5.6: Phase Shift of Sinusoidal Functions
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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 allows for different starting points since a sine wave does not have a beginning or an end.
What are five other ways of writing the function \(f(x)=2 \cdot \sin x ?\)
Phase Shift of Sinusoidal Functions
The general sinusoidal function is:
\(f(x)=\pm a \cdot \sin (b(x+c))+d\)
The constant \(c\) controls the phase shift. Phase shift is the horizontal shift left or right for periodic functions. 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.
To graph a function such as \(f(x)=3 \cdot \cos \left(x-\frac{\pi}{2}\right)+1,\) first find the start and end of one period. Then sketch only that portion of the sinusoidal axis. Finally, plot the 5 important points for a cosine graph while keeping the amplitude in mind. The graph is shown below.
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.
\(\cos (-x)=\cos (x)\)
\(\sin (-x)=-\sin (x)\)
Examples
Earlier, you were asked to write \(f(x)=2 \cdot \sin x\) in five different ways. 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.
Given the following graph, identify equivalent sine and cosine algebraic models.
Either this is a sine function shifted right by \(\frac{\pi}{4}\) or a cosine graph shifted left \(\frac{5 \pi}{4}\).
\(f(x)=\sin \left(x-\frac{\pi}{4}\right)=\cos \left(x+\frac{5 \pi}{4}\right)\)
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.
Since the period is 60 which works extremely well with the \(360^{\circ}\) in a circle, this problem will be shown in degrees.
\(
\begin{array}{|l|l|}
\hline \text { Time (minutes) } & \text { Height (feet) } \\
\hline 5 & 2 \\
\hline 20 & 42 \\
\hline 35 & 82 \\
\hline 50 & 42 \\
\hline 65 & 2 \\
\hline
\end{array}
\)
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.
\(60=\frac{360}{b}\)
Thus one equation would be:
\(f(x)=-40 \cdot \cos (6(x-5))+42\)
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.
\(
\begin{array}{|c|c|c|}
\hline 10: 15 \mathrm{AM} & 9 \mathrm{ft} & \text { High Tide } \\
\hline 4: 15 \mathrm{PM} & 1 \mathrm{ft} . & \text { Low Tide } \\
\hline 10: 15 \mathrm{PM} & 9 \mathrm{ft} & \text { High Tide } \\
\hline
\end{array}
\)
Find an equation that predicts the height based on the time. Choose when \(t=0\) carefully.
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.
\(
\begin{array}{|l|l|l|}
\hline \text { Time (hours : minutes) } & \text { Time (minutes) } & \text { Tide (feet) } \\
\hline 10: 15 & 615 & 9 \\
\hline 16: 15 & 975 & 1 \\
\hline 22: 15 & 1335 & 9 \\
\hline & \frac{615+975}{2}=795 & 5 \\
\hline & \frac{1335+975}{2}=1155 & 5 \\
\hline
\end{array}
\)
These numbers seem to indicate a positive cosine curve. The amplitude is 4 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:
\(f(x)=4 \cdot \cos \left(\frac{\pi}{360}(x-615)\right)+5\)
Use the equation from Example 4 to find out when the tide will be at exactly \(8 \mathrm{ft}\) on September \(19^{t h}\).
This problem gives you 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)
Review
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. \(k(x)=-2 \sin (2 x-\pi)+1\)
5. \(j(x)=-\cos \left(x+\frac{\pi}{2}\right)\)
Give one possible sine equation for each of the graphs below.
6.
7.
8.
Give one possible cosine function for each of the graphs below.
9.
10.
11.
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 \(22^{\circ} \mathrm{F}\). At \(15: \mathrm{OO}\), 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 \(4: 00 \mathrm{PM}\).
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 \(32^{\circ} \mathrm{F}\).
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