Phase Shifts - Online Tutor, Practice Problems & Exam Prep

Phase shifts in sine and cosine functions represent horizontal shifts of the graph. A phase shift occurs when a number is added or subtracted inside the function, affecting the graph's position. For example, in $\cos (x-\frac{\pi }{2})$, the graph shifts right by $\frac{\pi }{2}$ units. Understanding the relationship between the coefficients and shifts is crucial for graphing these functions accurately.

Phase Shifts

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 1, and then kind of makes this wave-like 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. An example of this would be, say, the cosine of \( x - \frac{\pi}{2} \). Now to see how this phase shift affects the graph, we can go ahead and plot some points. And the way that I'm going to plot these points is I'm going to take all of these values, and I'm going to subtract \( \frac{\pi}{2} \) from them, and then I'll evaluate the cosine. So the cosine of \( 0 - \frac{\pi}{2} \) is going to be the cosine of \( -\frac{\pi}{2} \). And we can see that our cosine graph is 0 at that point, so that means we're going to start at a value of 0.

Next we're going to have the cosine of \( \frac{\pi}{2} - \frac{\pi}{2} \). That's going to be the cosine of 0, which is just going to be 1. Next we'll have the cosine of \( \pi - \frac{\pi}{2} \). That's going to be the cosine of \( \frac{\pi}{2} \), which is back at 0. Now for \( \frac{3\pi}{2} - \frac{\pi}{2} \), 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 \( 2\pi - \frac{\pi}{2} \), which turned out to just be 0. So you're going to be back up here. Now if you go ahead and connect these points with a smooth curve, you're going to notice something kind of interesting about this graph. 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.

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 sine 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 has 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 \( b\cdot x - \text{some number} \), then that means that your \( h \) value is positive, and your graph is going to shift to the right by \( \frac{h}{b} \) units.

Now if you instead had a situation where the inside of your trig function was \( b\cdot x + \text{some number} \), that means your \( h \) value is negative, and the graph is going to shift left by \( \frac{h}{b} \) units. Now in the situation we had right here, we had the cosine of \( 1\cdot x - \frac{\pi}{2} \). And because the inside of this function is in the form \( b\cdot x - \text{a number} \), that means that our \( h \) value was positive and our graph shifted to the right. Now finding \( \frac{h}{b} \) for this graph would be our \( h \) value, which is \( \frac{\pi}{2} \), divided by our \( b \) value, which is 1, and \( \frac{\pi}{2} \) divided by 1 is the same thing as just \( \frac{\pi}{2} \). 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 0, and then we go up and down and complete a period at \( 2\pi \). This is actually very similar to the sine graph. In fact, it's the same thing. And this is the interesting thing 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 that this concept makes sense, let's try an example where we have to deal with a sine 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 \( 2x + \frac{\pi}{1} \) 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 \( 2x \). Now recall that this graph is going to have a bit of a different period than we're used to because we have this 2 in front. And the period for a sine or cosine function is equal to \( \frac{2\pi}{b} \). Well the \( b \) value we can see in front of the \( x \) is 2, 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 0 and \( \frac{\pi}{2} \). We're going to dip here through \( \pi \), and then 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 \( 2x + \pi \). And to figure this out, well, I can see that there's some kind of phase shift that happened. And recall that when you have a situation where the inside of your function is \( b\cdot x + \text{some number} \), that means the \( h \) value is negative and will shift to the left by \( \frac{h}{b} \) units. Now what is \( \frac{h}{b} \)? Well, \( \frac{h}{b} \) is going to be this \( h \) value which is \( \pi \) divided by this \( b \) value which is 2. So this graph is going to shift to the left by \( \frac{\pi}{2} \) units. So that means this point is actually going to start over here, meaning our graph is going to actually dip down first, and then we're going to cross through \( \frac{\pi}{2} \), then we're going to go up and then back down when we get to \( \pi \). So this is what the graph is going to look like over one full period, and that would be the solution to this problem.

If this seems 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 \( 2x + \pi \), could also be written as \( y \) is equal to \( 2 \cdot (x + \frac{\pi}{2}) \). So notice how this 2 took our graph and changed the period of it, and then this portion of the function shifted our graph to the left by \( \frac{\pi}{2} \) 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 phase 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.

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

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

Example 1

Hey, everyone. So let's give this problem a try. Here we're asked to graph the function y=3∗sin⁡(x+π). Okay. So how can we solve this problem? Well, what I'm first going to do is write out the most general form for the sin function, which is y=a∗sin⁡(bx-h+k). So this is our most general form. Now something that I notice is that we don't have a 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 3. Now what I also notice is for b x minus h, what we have in the parenthesis, we don't have a specific b value out front here, so we could just say that b is equal to 1. Right? And you could also say that, like, k is equal to 0, 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 number is 1 because one times x would just be x. Now I also notice that we have a π here, but notice that we have a plus π. 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=3∗sin⁡(x-(-π). 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 -π. 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 sine 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=3∗sin⁡(x). And on our graph over here, that's going to be a standard sine graph with an amplitude that peaks at 3 on the y-axis and valleys down at negative 3 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 π over 2, which will be up here at a y value of 3. We're going to cross through π, and then we're going to reach a valley at 3π over 2, which will be at a y value of negative 3. This is going to be the valley, and then this graph will keep waving. And this graph is going to keep going to the left as well. So we're going to see a valley here at negative π over 2, we're then going to cross through negative π, and then we're going to see a peak at negative 3π over 2, and then this graph is gonna keep waving. So this is what the graph for 3 times the sine 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, there's going to be some kind of phase shift. But 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 going to 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 π, but I'm just going to write π here because we already have this going to the left, And that's going to be divided by b, which is 1. And π over 1 is just π. So we can see that this graph is going to shift π units to the left. So this unit over here, π, is going to go all the way over here. So to rewrite this graph we'll 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 π over 2, which reaches 3, then we're going to see that this graph crosses through negative π, and we reach a valley down

Here’s what students ask on this topic:

What is a phase shift in trigonometric functions?

A phase shift in trigonometric functions, such as sine and cosine, refers to a horizontal shift of the graph. This occurs when a number is added or subtracted inside the function's argument. For example, in the function $\cos (x-\frac{\pi }{2})$, the graph shifts to the right by $\frac{\pi }{2}$ units. Understanding phase shifts is crucial for accurately graphing these functions.

How do you determine the direction of a phase shift?

The direction of a phase shift is determined by the sign of the number added or subtracted inside the trigonometric function. If the function is of the form $f(x-h)$, the graph shifts to the right by $h$ units. If the function is of the form $f(x+h)$, the graph shifts to the left by $h$ units.

How do you calculate the phase shift of a sine or cosine function?

To calculate the phase shift of a sine or cosine function, identify the form $f(bx-h)$ or $f(bx+h)$. The phase shift is given by $\frac{h}{b}$. If $h$ is positive, the graph shifts right; if negative, it shifts left.

What is the phase shift of the function y = sin(2x + π)?

For the function $y=\sin (2x+\pi )$, the phase shift can be calculated by rewriting it in the form $\sin (2x+\pi )$. Here, $h=\pi$ and $b=2$. The phase shift is $\frac{\pi }{2}$ units to the left.

How does a phase shift affect the period of a sine or cosine function?

A phase shift does not affect the period of a sine or cosine function. The period is determined by the coefficient $b$ in the function $f(bx-h)$. The period is calculated as $\frac{2}{\pi }/b$. The phase shift only moves the graph horizontally without changing its period.