Graphs of the Sine and Cosine Functions - Online Tutor, Practice Problems & Exam Prep

Hey everyone. So at this point in the course, we should be very familiar with the sine and cosine trigonometric functions, and we've recently talked about how these trig functions relate to the unit circle. Now, what we're going to be talking about in this video is how you can take the sine and cosine functions and graph them on a standard xy graph. Now, this might sound a bit intimidating since we don't know at all what these graphs are going to look like, but don't sweat it because it turns out what we're going to learn in this video is that using the values that we're familiar with on the unit circle, we can actually graph these in a very straightforward way using these values. Without further ado, let's get right into things. Now I want you to recall how the sine is associated with the y coordinates when you look at a unit circle, and the cosine is associated with the x coordinates. Now what you'll see when looking at these values is these values will repeat as you go around the unit circle. So let's say for example we're looking at the sine. We can see that our sine value starts at 0, and then it goes to 1, and then to 0, then to negative one, then it starts repeating. So it goes back to 0, then to 1, then to 0, and it just repeats as you go around. So you'll notice for the sine and cosine, these values repeat around the unit circle. And because of this, you're going to see some interesting behavior in their graphs.

Let's say we're looking at the sine graph where y is equal to the sine of x. We can figure out what this graph looks like by plotting some points. So let's say we want to find the sine of 0. Well, if we're looking at an x value of 0, the sine of 0 based on our unit circle is just going to be 0, because it's going to be this y coordinate. So that means we're going to be at a value of 0 starting right there on our graph. Now let's say we're looking at the sine of π/2. Well, the sine of π/2 according to our unit circle is 1, meaning we're going to be up here at π/2. Now let's say we're looking at the sine of π. Well according to our unit circle the sine of π is 0, meaning we're back down there on our graph. And based on the unit circle, if we go to 3π/2, the sine of 3π/2 is negative 1. Meaning we're going to be down there on our graph. You can keep plotting these points. So the sine of 2π would be a full trip around the circle right back to 0, and the sine of 5π/2 would be a full trip around the circle, and then all the way back up to 1. Meaning that your graph is going to look something like this. And if I go ahead and connect these points with the smooth curve, you'll notice something kinda interesting happens in our graph. We get this kind of wave that shows up, and that's what happens when you have these repeating values. So both the sine and cosine are going to be graphs that look like waves.

Now to prove this to ourselves, let's take a look at the cosine graph as well. So the cosine is going to be associated with the x coordinates, and if we look at the cosine of 0, we can see that that's right here at 1. For the cosine of π/2, that's going to be at 0. For the cosine of π, that's going to be there at negative 1. For the cosine of 3π/2, that's going to be back at 0, and then you can keep plotting these points. So for 2π, we're going to be back up at 1. And then for cosine of 5π/2, we're going to be back at 0, meaning your graph is going to look something like this. Now notice how we get the same kind of wave pattern that we got for the sine, except this graph starts a bit differently. Notice how our graph starts at an output of 1 for the cosine, and our graph starts at an output of 0 for the sine. But if you look at the unit circle, this should make perfect sense because we can see at 0 radians, our cosine starts at 1 and our sine starts at 0. So that's why we have the graphs that look like this, and then show this waving pattern based on the repeating values that we saw.

Now, some other terminology you should be familiar with are the high points and low points of our wave. The high points are known as crests, or they're also called peaks. So the peaks of the wave are going to be all these places that we see our high values at, our highest values on the wave. And another thing you should be familiar with is the low points, which are known as troughs or valleys. The valleys are going to be these lowest points on each of the waves. Now it turns out these peaks and valleys are not always going to be at positive one and negative one for the sine and cosine, because recall that we learned back in algebra that you can actually have transformations on your function. And it turns out that the sine and cosine graphs can also go through transformations. Now one of the simplest transformations we learned about was something called the vertical shift. And this occurs when you go ahead and take a function and you add some constant k to it, which will cause it to vertically shift up or down. Well, it turns out for these sine and cosine graphs, you could also add a constant k, which vertically shifts the function. So adding a constant to the sine or cosine will shift the wave up or down in a certain way. Now if the k value that you have is positive, the graph is going to shift up, whereas if the k value is negative, the graph will shift down.

Now to understand this a bit better, let's actually try an example where we put both these concepts together. So here we have a situation where we are asked to graph the function y is equal to the sine of x plus 1. Notice we have a sine function we're dealing with, and it's been shifted in some kind of way. Now to solve this problem, what I'm first going to do is try graphing the function y is equal to the sine of x. So I'm going to go ahead and ignore this plus one for now. Now recall for the sine of x, we start at a value of 0, at an output of 0, and then we reach a peak at π/2, then we go back to 0 through π, and then we reach a valley at 3π/2, then we go back to 0 over 2π, and we reach another peak at 5π/2. But how exactly could we graph the function y is equal to the sine of x plus 1? Well, this is what we're ultimately looking for in this problem. And because we have a plus 1, this tells us that we're going to have some kind of vertical shift. It's a positive number, meaning it's going to shift up. So all we need to do is take this entire graph and shift it up by 1. So this point is going to be at an output of 1. This peak would be at an output of 2, and then this point would be at an output of 1, and then this valley at negative one would now be at an output of 0. So what we're going to do is trace a curve that looks like this. We'll have a peak at an output of 2, then we'll go down to 1, and then we'll reach our valley at an output of 0, then we'll go back up to an output of 1, and then we'll reach a peak at an output of 2. So this would be the graph for y equals the sine of x plus 1, and that is the answer to this example. So hopefully, this gave you a good understanding of how to graph sine and cosine functions and deal with any kind of vertical shifts if you have them. So I hope you found this video helpful. Thanks for watching.

