Phase Shifts - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Phase Shifts

Video duration:

6m

Play a video:

Hey, everyone. So in recent videos, we've been talking about graphs of the sine and cosine functions and we've been discussing how these graphs can be transformed, maybe they'll be shifted up or down or stretched in some kind of way. Well, what we're now going to be taking a look at is another kind of transformation known as a phase shift. Now, this might sound really complicated and technical but don't sweat it because we're going to learn in this video that a phase shift is really just a horizontal shift which can occur for your sine or cosine. And just like how a vertical shift took your graph and shifted it up or down. A phase shift can take your graph and shift it to the left or right. So without further ado, let's take a look at some situations that you'll see in this course associated with a phase shift. Now, we should be familiar with this graph, which is the graph for the cosine of X. It starts here at a high value of one and then kind of makes this wave length pattern that extends on both sides of the graph. Now, a way that you could see a phase shift is if you had some kind of number inside the parentheses that was being added or subtracted to the X within your cosine. And an example of this would be say the cosine of X minus pi over two. Now to see how this phase shift affects the graph, we can go ahead and plot some points and the way that I'm gonna plot these points and I'm going to take all of these values and I'm going to subtract pi over two from them and then I'll evaluate the cosine. So the cosine of zero minus pi over two is going to be the cosine of negative pi over two. And we can see that our cosine graph is zero at that point. So that means we're going to start at a value of zero. Now, next, we're going to have the cosine of pi over two minus pi over two. That's gonna be the cosine of zero, which is just going to be one. Next, we'll have the cosine of pi minus pi over two. That's gonna be the cosine of pi over two, which is back at zero. Now, for three pi over two minus pi over two, this cosine is going to evaluate to negative one, meaning you'll be down there on your graph and then we'll take a look at the cosine of two pi minus pi over two, which turned out to just be zero. So you're going to be back up here. Now, if you go ahead and connect to these points with a smooth curve, you're going to notice something kind of interesting about this graph and we can go ahead and extend this back here as well notice how we have a very similar wave that we started with, except it appears that this wave is kind of out of phase with the original wave. And that's exactly what happens with the phase shift. In fact, you'll notice we actually have very similar values except all of these have been shifted over to the right. We have this one that went there, the zero that went there, this negative one that went there and then the zero that went here. So subtracting pi over two from the inside of this cosine function cause every point on the original cosine graph to be shifted over to the right by pi over two units. We saw it here, we saw it there and we saw that at every point on this graph. So whenever you have some kind of number subtracted inside the sine or cosine, we'll call that number H this number is going to cause a phase shift in your graph. Now, I will mention here that not every textbook you see is going to use the letter H, some textbooks will use other variables or letters. But as long as you see it added or subtracted from the inside of the sin or cosine function. It means the same thing. Now all of this really begs some interesting questions like how do we know how much our graph is shifted or how do we know what direction our graph has shifted? Well, it turns out all of that can be discovered by looking at the inside of the trig function because if you have a situation where you have a sine or cosine of BX minus sum number, then that means that your H value is positive and your graph is going to shift to the right by H over B units. Now, if you instead have a situation where the inside of your trip function was BX plus sum number, that means your H value is negative and the graph is going to shift the left by H over B units. Now, in the situation we had right here, we had the cosine of one X minus pi over two. And because the inside of this function is in the form BX minus a number that means that our H value is positive. And our graph shifted to the right now finding H over B for this graph would be our H value which is pi over two divided by RB value, which is one and pi over two divided by one is the same thing as just pi over two. So this shows us how much our graph shifted by. Now, another thing that you might notice is we have actually kind of a familiar looking wave notice we start at a value of zero and then we go up and down and complete a period at two pi. This is actually very similar to the sine graph. In fact, it's the same thing and this is the interesting about phase shifts. It can actually make your cosine graph look like a sine graph or the other way around now to really make sure this concept makes sense. Let's try an example where we have to deal with a sin or cosine function which has gone through a phase shift. So in this example, we are asked to graph, the function Y is equal to the sine of two X plus pi over one full period. Now to solve this problem, what I'm first going to do is draw a graph which we're more familiar with. So I'm going to ignore this pi for now and I'm just going to draw the graph for Y is equal to the sine of two X. Now recall that this graph is going to have a bit of a different period than we're used to because we have this in front and the period for a sine or cosine function is equal to two pi over B. Well, the B value we can see in front of the X is two and then these twos will cancel giving us just pi. So this sine graph is going to start here at the origin and then we're going to reach a peak between zero and pi over two. We're going to dip here through PI and we're going to reach a full period at pi. Now, from here, what I'm going to do is I'm going to use this graph to draw the graph of this function which is Y is equal to the sine of two X plus pi. And to figure this out, well, I can see that there's some kind of face shifts that happened. And recall that when you have a situation where the inside of your function is BX plus some number, that means the H value is negative and will shift to the left by H over B units. Now, what is H over B? Well, H over B is going to be this H value which is pi divided by this B value, which is two. So this graph is going to shift to the left by pi over two units. So that means this point is actually going to start over here. I mean, our graph is going to actually dip down first and then we're going to cross through pi over two, then we're going to go up and then back down when we get the pi. So this is what the graph is going to look like over one full period. And that would be the solution to this problem. Now, if this seem kind of confusing to you, there's actually another way that you can look at this function that we have Y equals the sine of two X plus pi could also be written as Y is equal to two times X plus pi over two. So notice how this two took our graph and changed the period of it. And then this portion of the function shifted our graph to the left by pi over two units. So this is another way that you could write the function if this helps you to visualize things better. So this is the main idea of face shifts and what they look like in the equation and how they affect the graph. So hope you found this video helpful. Thanks for watching.

