Graphs of Sine & Cosine: Videos & Practice Problems

Understanding the sine and cosine functions involves recognizing their wave patterns, which can be transformed through vertical shifts, amplitude changes, reflections, and phase shifts. The amplitude indicates the height of the peaks, while the period defines the width of the wave, calculated as $P=\frac{2}{b}$. A phase shift occurs when a constant is added or subtracted within the function, shifting the graph horizontally. Mastering these concepts is essential for graphing and analyzing trigonometric functions effectively.

Graph of Sine and Cosine Function

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 will 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 x y 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. So 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 zero, and then it goes to one, and then to zero, then to negative one, and then it starts repeating. So it goes back to zero, then to one, then to zero, 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=sin⁡(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 zero. Well, if we're looking at an x value of zero, the sine of zero based on our unit circle is just going to be zero, because it's going to be this y coordinate. So that means we're going to be at a value of zero starting right there on our graph. Now let's say we're looking at the sine of π2. Well, the sine of π2 according to our unit circle is one, meaning we're going to be up here at π2. Now let's say we're looking at the sine of π.

Well according to our unit circle the sine of π is zero, meaning we're back down there on our graph. And based on the unit circle, if we go to 3π2, the sine of 3π2 is negative one. 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 zero, and then the sine of 5π2 would be a full trip around the circle, and then all the way back up to one, 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 kind of 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. The cosine is going to be associated with the x coordinates. If we look at the cosine of zero, we can see that that's right here at one. For the cosine of π2, that's going to be at zero. For the cosine of π, that's going to be there at negative one. For the cosine of 3π2, that's going to be back at zero. And then you can keep plotting these points.

So for 2π, we're going to be back up at one. And then for cosine of 5π2, we're going to be back at zero. 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 one for the cosine, and our graph starts at an output of zero for the sine.

But if you look at the unit circle, this should make perfect sense because we can see at zero radians, our cosine starts at one and our sine starts at zero. 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 a 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=sin⁡(x)+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=sin⁡(x). So I'm gonna go ahead and ignore this plus one for now. Now recall for the sine of x, we start at a value of zero, at an output of zero. And then we reach a peak at π2, then we go back to zero through π, and then we reach a valley at 3π2, then we go back to zero for 2π, and we reach another peak at 5π2. But how exactly could we graph the function y=sin⁡(x)+1?

Well, this is what we're ultimately looking for in this problem. And because we have a plus one, this tells us that we're gonna 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 one. So this point is going to be at an output of one. This peak would be at an output of two, and then this point would be at an output of one, and then this valley at negative one would now be at an output of zero. So what we're going to do is trace a curve that looks like this. We'll have a peak at an output of two, then we'll go down to one, and then we'll reach our valley at an output of zero, then we'll go back up to an output of one, and then we'll reach a peak at an output of two. So this would be the graph for y=sin⁡(x)+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.

Sketch the function $y=\cos\left(x\right)-1$ on the graph below.

• A.
• B.
• C.
• D.

Answer: (B)

Determine the value of $y=\sin\left(-\frac{\pi}{2}\right)+50$ without using a calculator or the unit circle.

• A. $y=50$
• B. $y=51$
• C. $y=49$
• D. $y=50+\frac{\sqrt3}{2}$

Answer: (C)

Graph of Sine and Cosine Function Example 1

Hey everyone. Let's see how we can solve this example. So in this problem, we're asked to graph the function \( y = \sin(x) + 3 \) 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 \( y = \sin(x) \) is going to look like.

So I'm going to go ahead and ignore this plus three for now and just graph the sine of x, because we can incorporate this plus three later. So recall that the sine of x graph is a value that starts at the center, and what you can do is go to the right; this is going to be a wave. Now our wave is going to reach a peak at \( \pi/2 \), and then we're going to cross down here through \( \pi \), and then we're going to reach a valley at \( 3\pi/2 \) 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 two outputs on the y-axis. Now we can continue this wave going back on our graph as well.

So we're going to reach a valley as we get to \( -\pi/2 \), and we're going to come and cross through \( -\pi \), then we're gonna reach a peak at \( 3\pi/2 \), 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 \( y = \sin(x) + 3 \), 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 three units.

