Graphs of Trigonometric Functions - Online Tutor, Practice Problems & Exam Prep

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 xy graph. 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 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. You'll see that these values will repeat as you go around the unit circle. For example, if we're looking at the sine, we can see that our sine value starts at 0, then it goes to 1, then to 0, then to negative one, and then it starts repeating. So it goes back to 0, then to 1, then 0, and it just repeats as you go around.

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. For the sine of 0, based on our unit circle, the sine of 0 is just going to be 0. Now, for the sine of π2, according to our unit circle, it is 1. For the sine of π, the sine of π is 0. If we go to 3π2, the sine of 3π2 is negative 1. So the sine of 2π would be a full trip around the circle right back to 0, and then the sine of 5π2 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. 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, and when we look at the cosine of 0, we can see that it starts right here at 1. For the cosine of π2, it's going to be at 0. For the cosine of π, that's going to be at negative one. For the cosine of 3π2, it'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 the cosine of 5π2, we're going to be back at 0, meaning your graph is going to look something like this. 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 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 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 0, at an output of 0, and then we reach a peak at π2, then we go back to 0 through π, and then we reach a valley at 3π2, then we go back to 0 over 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 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=sin⁡x+1, and that is the answer to this example.

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. 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.

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. $y=50+\frac{\sqrt3}{2}$

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 \( y \) 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 am 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 then 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 1, 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 \( \frac{-\pi}{2} \), and we're going to come and cross through \( -\pi \). 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 going to be our coordinate there. So we're going to be 3 units up for where our graph starts, then this peak here is going to be 3 units up as well, which is going to be at positive 4.

This zero 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 \( \frac{-\pi}{2} \), we're also going to be at positive 2 for our output. At \( -\pi \), we're at an output of 0, so our output is going to be up here at 3, and then at \( \frac{-3\pi}{2} \), 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. So hope you found this video helpful. Thanks for watching.

Graphs of Secant and Cosecant Functions

Hey, everyone. So up to this point, we've spent a lot of time dealing with the sine and cosine functions, and we've discussed their graphs and different ways that the graphs can be transformed, like being stretched or shifted in some way. Well, in this video, we're going to learn that we can graph more trigonometric functions such as the secant and cosecant. And this might sound a little bit scary at first, but don't sweat it because it turns out we can actually use the graphs that we learned about for sine and cosine to graph the cosecant and secant. So without further ado, let's get right into this.

Now, recall that the cosecant and secant are reciprocals; they're reciprocal identities for the sine and cosine. So cosecant is 1 over the sine, and secant is 1 over the cosine, and you can use these relationships to figure out what their graphs look like. So I'm just going to jump right into the cosecant graph. And if I go ahead and take the reciprocal for all these outputs that I see, the reciprocal of 1 is just 1. The reciprocal of negative one is negative one because 1 divided by negative one would be negative one, and then the reciprocal of 1 is just 1 there.

So that means at an x value of pi over 2, we'll have 1. For 3 pi over 2, we're going to have negative 1, and then for 5 pi over 2, we're going to have 1 again. Now I'm also going to need to take the reciprocal of these values, but notice that all of these values are 0. And if I take the reciprocal of 0, that's just going to be 1 over 0. But this is actually a problem, because recall that in math, it's a fundamental rule that we cannot divide by 0.

So what does this mean? Well, that means that every place we see 0 for our sine value, our cosecant is going to be undefined. So all of these values will be undefined. And for undefined values, what we're basically saying in this instance is that they're approaching infinity. And infinity is not a number, so we can't define it.

So what we can do is we can take asymptotes, and we can draw them where we see the sine function reach 0. So we're going to have an asymptote at an x value of 0. We'll have another asymptote at an x value of pi and we'll have another asymptote at an x value of 2 pi. Now from here, I need to figure out how the rest of the graph is going to behave.

And the way that I can do this is by simply observing how the sine function behaves. Notice how the sine function gets smaller and smaller as we go to the left and as we go to the right. And what happens if you take the reciprocal of a number that gets smaller and smaller? Well, the whole fraction is going to get bigger and bigger. And because of this, our cosecant function is going to blow up as we go to the left and as we go to the right.