2

Problem

Problem

Describe the phase shift for the following function:

$y=\cos\left(5x-\frac{\pi}{2}\right)$

A

$\frac{\pi}{2}$ to the right

B

$\frac{\pi}{2}$ to the left

C

$\frac{\pi}{10}$ to the right

D

$\frac{\pi}{10}$ to the left

3

Problem

Problem

Describe the phase shift for the following function:

$y=\cos\left(2x+\frac{\pi}{6}\right)$

A

$\frac{\pi}{6}$ to the right

B

$\frac{\pi}{6}$ to the left

C

$\frac{\pi}{12}$ to the right

D

$\frac{\pi}{12}$ to the left

4

example

Example 1

Video duration:

3m

Play a video:

Hey, everyone. So let's give this problem a try here. We're asked to graph the function Y is equal to three times the sin of X plus pi. OK. So how can we solve this problem? Well, what I'm first going to do is write out the most general form for the sine function, which is Y is equal to A times the S of BX minus H plus K. So this is our most general form. Now, something that I noticed is that we don't have a plus K out front here. So we can ignore any kind of vertical shift in our graph, but we do have an A value and our A value or amplitude is equal to three. Now, what I also noticed is for BX minus H what we have in the parentheses, we don't have a specific B value out front here. So we could just say that B is equal to one, right? And you could also say that like K is equal to zero, for example, but that's not really necessary for this problem. But if we don't see a number in front of the X, we can just say that numbers one because one times X would just be X. Now, I also noticed that we have a pi here, but no, that we have a plus pi and the general form of this equation asks for a minus sign. So what we're going to need to do is find a way to get a minus sign in this equation. Well, the way that we can do that is by changing this plus to minus a negative. So we're going to have that Y is equal to three times the sign of X minus negative pi and this is a perfectly valid way to write this because these negatives would normally just cancel. So what that does is shows us what our H value is, our H value is negative pi. So this is all the details that we have for this trig function. So what we can do is we can try to graph this. Well, since we're dealing with a sign function and what I'm first going to do to graph this is actually I'm going to ignore the H value that we have in here. So I'm going to see what I can do to just draw a graph for Y is equal to three times the sine of X. And on our graph over here, that's going to be a standard sine graph with an amplitude that peaks at three on the Y axis and valleys down at negative three on the Y axis. So we'll start here at the center like we do for a sine graph and we're going to reach a peak at pi over two, which will be up here at a Y value of three. We're going to to cross through PI and then we're going to reach a valley at three pi over two, which will be at A Y value of negative three. This is going to be the valley and then this graph will keep waving and this graph is gonna keep going to the left as well. So we're going to see a valley here at negative pi over two, we're then going to cross through negative pi and then we're going to see a peak at negative three pi over two. And then this graph is gonna keep wing. So this is what the graph for three times the sin of X would look like. Now what I need to do from here is incorporate the phase shift because we can see that because we have this H value here, it's gonna be some kind of phase shift. But the way that I can find that phase shift and I should actually say the H value is negative pi and the way that I can find that phase shift is by recognizing the phase shift is going to be H over B units. And since I can see we have a negative H value, it's gonna be H over B units to the left. So what I can do is take our H value to find H over B, our H value is negative pi. But I'm just gonna write pi here because we already have this going to the left. And that's going to be divided by B which is one and pi over one is just pi. So we can say that this graph is going to shift pi units to the left. So this unit over here at PI is going to go all the way over here. So to rewrite this graph will start here at the center. But this whole portion of the graph is going to be moved over. So we're actually going to see a peak at negative pi over two which reaches three. Then we're going to see that this graph crosses through negative pi and we reach a valley down at negative three pi over two and we keep waving to the left and then this graph is going to dip down here in valley at pi over two. Then we're going to cross through pi and then reach a peak as we get 23 pi over two. And then this graph is going to keep waving as we go to the right. So this right here is what the graph would look like for three times the sine of X plus pi. This is what our graph would look like and that would be the solution to this problem. So hope you found this video helpful Thanks for watching.