Hey, everyone. So up to this point, we've been talking a lot about the sine and cosine functions, as well as what their graphs look like. And we've been talking about various ways that these graphs can be transformed. Recall in a previous video, we talked about the amplitude, which is a number that tells you how tall or short your graph is. Now, what we're going to be talking about in this video is the period. And this might sound a bit complicated just because we haven't really heard of the period before, but it turns out, just like how the amplitude tells you how tall your graph is, the period tells you how wide your graph is. And I think you'll find the problems you come across in this course, what you learn about the period is actually quite straightforward. So without further ado, let's get right into this.
Now we should be familiar with the graph for the sine of \( x \). The sine of \( x \) creates this graph, which has this kind of wave pattern. And what we can do to find the period of this graph is to figure out the distance along the x-axis that it takes to complete a full wave or cycle. So if we look, we can see that one full cycle has been completed. We go up and then down and then back up again. And to find the distance along the x-axis, we can just start at the origin of this wave. We can travel along the x-axis until we reach the end of the cycle. And as you can see, this is going to be at \( 2\pi \), meaning that our period is \( 2\pi \) for this graph.
Now the question becomes, what happens if we had a different graph? Let's say we had the sine of \( 2x \). Well, let's figure out what this graph is going to look like and see if it has the same period. So I'm going to plot some points on this graph, and I'm going to do this by taking all the inputs, the x values, and multiplying them by 2. So we'll have the sine of \( 2 \times 0 \), which is just the sine of 0 or 0. We'll then have the sine of \( 2 \times \frac{\pi}{4} \), and \( 2 \times \frac{\pi}{4} \) is \( \frac{\pi}{2} \). And the sine of \( \frac{\pi}{2} \) is 1, giving us a point right there. Now we'll have the sine of \( 2 \times \frac{\pi}{2} \), which comes out to the sine of \( \pi \), which is just 0, so we're going to have another point right there. We'll have the sine of \( 2 \times \frac{3\pi}{4} \), which actually turns out to be negative one. And then we'll have the sine of \( 2 \times \pi \), which the sine of \( 2\pi \) is just 0. So that means that your graph is going to look something like this.
Now I want you to notice something about this graph. Notice how we got a very similar wave pattern that we got for the other graph. But for this function, it looks like our graph has been horizontally shrunk and that's the entire idea of changing the period because what happened is this graph has a shorter period. Notice that rather than going all the way over to \( 2\pi \) like we did for the original function, the sine of \( x \), we now only go to just \( \pi \) for the sine of \( 2x \). So we could say that the period for this graph is \( \pi \). So notice how having a number inside of the sine function in front of the \( x \) changed the period of our graph. And this number that modifies the period, we typically will represent this with the letter \( b \). So, if you have this number inside the sine or cosine function, it's going to modify or change the period depending on what that number is. Now, if \( b \) is a value greater than 1, like we saw in this example, we had 2, then the graph is going to horizontally shrink, and we saw that this graph shrank. But if \( b \) is a value between 0 and 1, then the graph is actually going to horizontally stretch.
And if you want a straightforward way to calculate how much your graph shrinks or stretches when you have a sine or cosine function, what you can do is use this equation for the period. The period is equal to \( \frac{2\pi }{b} \). So if we wanted to use this equation for the sine of \( x \), what we can see is that the sine of \( x \) does not have a number in front of the \( x \). So that means that \( b \) would just be 1, this value here. So what our period would be is \( \frac{2\pi}{b} \), which is 1. And \( \frac{2\pi}{1} \) is just \( 2\pi \), which is what we figured out. Now we can also use this equation on the other situation we had where we had the sine of \( 2x \). Now recall that the \( b \) is going to be the value in front of the \( x \), which is 2. So if we wanted to calculate the period, we'd have \( \frac{2\pi }{b} \), where \( b \) is 2. The 2's here are going to cancel, leaving us with just \( \pi \). So our period is \( \pi \), which is what we had for this situation.
So notice how this confirms how wide both of our graphs are. We can see that the period for the sine of \( 2x \) is all the way over to \( \pi \), and we can see the period for the sine of \( x \) goes all the way over to \( 2\pi \). Now to make sure we're fully understanding this concept, let's see if we can try an example. And in this example, we're asked without graphing to calculate the period of the following functions. Well, notice in both these examples we're dealing with sines and cosines, and so we know that the period is going to be \( \frac{2\pi}{b} \). Now we'll go ahead and start with example a. And for example a, I can use this equation to find the period. So the period is going to be \( \frac{2\pi}{b} \). Now recall that \( b \) is the number that's in front of the \( x \) inside of the trig function. So \( b \) is going to be \( \frac{1}{2} \). Now we have \( \frac{2\pi }{\frac{1}{2}} \), and what I can do is take this one-half of the denominator, and I can flip it and bring it to the numerator. So you bring this 2 up to the top, which is going to give me \( 2 \times 2\pi \) divided by 1. Now anything divided by 1 is just going to be that number, and \( 2 \times 2 \) is 4. So our period is going to be \( 4\pi \), and this right here is the answer for example a. But what about for example b? Well, we can use the same equation. So the period is going to be \( \frac{2\pi }{b} \). Now we have the cosine of \( 4\pi x \) and we're going to have our \( b \) value be everything that's in front of the \( x \), so in this case, it's going to be \( 4\pi \). Now 4 is the same thing as \( 2 \times 2 \), and then we'll still have \( \pi \). And the reason I'm writing it like this, is notice how one of the twos are going to cancel, as well as the \( \pi \)s are going to cancel. So all we're going to be left with for the period is \( \frac{1}{2} \). And so that means that the period, for example, b, is equal to \( \frac{1}{2} \). So this is how you can solve problems and deal with situations where the period changes for the sine and cosine. I hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.