So we, in essence, end up with this kind of smiley face thing happening right here. And it turns out that this logic stays true for all the other points. So at this point right here, we can see that our value is actually getting bigger and bigger as we go to the left and right. If the values are getting bigger, that means that the reciprocal is going to get smaller and smaller. And likewise, we can see right here that these values get smaller here, so it's going to get bigger and bigger as you blow up to the left and to the right at this point.

So notice how we end up with these smiley and frowny faces where we have the peaks and valleys respectively. Now let's also take a look and see how the secant function behaves. Well, I can use the same logic by taking the reciprocal of all these values for the cosine. So the reciprocal of 1 is 1, and then we have the reciprocal of negative one and the reciprocal of 1. So going to our graph, our points are going to be at 1 at the peak there, at negative one for the valley at next value of pi, and then at another peak, we're going to have 1 as well.

Now because secant is a reciprocal of the cosine, all the places that we see 0 for the cosine are going to be undefined for the secant. So when doing this, we can actually draw asymptotes at all these undefined values. So at an x value of pi over 2, we're going to have an asymptote. At 3 pi over 2, we're going to have another asymptote. And then at 5 pi over 2, we're going to have another asymptote.

Now just like we saw with the cosecant, this graph is a reciprocal of the cosine. So just like with the cosecant graph, we're going to see these kind of smiley faces and frowny faces where the peaks and valleys are. So notice how these two graphs end up looking very similar. The really key difference between these two graphs is where the asymptotes are located. Because notice for the cosecant, we ended up with asymptotes at 0, pi, 2 pi, basically just integer multiples of pi.

And for the secant, we ended up with asymptotes at pi over 2, 3 pi over 2, 5 pi over 2, or basically just odd multiples of pi over 2. Now something that's really nice about the cosecant and secant graphs is it turns out that you can use all the same transformation rules that you use for the sine and cosine. So all the stretches and shifts for the cosecant and secant function that we learned about are going to hold true. And to see this, let's actually take a look at an example. So in this example, we are asked to graph the function y is equal to the cosecant of 2x.

Now you may be a little bit curious where we need to start with this example or how we could even do this, but what I personally like to do when dealing with these types of problems is to first figure out what I know and then build off of that. Now one of the things I know about the cosecant is I know the cosecant is a reciprocal of the sine. And so what I'm actually going to do is first find the graph of y equals the sine of 2x. Now this is not the graph that we're looking for, but this will give me an idea as to what the cosecant is going to look like since I know their relationship. So recall that for the sine of 2x, well, we first need to figure out what the period is.

And the period for a sine or cosine graph is 2 pi divided by b. Now the b value that we have in front is 2. So we're going to have 2 pi divided by 2 where the twos will cancel and give us pi. So that means that we're going to have a sine graph that has its period at pi. So what that means is that our sine graph will start at the origin, and we're going to reach a peak right about there, then we're going to cross through pi over 2 and complete a full wave at pi.

Now what we're going to do is have another cycle that reaches a peak there, goes through 3 pi over 2, and then goes here to 2 pi, and then we're going to reach another peak at 5 pi over 2, and that's what our sine of 2x graph is going to look like. Now to figure out what the cosecant is going to look like, well, we can just recognize that for y equals the cosecant of 2x, this is a reciprocal for the graph that we had here. So just like we saw up above, we're going to have points at all the peaks and valleys for this graph. And we will have asymptotes for every place where this graph reaches 0 or crosses the x axis. So we'll have an asymptote here at 0, we'll have an asymptote at pi over 2, we'll have another asymptote at 3 pi over 2, and at 2 pi.

And then these are all the asymptotes that we can draw for as far as we've seen this graph. Now all I need to do from here is draw the little smiley and frowny faces, and I know that those go between each of the asymptotes. So I'll get a smiley face there. I'll get a frowny face right there. I'll get another smiley face up here. I'll get a frowny face there, and then I'll get another smiley face which would peak right about there. So this is what our graph is going to look like, and that is the solution to this problem. So this is how you can deal with graphs for the secant and cosecant. Let me know if you have any questions. Thanks for watching.

Below is a graph of the function $y=\sec\left(bx-\pi\right)$. Determine the value of b.



• A. $b=2$
• B. $b=4$
• C. $b=\frac{\pi}{2}$
• D. $b=\pi$

Below is a graph of the function $y=\csc\left(bx\right)$. Determine the value of b.