So what that means is I can take every point and shift it up by three. So the point that we had that started at the center is now going to be up here at a value of zero three. That's going to be our coordinate there. So we're going to be three units up for where our graph starts. Then this peak here is going to be three units up as well, which is going to be at positive four.

This zero point that we have at \( \pi \), this is going to be three units up so it's going to also be at three. And then this valley that we have at negative one is going to be up three units at positive two. 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 \( -\pi/2 \), we're also going to be at positive two for our output.

At \( -\pi \), we're at an output of zero, so our output is going to be up here at three. And then at \( -3\pi/2 \), we can see their output is one, so that means our output will be up here at four. 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 \( y = \sin(x) + 3 \), and that is the solution to this problem. Hope you found this video helpful.

Thanks for watching.

Amplitude and Reflection of Sine and Cosine

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. We can see that that distance is one unit, meaning the amplitude of this sine function is one. 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 one in front of our sine function, we have two. 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 two.

So we would have two times zero, which is zero, meaning we would start here on our graph. Now when we get to π/2 on the x axis, we'd have two times one, which is two. When we get to π on the x axis, we'd have two times zero, which is again just zero. Then we would have two times negative one, which is negative two. Then we'd have two times zero, which is zero.

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 two 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 two, so our amplitude for this function was two. And likewise, when we just had a one in front of the sine function, our amplitude was one. 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. If we want to find what this looks like, we can take all the outputs for the sine of x and make them negative. Now negative zero would just be zero.

Negative one would look something like that. We then have negative zero, once again, which is just zero. 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 zero, which is again just zero 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 sign 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 three halves. So basically, our amplitude is one and a half. And if I go to our graph, 1.5 would be right between one and two here, and then we would have negative 1.5, which is right down there between negative one and negative two. Now recall for the cosine function, the cosine typically starts at a high value, then goes through π/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, and we're going to reach a peak at 3π/2. Then we're going to go down through 2π, reach a valley at 5π/2. Then we're going to come back up here and touch π on the x axis, and the graph is going to keep waving like that. So this is what the graph will look like, and that's the answer to this problem.

So hopefully, this gives you a better understanding of how to deal with changes in the amplitude or a reflection for the sine or cosine functions. Hope you found this video helpful. Thanks for watching.

Determine the value of $y=-2\cdot\sin\left(-\frac{3\pi}{2}\right)+10$ without using a calculator or the unit circle.

• A. $y=8$
• B. $y=10$
• C. $y=-2$
• D. $y=12$

Answer: (A)

Graph the function $y=-3\cdot\cos\left(x\right)$.

• A.
• B.
• C.
• D.

Answer: (B)

Amplitude and Reflection of Sine and Cosine Example 2

Hey everyone. So in this example, we are asked to graph the function, y=2×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×sin⁡(x). So we're going to ignore this negative one for now, but we'll get to that in a moment.

To graph this function, we'll consider sin⁡(x). We start at our graph's center. We reach a peak at an output of one, then we dip through π, and then this keeps waving. However, we have a multiplier of two here, which is going to change the amplitude. So we still start at an output of zero, but now our peaks are going to reach an output of two, and our valleys are going to reach an output of negative two.

Our graph will start at the center of our graph, then we're going to reach a peak at π/2. So we reach our peak right about there, then we're going to dip through π on our x-axis, then we're going to 3π/2, we'll reach a valley, and then we'll continue waving to the right. Now, we can also extend this graph to the left. We're going to reach a valley at −π/2, we're going to reach a peak as we go to −3π/2, and we're going to keep waving to the left.

This is what our sine graph is going to look like with 2×sin⁡(x). Now to graph y=2×sin⁡(x)−1, this minus one is going to cause a vertical shift. Since it's a negative value, we're going to have a vertical shift down by one unit. When we shift down, all of these points are going to move down by one unit. So, the peak here will go one unit down.

This valley will go one unit down. This peak will go one unit down, and this valley will go one unit down. And then the center here, where we started, is going to start at an output of negative one. 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 down by one unit, and the valleys of this graph are going to be at negative three. To draw this curve, we'll start here at negative one, and then we'll reach our peak right about there when we get to π/2.

