sin\left(3x - \frac{\pi}{4}\right) equals 1.
10. Find the smallest positive x-intercept of \displaystyle y = \sin\left(2x + \frac{\pi}{3}\right).

From the slide to the formula

11. The graph of y = \sin 2x is slid \displaystyle \frac{\pi}{8} to the right. Write its equation with the input multiplied out.
12. The graph of y = \cos 3x is slid \displaystyle \frac{\pi}{9} to the left. Write its equation with the input multiplied out.
13. Write a sine function with b = 2 whose cycle starts at \displaystyle x = \frac{\pi}{3}, with the input multiplied out.

The tandem bicycle

These use Example 4, where Bekzod’s pedal height is h_B = 7\sin(2\pi t).

14. On another day Jasur takes Shahzoda’s seat, and his pedal height is \displaystyle h_J = 7\sin\left(2\pi t - \frac{\pi}{3}\right). How many seconds behind Bekzod’s pedal is his?
15. When does Jasur’s pedal from Exercise 14 first reach the top, for t \ge 0?
16. Bekzod’s left pedal is at the other end of the same crank: h_L = 7\sin(2\pi t - \pi). Find its lag behind his right pedal in seconds, and show that h_L = -h_B.
17. Climbing a hill, they pedal at two turns a second: h_B = 7\sin(4\pi t) and \displaystyle h_S = 7\sin\left(4\pi t - \frac{\pi}{2}\right). How many seconds behind is Shahzoda’s pedal now?
18. A pedal with height h = 7\sin(2\pi t + d), where 0 \le d < 2\pi, first reaches the top at \displaystyle t = \frac{1}{10} second. Find d.

Reasoning

19. Jasur says \displaystyle y = \sin\left(3x - \frac{\pi}{3}\right) is the graph of y = \sin 3x slid \displaystyle \frac{\pi}{3} to the right. Test his claim at \displaystyle x = \frac{\pi}{9} and give the correct slide.
20. The phase shifts of y = \cos(2x - \pi) and y = \cos(2x + \pi) point in opposite directions. Explain why the two graphs are nevertheless the same.
21. Take b > 0 and d > 0. Explain why the phase shift of \sin(bx - d) is smaller than d when b > 1 and larger than d when 0 < b < 1.
22. Explain why solving “input = 0” always gives the same phase shift as factoring the input.