• A. $b=\frac12$
• B. $b=3$
• C. $b=\frac43$
• D. $b=\frac{3\pi}{4}$

Introduction to Tangent Graph

Hey, everyone. So in the last few videos, we've been on this journey of looking at the various graphs of trigonometric functions. We've already seen what the sine and cosine look like, as well as what the secant and cosecant look like. Well, in this video, we're now going to take a look at the graph of the tangent function. If this sounds intimidating, don't sweat it because we're going to learn in this video that just like the cosecant and secant function, the tangent function will also build off of the knowledge that we've already learned about these graphs, like the sine and cosine, for example.

So without further ado, let's take a look at what some of these problems might look like in this course where we have to deal with the tangent function as a graph. Now I want you to recall this identity, which we learned about early on, that says that the tangent of some value is equal to the sine over the cosine. So what we can do is use this to graph what the tangent looks like by taking the sine and dividing it by the cosine values. Now I'm going to go ahead and start with 0. And I know that I'm doing this a bit out of order.

I'm not starting from the left side this time, but you're going to see why. So if I start with 0, notice that we're going to have at an x value of 0 that our tangent is sine over cosine. Well, that's just going to be 0 divided by 1, which is 0. So that means we're going to start right about there on our graph. Now what I'm going to do from here is go to the right and the left on this graph.

So start by going to the right at an x value of π/4. This is going to give you 2/2 divided by 2/2, which is just a value of 1. So at π/4, we have an output of 1 for our tangent graph. Now if I go to the left, we're going to have negative 2/2 divided by that same value but positive. And doing this will give you a value of negative one, meaning we're going to be down here on our graph.

Now what happens if I go to π/2 and negative π/2 on the x axis? Well, I want you to notice something in both these situations. What we end up having is this case where we have to divide by 0. We're going to have negative one over 0 and one over 0. We learn in math that you cannot divide by 0, so that means that these values are undefined.

And because we get undefined values at negative xmlns="http://www.w3.org/1998/Math/MathML"π/2 and at π/2, that means we're going to have asymptotes show up. So we're going to have an asymptote right about here, and an asymptote right about there. And recall that an asymptote just means the function is approaching infinity at that point. So that means our graph is going to look something like this. We're going to go down and to the left, and then we're going to go up and to the right.

So this would be the curve for our tangent function. Now there's a couple of things you should know about the tangent graph. The tangent graph actually has some similarities to the secant graph, which we saw in the previous video. And one of the similarities is that the asymptotes are at the same values on the x axis. Notice how we have asymptotes at negative 3 π/2, negative π/2, π/2, and 3 π/2.

Well, for our tangent graph, we can see these asymptotes are the same, and it turns out that for the tangent graph, all of the asymptotes or places where the curve repeats are going to be when the cosine of x is equal to 0, which are going to be at odd multiples of π/2. So because of this, on our tangent graph, we're going to see another asymptote at negative 3 π/2, and an asymptote at positive 3 π/2. Now the way that the tangent graph is different than the secant graph is, well, obviously, this curve is different, but it turns out that for the tangent graph, the tangent graph actually has a period of only π. Right? For the secant graph, we can see that what happens is we get this smiley face, then this frowny face, and then the curve is going to repeat.

So we'll get another smiley, then another frowny. So it repeats after 2 π units because the distance from here to there on the x axis would be 2 π. Well, it turns out that the tangent graph will actually repeat every π units. So rather than getting some kind of different curve, you're going to get this same curve right about here, and you're going to get the same curve right about there. So every time you travel a distance of π on the x axis, you're going to get your curve repeating.

And what this does is actually changes the period for the tangent. So in the past, we've seen that the period for the sine, cosine, cosecant, and secant is 2π over b. Well, it turns out when dealing with the tangent, the period is only π over b. So that's something you have to watch out for when dealing with the tangent graph, but the nice thing about the tangent graph is that all of the same transformation rules apply that have been used for sine and cosine aside from the period change. Now to make sure that we're understanding this, let's try an example where we have to graph a function which deals with the tangent.

So here we're asked to graph the function y is equal to the tangent of π/2 times x. And the way that I'm going to do this is first figuring out what the period of our graph is going to be. Well, recall that the period for tangent, as we just learned, is π over b, rather than 2 π over b. And the b value is what we have in front of the x, so when we have π divided by π/2. Now what I can do is take this fraction we have in the denominator and flip it and bring it to the numerator.