Hey, everyone. Let's see how we can solve this example. So in this problem, we're asked to graph the function \( y \) is equal to the sine of \( x \) plus 3 on the graph below. Now in order to solve this problem, I always like to start off by graphing what I'm already familiar with. And something that I'm familiar with is what the graph \( y \) is equal to the sine of \( x \) is going to look like. So I'm going to go ahead and ignore this plus 3 for now and just graph this sine of \( x \), because we can incorporate this plus 3 later.

So recall that the sine of \( x \) graph is a value that starts at the center, and what you can do is going to the right; this is going to be a wave. Now our wave is going to reach a peak at \( \frac{\pi}{2} \), and then we're going to cross down here through \( \pi \), and we're going to reach a valley at \( \frac{3\pi}{2} \), and keep waving as we go to the right. Now notice that the peak is at positive one, and the valley is at negative one for the 2 outputs on the \( y \)-axis.

Now we can continue this wave going back on our graph as well. So going to reach a valley as we get to negative \( \frac{\pi}{2} \), and we're going to come and cross through negative \( \pi \). We're going to reach a peak at \( \frac{3\pi}{2} \), and then this graph will keep waving. So this is what the sine of \( x \) graph is going to look like.

Now in order to graph this function, which is \( y \) is equal to the sine of \( x \) plus 3, what I can recognize is that we have a number being added to our sine of \( x \). This is going to cause a shift. And because we have a positive number, this graph is going to shift up by 3 units. So what that means is I can take every point and shift it up by 3. So the point that we had that started at the center is now going to be up here at a value of 3. That's gonna be our coordinate there. So we're gonna be 3 units up for where our graph starts, then this peak here is going to be 3 units up as well, which is gonna be at positive 4. This 0 point that we have at \( \pi \), this is going to be 3 units up, so it's going to also be at 3, and then this valley that we have at negative one is going to be up 3 units at positive 2.

So that's what the graph is going to look like as we go to the right, and this will also be the same as we go to the left. So for negative \( \frac{\pi}{2} \), we're also going to be at positive 2 for our output, and negative \( \pi \) we're at an output of 0, so our output is going to be up here at 3, And then at negative \( \frac{3\pi}{2} \), we can see their output is 1, so that means our output will be up here at 4. So connecting these points, the graph is going to look something like this, where we have this wave-like behavior that goes to the left and right side of this graph.

So this is our sketch for the function \( y \) is equal to the sine of \( x \) plus 3, and that is the solution to this problem. I hope you found this video helpful. Thanks for watching.

Welcome back, everyone. So in the last video, we got introduced to the sine and cosine functions as graphs. And we also saw ways that these functions could be transformed by a vertical shift. Now, what we're going to be talking about in this video is new ways to transform the sine and cosine function by using the amplitude or reflection. Now, this might sound a bit complicated, but don't sweat it because what we're going to learn in this video is that these types of transformations you can do to the sine and cosine are actually very similar to transformations we've already learned about for functions back in algebra. So, you're going to find, I think, that a lot of these graphs are actually very straightforward. So without further ado, let's get right into this.

Now when it comes to the amplitude, the amplitude is a number, and it's a number that affects how tall the peaks of your graph are. Now I want you to recall that the sine and cosine function are both repeating waves. And if you find the distance from the midline to the peaks or valleys of that wave, that is going to tell you the amplitude. So, let's say, for example, that we're looking at a sine graph, and we take a midline and we draw it right through the middle of this graph. Well, if we draw a midline horizontally through this graph, we can find the amplitude by looking at the distance from this midline to either a peak or a valley to this wave. And we can see that that distance is 1 unit, meaning the amplitude of this sine function is 1.

But what would happen if we had a different amplitude, or how could we have a different amplitude? Well, let's take a look at a different situation where rather than having a 1 in front of our sine function, we have 2. So to figure out what this looks like I can plot some points on this graph by taking all the outputs of the sine and multiplying them by 2. So, we would have 2 times 0, which is 0 meaning we would start here on our graph. Now, when we get to π/2 on the x-axis we would have 2 times 1, which is 2. When we get to π on the x-axis, we'd have 2 times 0, which is again just 0. Then we would have 2 times -1, which is -2. Then we would have 2 times 0, which is 0. So if you go ahead and draw this graph by connecting these points with a smooth curve, you're going to get a function that looks something like this.

