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Horizontal Translations or Phase Shifts

Practice Horizontal Translations or Phase Shifts
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Phase Shift of Sinusoidal Functions

A periodic function that does not start at the sinusoidal axis or at a maximum or a minimum has been shifted horizontally.  This horizontal movement invites different people to see different starting points since a sine wave does not have a beginning or an end. 

What are five other ways of writing the function f(x)=2 \cdot \sin x ?

Watch This

http://www.youtube.com/watch?v=wUzARNIkH-g James Sousa: Graphing Sine and Cosine with Various Transformations


The general sinusoidal function is:

f(x)=\pm a \cdot \sin (b(x+c))+d

The constant c  controls the horizontal shift.  If c=\frac{\pi}{2}  then the sine wave is shifted left by \frac{\pi}{2} .  If c=-3  then the sine wave is shifted right by 3.  This is the opposite direction than you might expect, but it is consistent with the rules of transformations for all functions

Generally b  is always written to be positive.  If you run into a situation where  b is negative, use your knowledge of even and odd functions to rewrite the function.

\cos (-x) &= \cos (x)\\\sin (-x) &= -\sin (x)

Example A

Graph the following function: f(x)=3 \cdot \cos \left(x-\frac{\pi}{2}\right)+1 .

Solution:  First find the start and end of one period and sketch only that portion of the sinusoidal axis.  Then plot the 5 important points for a cosine graph while keeping the amplitude in mind.

Example B

Given the following graph, identify equivalent sine and cosine algebraic models.

Solution:   Either this is a sine function shifted right by \frac{\pi}{4}  or a cosine graph shifted left \frac{5 \pi}{4} .

f(x)=\sin \left(x-\frac{\pi}{4}\right)=\cos \left(x+\frac{5 \pi}{4}\right)

Example C

At t=5  minutes William steps up 2 feet to sit at the lowest point of the Ferris wheel that has a diameter of 80 feet.  A full hour later he finally is let off the wheel after making only a single revolution.  During that hour he wondered how to model his height over time in a graph and equation. 

Solution:  Since the period is 60 which works extremely well with the 360^\circ  in a circle, this problem will be shown in degrees.

Time (minutes) Height (feet)
5 2
20 42
35 82
50 42
65 2

William chooses to see a negative cosine in the graph.  He identifies the amplitude to be 40 feet.  The vertical shift of the sinusoidal axis is 42 feet.  The horizontal shift is 5 minutes to the right. 

The period is 60 (not 65) minutes which implies b=6  when graphed in degrees.


Thus one equation would be:

f(x)=-40 \cdot \cos (6(x-5))+42

Concept Problem Revisited

The function f(x)=2 \cdot \sin x  can be rewritten an infinite number of ways.

2 \cdot \sin x=-2 \cdot \cos \left(x+\frac{\pi}{2}\right)=2 \cdot \cos \left(x-\frac{\pi}{2}\right)=-2 \cdot \sin (x- \pi)=2 \cdot \sin (x-8 \pi)

It all depends on where you choose start and whether you see a positive or negative sine or cosine graph.


Phase shift is a typical horizontal shift left or right that is used primarily with periodic functions. 

Guided Practice

1. Tide tables report the times and depths of low and high tides.  Here is part of tide report from Salem, Massachusetts dated September 19, 2006. 

10:15 AM

9 ft.

High Tide

4:15 PM

1 ft.

Low Tide

10:15 PM

9 ft.

High Tide

Find an equation that predicts the height based on the time.  Choose when t=0  carefully. 

2. Use the equation from Guided Practice #1 to predict the height of the tide at 6:05 AM. 

3. Use the equation from Guided Practice #1 to find out when the tide will be at exactly 8 ft on September 19^{th}


1. There are two logical places to set t=0 .  The first is at midnight the night before and the second is at 10:15 AM.  The first option illustrates a phase shift that is the focus of this concept, but the second option produces a simpler equation.  Set t=0  to be at midnight and choose units to be in minutes. 

Time (hours : minutes) Time (minutes) Tide (feet)
10:15 615 9
16:15 975 1
22:15 1335 9
\frac{615+975}{2}=795 5
\frac{1335+975}{2}=1155 5

These numbers seem to indicate a positive cosine curve.  The amplitude is four and the vertical shift is 5.  The horizontal shift is 615 and the period is 720.

720=\frac{2 \pi}{b} \rightarrow b=\frac{\pi}{360}

Thus one equation is:

f(x)=4 \cdot \cos \left(\frac{\pi}{360} (x-615)\right)+5

2. The height at 6:05 AM or 365 minutes is: f(365) \approx 2.7057 \ \text{feet} .

3. This problem is slightly different from question 2 because instead of giving  x and using the equation to find the y , this problem gives the  y and asks you to find the x .  Later you will learn how to solve this algebraically, but for now use the power of the intersect button on your calculator to intersect the function with the line y=8 .  Remember to find all the  x values between 0 and 1440 to account for the entire 24 hours.

There are four times within the 24 hours when the height is exactly 8 feet.  You can convert these times to hours and minutes if you prefer.

t \approx 532.18 (8:52), 697.82 (11:34), 1252.18 (20 : 52), 1417.82 (23:38)


Graph each of the following functions.

1. f(x)=2 \cos \left(x-\frac{\pi}{2}\right)-1

2. g(x)=-\sin (x-\pi)+3

3. h(x)=3 \cos (2 (x-\pi))

4. k(x)=-2 \sin (2x -\pi)+1

5. j(x)=-\cos \left(x+\frac{\pi}{2}\right)

Give one possible sine equation for each of the graphs below.




Give one possible cosine function for each of the graphs below.




The temperature over a certain 24 hour period can be modeled with a sinusoidal function.  At 3:00, the temperature for the period reaches a low of 22^\circ F .  At 15:00, the temperature for the period reaches a high of 40^\circ F .

12. Find an equation that predicts the temperature based on the time in minutes.  Choose t=0  to be midnight.

13. Use the equation from #12 to predict the temperature at 4:00 PM. 

14. Use the equation from #12 to predict the temperature at 8:00 AM. 

15. Use the equation from #12 to predict the time(s) it will be 32^\circ F .

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