So I can flip this right here, and we're going to get 2 π divided by π, because this will stay in the denominator. And then those pi's are going to cancel giving us a period of 2. So if our period is 2, that means that every 2 units we're going to have our curve repeat. So what I'm going to do is label the x axis like this. We're going to see that we have points here 1, 2, 3 on the x axis, and then we have negative 1, negative 2, negative 3.

Now we've learned that how the tangent graph starts right here at the middle, and then it curves up into the right and down into the left. And we also know that this tangent graph is going to have a period of 2. So what I can do is draw asymptotes here at negative one, and I can draw another asymptote at positive one, because this would be a period of 2 on the x axis. So what I can do is draw a curve that looks something like this. So this would be the tangent graph.

Now because the period is 2, if I start here where this asymptote is and I go 2 units to the right, we're going to have another asymptote at positive 3. And if I go to negative one, we can go 2 units to the left and get another asymptote at negative 3. And this curve is going to repeat between each of these asymptotes, so we're going to get the same curve right about there and the same curve back here. So this is what the graph would look like, and that is the solution to the problem. So hopefully this gives you a better understanding of how to work with graphs when you're dealing with the tangent function.

Thanks for watching, and please let me know if you have any questions.

Below is a graph of the function $y=\tan\left(bx\right)$. Determine the value of b.

• A. $b=\frac14$
• B. $b=\pi$
• C. $b=2$
• D. $b=\frac12$

Introduction to Tangent Graph Example 1

Hey, everyone. So here we have an example where we are asked to graph the function y is equal to 12tan(x−π2). Now to solve this problem, what I'm going to do is I'm first going to graph the function y is equal to 12tan(x). So I'm going to ignore this π2 for now, but we will get to this later. Let's start with this function.

To find y is equal to one half the tangent of x, well, let's just recall what the tangent graph looks like. I know for the tangent graph we have asymptotes at negative π2, and at π2. Now we do have more asymptotes on this graph, we're going to have other asymptotes at odd multiples of π2, but for now I'm just going to draw these two. As for the curve, I know the curve starts here in the middle, and then it goes up to the right, and then down to the left. Another thing to recall for this tangent is that we have points at π⁄4, comma 1, and then at negative π⁄4 comma negative 1.

The graph ends up looking something like this. But notice that we have a one-half in front of our tangent, and this one-half, as you may recall, changes the amplitude or tallness of our graph. So rather than having these points at 1 and negative one on the y-axis, these are going to be reduced. So we're then going to have a point instead at 12, and then at negative 12 for these two values of π⁄4 and negative π⁄4 respectively. So the graph is going to look a little bit more like this, where we can see that it's a little bit shorter than we were before.

So this right here would be the curve for y equals one-half the tangent of x. Now our last step for solving this problem is going to be to incorporate this π2. We need to figure out what y equals one half times the tangent of x minus π2 looks like. Well, something that I can see is that we have a π2 here, which is going to shift our graph in some sort of way. And to find the shift, well, recall that we can find the h over b ratio which tells us how much we've shifted to the left or to the right.

Now see that we have a minus sign here which means that this is actually a positive h value. So we have positive π⁄2 divided by b, which in this case is any number in front of the x, but since there's nothing there, we just write 1. So we're going to have h over b is equal to just π⁄2, because this one in the denominator is just going to keep this the same. So that means that our graph is going to be shifted π⁄2 units to the right, because we got a positive h over b. So what I can do is take this point right here, and I can shift it π⁄2 units to the right.

In fact, I can take this asymptote, and shift it π⁄2 units to the right, which would put us at π. Then I can take this asymptote, and shift it π⁄2 units to the right, which would put us at 0. So this is what the graph is going to look like, and because we have asymptotes at π and then at 0, we're going to have another asymptote at negative π on the x-axis. And what I can do from here is continue drawing these curves since they're going to repeat every π amount of units. So we'd have another curve which is right back here.

It's going to go look like this, and then like that. It's going to be the same as the curve that we had before, because this curve just continues repeating. So the curve is going to be back here as well. It's going to go up and then down, and then this curve is going to repeat over here going up and down to the left as well. So this is what our curves are going to look like.

This is the graph for the tangent, and that is the solution to this problem. So I hope you found this video helpful. Thanks for watching.