And I want you to notice what this function looks like for 2 times the sine of x. Notice it's very similar to what we had for the sine of x, except it appears that our graph has been vertically stretched in some kind of way, like it's taller. Well, that's the idea of changing the amplitude. It's literally just like a vertical stretch that we learned about back in algebra. And just like we had a number in front of the function, we now have a number in front of the sine. So if you want to change the amplitude of your graph, you just need to change this number in front of the sine or cosine function, and that's really the main idea of the amplitude. We can see that we multiply the sine function by 2, so our amplitude for this function was 2. And likewise, when we just had a 1 in front of the sine function, our amplitude was 1.

Now, another thing that you'll frequently see in this course is situations where you have a negative amplitude. And when the amplitude is negative, you're going to get a graph that is reflected over the x-axis. So to see what this looks like, well, let's plot another graph. So let's say that we have negative sine of x. Well, if we want to find what this looks like, we can take all the output for the sign of x and make them negative. Now negative 0 would just be 0, negative one would look something like that. We then have negative 0, once again, which is just 0. And if we had a negative negative one, that's the same thing as canceling the negative signs giving you a positive one. And then you'd have negative one times 0, which is again just 0 on the graph, meaning that it's going to look something like this when you connect these points with a smooth curve.

Now notice for the negative sine graph, it literally looks like we took the sine function and just flipped it upside down over the x-axis, and that's exactly what happened. And you may recall back in algebra that a negative number in front of the function would flip the function upside down, and that's exactly what happens with the sine and cosine as well. So this is really the main idea behind changing the amplitude or reflecting your graph over the x-axis.

Now to make sure that we're really understanding this, let's actually try an example where we have both a change in the amplitude and a reflection. So here we're asked to graph the function y is equal to negative three halves times the cosine of x. Now, this function right here could also be written as y equals negative 1.5 times the cosine of x. 1.5 is the same thing as 3 halves. So basically, our amplitude is 1 and a half. And if I go to our graph, 1.5 would be right between 1 and 2 here, and then we would have negative 1.5, which is right down there between negative one and negative 2.

Now recall, for the cosine function, the cosine typically starts at a high value, then goes through 3π/2, reaches a valley, then goes back up through 3π/2, reaches a peak, and then comes back down through 5π/2 and keeps waving like that. But recall that we have a negative sign in front of this function. So what that's going to do is take our little cosine graph, and it's going to flip it over the x-axis. So what we're actually going to do for the cosine is we're going to start at a low value, then we're going to go up through π/2, and we're going to reach a peak at π. Then we're going to go down through 3π/2, reach a valley at 2π. Then we're going to come back up here and touch 5&...

Hey everyone. So in this example, we're asked to graph the function, \( y = 2 \cdot \sin(x) - 1 \). Now whenever I'm dealing with these types of problems, I like to start off by solving what I'm familiar with and graphing it, and then building off the graph from there. So what I'm going to do in this problem is I'm going to start by graphing \( y = 2 \cdot \sin(x) \). We're going to ignore the negative one for now, but we'll get to that in a moment. To graph this function, we'll consider the sine of x. What we do is we start here at our graph. We reach a peak at an output of 1, then we dip through \( \pi \), and this pattern continues. However, we can't keep the same pattern for this graph, because we have a 2 here, and 2 changes the amplitude. So we're still going to start at an output of 0, but now our peaks will reach an output of 2, and our valleys will reach an output of -2. Our graph is going to look something like this:

We'll start here at the center of our graph, then we'll reach a peak when we get up here to \( \frac{\pi}{2} \). Our peak is right about there, and then we're going to dip through \( \pi \) on our x-axis, go to \( \frac{3\pi}{2} \), reach a valley, and continue waving to the right. We can also extend this graph to the left. We'll reach a valley at \( -\frac{\pi}{2} \), a peak as we get to \( -\frac{3\pi}{2} \), and continue waving to the left. This is what our sine graph will look like for \( y = 2 \cdot \sin(x) \).

Now, to graph \( y = 2 \cdot \sin(x) - 1 \), this minus one will cause a vertical shift. Since it's a negative value, we'll have a vertical shift down by 1 unit. When we shift down, all of these points will be 1 unit lower. This means this peak right here will move 1 unit down, this valley will move 1 unit down, this peak and this valley will also move 1 unit down. The center, where we started, will actually start at an output of negative one.

So, we can adjust where the peaks and the valleys are going to be, recognizing that the peaks of this graph are actually going to be at positive one when we shift one unit down, and the valleys of this graph are going to be down here at negative 3. Let's go ahead and draw this curve. We'll start here at negative 1, and then we're going to reach our peak right about there when we get to \( \frac{\pi}{2} \). Then we'll cross back down to negative 1 when we get to \( \pi \) on the x-axis, and reach our valley at \( \frac{3\pi}{2} \), which will be at an output of negative 3, and then we're going to keep waving to the right. Likewise, to the left, we'll go down here and reach a valley at \( -\frac{\pi}{2} \) on the x-axis. We'll go back up when we get to \( -\pi \), reach our peak when we get to \( -\frac{3\pi}{2} \), and keep waving as we go to the left. So this is what the graph will look like for \( y = 2 \cdot \sin(x) - 1 \). That concludes this explanation for the graph of this function. Thanks for watching.

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.