Then we'll cross back down to negative one when we get to π on the x-axis, and then we're going to reach our valley at 3π/2, which is going to be at an output of negative three, and then we're going to keep waving to the right. Likewise, to the left, we're going to go down here and reach a valley at −π/2 on the x-axis. We're going to go back up to negative pi, and we're going to reach our peak when we get to −3π/2, and we're going to keep waving as we go to the left. So, this right here is what the graph is going to look like for y=2×sin⁡(x−1). It's going to be this red graph right here, and that is the solution to this problem.

Hope you found this video helpful. Thanks for watching.

Period of Sine and Cosine Functions

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 once 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 till we reach the end of the cycle.

And as you can see, this is going to be at two pi, meaning that our period is two pi for this graph. Now the question becomes, what happens if we have a different graph? Let's say we had the sine of two x. 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 two.

So we'll have the sine of two times zero, which is just the sine of zero or zero. We'll then have the sine of two times pi over four, and two times pi over four is pi over two. And the sine of pi over two is one, giving us a point right there. Now we'll have the sine of two times pi over two, which comes out to the sine of pi, which is just zero. So we're going to have another point right there.

Well, the sine of two times three pi over four, which actually turns out to be negative one when you do that. And then we'll have the sine of two times pi, which the sine of two pi is just zero. 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 two pi, like we did for this original function, the sine of x, we now only go to just pi for the sine of two x. 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 one, like we saw in this example, we had two, then the graph is going to horizontally shrink, and we saw that this graph shrank. But if b is a value between zero and one, 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 2πb.

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 one, this value here. So what our period would be is 2π1, which is just two 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 two x.

Now recall that the b is going to be the value in front of the x, which is two. So if we wanted to calculate the period, we'd have 2π2. The twos 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 two x is all the way over to pi, and we can see the period for the sine of x goes all the way over to two 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 2πb.

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 2πb. 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 one half.

Now we have 2π1/2, 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 two up to the top, which is going to give me two times two pi divided by one. Now anything divided by one is just going to be that number, and two times two is four. So our period is going to be four 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 2πb. Now we have the cosine of four 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 four pi.

Now four is the same thing as two times two, 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 are going to cancel. So all we're going to be left with for the period is one over two. And so that means that the period, for example, b, is equal to one half. 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.

Given below is the graph of the function $y=\sin\left(bx\right)$. Determine the correct value for b.

• A. $b=\pi$
• B. $b=2$
• C. $b=\frac12$
• D. $b=4$

Answer: (D)

The Period for the function $y=\cos\left(bx\right)$ is $T=20\pi$. Determine the correct value of b.

• A. $b=\frac{1}{10}$
• B. $b=10$
• C. $b=20$
• D. $b=\frac{1}{20}$

Answer: (A)

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 one, and then kinda 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 parenthesis that was being added or subtracted to the x within your cosine.

And an example of this would be, say, the cosine of x-π2. 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 is I'm going to take all of these values and I'm going to subtract π2 from them, and then I'll evaluate the cosine. So the cosine of zero minus π2 is gonna be the cosine of -π2. 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 π2 minus π2. That's gonna be the cosine of zero, which is just going to be one. Next we'll have the cosine of π minus π2. That's gonna be the cosine of π2, which is back at zero. Now for 3π2 minus π2, 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 2π minus π2, which turned out to just be zero. 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 kinda 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, this zero that went there, this negative one that went there, and then this zero that went here. So subtracting π2 from the inside of this cosine function caused every point on the original cosine graph to be shifted over to the right by π2 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 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 bx-somenumber, then that means that your h value is positive, and your graph is going to shift to the right by hb units. Now if you instead had a situation where the inside of your trig function was bx+somenumber, that means your h value is negative, and the graph is going to shift left by hb units. Now in the situation we had right here, we had the cosine of one x minus π2. And because the inside of this function is in the form bx-anumber, that means that our h value is positive and our graph shifted to the right. Now finding hb for this graph would be our h value, which is π2 divided by our b value, which is one, and π2 divided by one is the same thing as just π2.