Introduction to Cotangent Graph

Welcome back, everyone. So at this point, we should be familiar with the graphs for most of the trigonometric functions. We've talked about the sine and cosine graphs, the secant and cosecant. We even looked at the tangent graph in the previous video. Now last but not least, we're going to be taking a look at the cotangent graph.

And this might sound a bit complicated just because learning about another trigonometric graph might feel kind of exhausting at this point, but what we're going to learn is that the cotangent graph is actually closely related to the tangent graph. And if you recognize how these two are related, then sketching this graph is actually pretty straightforward. So without further ado, let's get right into things. Now recall how based on reciprocal identities, the cotangent is just one over the tangent. So we can use this fact to draw the graph for the cotangent.

Now the first thing that I'm going to look for is where we can draw these asymptotes because this will tell us how wide each of the cycles of our graph is. Now if I look at the tangent graph, I can see that we reach 0 at negative pi, 0, and pi. So we reach 0 on the y-axis at all these points. And it is a fundamental math rule that you cannot divide by 0. So any place that we see 0 on the tangent is going to be a value that cannot exist on our cotangent graph, so that's going to be where we draw the asymptotes.

So there's going to be an asymptote right here at negative pi. There's also going to be an asymptote here at 0 on the x-axis, and there's going to be an asymptote over here at positive pi. Now from here, we need to draw the curves. And it turns out that the curves for the cotangent graph are similar to the tangent graph, but they're going to be flipped upside down. So rather than having curves that go from bottom left to top right, we're going to have curves that go from top left to bottom right.

So the graph is going to look something like this. Now I want you to notice all the points on the cotangent graph where we touch the x-axis. This is where our output is going to be equal to 0. And we can actually see that these values make sense when we look at the tangent graph because notice that we touch the x-axis at negative 3 pi over 2. And on our tangent graph, our graph gets very small when we get to negative 3 pi over 2 in the negative direction.

So basically we go to negative infinity. And whenever you divide a number that is approaching infinity, whether it's a negative or positive number, then your entire fraction or output is going to be equal to 0 approximately, or it's going to approach 0. So that is why we have 0 at negative 3 pi over 2, at negative pi over 2, at pi over 2, and at 3 pi over 2. So you can see at all of these points on our tangent graph, we're either getting really big there towards infinity or really small towards negative infinity. Now one of the main similarities between the tangent and cotangent graph is the distance on our x-axis, which we have the curves repeat.

Because notice for our tangent graph that we repeat from negative pi over 2 to pi over 2, which is just a distance of pi. And it is a distance of pi for each of these curves on this graph. So that is one of the nice things about the tangent graph is we repeat every pi units, and it turns out that the cotangent graph is the exact same thing. Because if you look, we can see here our graph goes from negative pi to 0, and that's just the distance of pi. And we also go from 0 to pi, which is also a distance of pi.

So it turns out that this graph repeats every pi amount of units as well. Now because these graphs have the same repeating behavior, it turns out their periods are also going to be the same. So just like the period of the tangent was pi divided by b, the period of the cotangent is also pi divided by b. Now the way these graphs are different from each other is the values of their asymptotes. So for example, if we take a look at the tangent graph, we can see that these asymptotes will always show up at odd multiples of pi over 2.

Notice we have an asymptote at negative 3 pi over 2, negative pi over 2. We have an asymptote at pi over 2, 3 pi over 2. Then if we kept extending this graph, we'd also have asymptotes at 5 pi over 2 and 7 pi over 2. Now notice for our cotangent graph, our asymptotes are at negative pi. They're at 0.

They're at pi. If we kept extending this graph, we would see asymptotes at 2 pi, 3 pi, 4 pi. So it turns out for the cotangent graph, the asymptotes are always going to be at integer multiples of pi. So the location of the asymptotes on the x-axis, as well as the fact that these curves are flipped upside down, are really the two main differences between the cotangent and tangent graphs. Now to make sure we understand this concept well, let's try an example where we have to deal with the cotangent.

Now to solve this problem, what I'm first going to do is figure out where the asymptotes are going to be located on this graph. And I can do this by looking at what our cotangent is. Now recall that the period of the cotangent is just pi divided by b. Now what I can see here is that our b value is the value we have in front of the x, and I can see that b value is equal to pi. So we're going to have pi over pi, and whenever you divide one number by itself, it's just going to be equal to 1, because they're both going to cancel and give you 1.