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 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+π 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 two in front. And the period for a sine or cosine function is equal to 2π over b. Well, the b value we can see in front of the x is two, and then these two's 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 π2. 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+π. 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 bx+somenumber, that means that h value is negative and will shift to the left by hb units. Now what is hb? 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 π2 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 gonna cross through π2, 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. Now 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+π could also be written as y is equal to two times x plus π2. 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 π2 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. 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

Answer: (C)

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

Answer: (D)

Phase Shifts Example 3

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 sine 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 three. Now what I also notice is for b x - 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 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 number is one because one times x would just be x. Now I also noticed 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 negative 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, so 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 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 π /2, which will be up here at a y-value of three.

We're going to cross through π, and then we're going to reach a valley at 3 π /2, 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 going to keep going to the left as well. So we're going to see a valley here at negative π / 2. We're then going to cross through negative π, and then we're going to see a peak at negative 3 π / 2.

Here’s what students ask on this topic:

What is the difference between the graphs of sine and cosine functions?

The sine and cosine functions both produce wave-like graphs, but they differ in their starting points. The sine graph, y = sin(x), begins at the origin (0, 0), while the cosine graph, y = cos(x), starts at its maximum value (0, 1). Both graphs have the same amplitude (height of the wave) and period (length of one complete cycle), typically 2π for their standard forms. The sine graph is symmetric about the origin, making it an odd function, while the cosine graph is symmetric about the y-axis, making it an even function. These differences arise because sine corresponds to the y-coordinate and cosine to the x-coordinate on the unit circle. Despite these distinctions, their wave patterns are identical, just horizontally shifted by π/2 radians.

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

The period of a sine or cosine function determines the length of one complete wave cycle along the x-axis. It is calculated using the formula:

$P=\frac{2}{b}$

Here, b is the coefficient of x inside the sine or cosine function. For example, in y = sin(2x), b = 2. Substituting into the formula, the period is:

$P=\frac{2}{2}=\pi$

This means the wave completes one cycle in π units. If b is less than 1, the graph stretches horizontally, increasing the period. If b is greater than 1, the graph compresses, decreasing the period.

What is the amplitude of a sine or cosine function, and how is it determined?

The amplitude of a sine or cosine function represents the height of the wave from its midline to its peak or trough. It is determined by the absolute value of the coefficient in front of the sine or cosine function. For example, in y = 3sin(x), the amplitude is |3| = 3. Similarly, in y = -2cos(x), the amplitude is |−2| = 2. The amplitude affects the vertical stretch or compression of the graph but does not change the period or phase shift. A larger amplitude results in taller waves, while a smaller amplitude produces shorter waves. If the coefficient is negative, the graph is reflected over the x-axis, but the amplitude remains positive.

What is a phase shift in the graphs of sine and cosine functions?

A phase shift is a horizontal shift of the sine or cosine graph along the x-axis. It occurs when a constant is added or subtracted inside the function's argument. For example, in y = sin(x − π/2), the graph shifts to the right by π/2 units. The general form for phase shift is determined by the equation:

$\mathrm{Phase\; Shift}=\frac{h}{b}$

Here, h is the constant added or subtracted, and b is the coefficient of x. If h is positive, the graph shifts to the left; if h is negative, it shifts to the right. For example, in y = cos(2x + π), the phase shift is:

$-\frac{\pi }{2}$

This means the graph shifts left by π/2 units.

How do vertical shifts affect the graphs of sine and cosine functions?

Vertical shifts move the sine or cosine graph up or down along the y-axis. This occurs when a constant k is added or subtracted to the function. For example, in y = sin(x) + 2, the entire graph shifts up by 2 units. Similarly, in y = cos(x) − 3, the graph shifts down by 3 units. The midline of the graph, which is usually y = 0, becomes y = k. Vertical shifts do not affect the amplitude, period, or phase shift of the graph; they only change the vertical position. To graph a vertically shifted function, plot the standard sine or cosine graph first, then move all points up or down by the value of k.

What happens to the sine or cosine graph when the amplitude is negative?

When the amplitude of a sine or cosine function is negative, the graph is reflected over the x-axis. For example, in y = −sin(x), the wave pattern of the sine graph flips upside down. Peaks become troughs, and troughs become peaks. However, the amplitude, which is the absolute value of the coefficient, remains positive. For instance, in y = −3cos(x), the amplitude is |−3| = 3, but the graph is inverted. This reflection does not affect the period, phase shift, or vertical shift of the graph; it only changes the orientation of the wave.