So that means that our period is equal to 1. So what I can do on our graph is I can draw points every one unit on our x-axis. So I'll go 1, 2, 3 to the right, and they'll go negative 1, negative 2, negative 3 to the left. Now the period of this graph is 1, meaning I can draw these asymptotes for every one unit. So we're going to have an asymptote here at 0, we're going to have it at 1, and then at 2, and then at 3.

And then these asymptotes are also going to repeat behind the graph as well. So I have an asymptote at negative 1, negative 2, negative 3, and this is what the asymptotes are going to look like. Now my last step for solving this problem is going to be to draw the curves. And recall that the curves go from top left to bottom right between these asymptotes. So what I'm going to do is have a curve that looks something like that, and these curves are just going to keep repeating for each period of this graph because that's what the repeating function is going to do.

So this right here is the graph of our function and the solution to this problem. So that is how you can deal with graphs when it comes to the cotangent function. Hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.

Below is a graph of the function $y=\cot\left(bx+\frac{\pi}{2}\right)$. Determine the value of b.

• A. $b=\frac14$
• B. $b=1$
• C. $b=2$
• D. $b=\frac12$

Introduction to Cotangent Graph Example 1

Hey, everyone. So let's see how we can go about solving this example. We are asked to graph the function \(y = -2 \cot\left(\frac{1}{4}x\right)\). Now, what I notice in this problem is we have quite a few things going on. We have a negative sign here, which is going to cause our graph to be flipped or reflected over the x-axis.

I also notice we have this 2 here, which is going to change the amplitude or tallness of our graph. And we have this \(\frac{1}{4}\) on the inside of our cotangent, which is going to change the period. So where do we even get started with this? Well, what I like to do is start with what I'm familiar with. And what I'm going to do is I'm just going to focus on \(y = \cot\left(\frac{1}{4}x\right)\).

The reason for focusing on this is because I know how to deal with the period of a cotangent, and I know what the general cotangent graph looks like, so I can build off of something. Now the period for a cotangent graph is going to be \(\frac{\pi}{b}\), where \(b\) is the number that is in front of the \(x\). Since I can see that that's \(\frac{1}{4}\), I get that we have \(\frac{\pi}{\frac{1}{4}}\). You can flip this fraction and bring it to the top, which is going to give you \(4\pi\) over 1, which is just \(4\pi\). So the period is \(4\pi\).

And something that I know about the cotangent graph is that it has an asymptote that starts at an \(x\) value of 0. So since we start at an \(x\) value of 0, and our period is \(4\pi\) when we have this \(b\) value in front of \(x\), that means that starting at 0, we're going to repeat every \(4\pi\) units. So I can draw another asymptote here at \(4\pi\), and I can draw another asymptote over there at \(8\pi\). Now something I know about the cotangent graph is that it goes up into the left, and then goes down into the right, and it reaches 0 on the \(x\) axis right in between the two asymptotes in the middle. Now what you should typically see with a cotangent graph is you're going to see points that are in between these two values, and in between those two values.

So this over here should be \(\pi\), and this should be \(3\pi\), and that makes sense we have \(\pi\), \(2\pi\), \(3\pi\), \(4\pi\), and at these values you should see an output of 1. So what we should have is an output right about there, which would be at 1, and we should see an output right here, which would typically be at negative one. So adjusting our curve a little bit, this is what the function would look like for the \(\cot\left(\frac{1}{4}x\right)\), and then you would get this curve continuously repeating as we go along. But this is not what we're trying to find. We're not trying to find the \(\cot\left(\frac{1}{4}x\right)\), we're trying to find \(-2 \cdot \cot\left(\frac{1}{4}x\right)\).

Now this number out in front here, the 2, this is going to change the amplitude or basically vertically stretch our graph. So rather than having an output of 1 when you're at \(\pi\), you should actually have an output of 2 at this value. And then rather than having an output at negative one for \(3\pi\), you should actually have an output at negative 2. So adjusting our curve, the graph should look a bit more stretched vertically when we incorporate this 2. But there's one more thing we need to change about our graph, and that's the negative sign.

This negative sign is going to take our whole graph, and it's going to flip it over the \(x\) axis. So doing this, we actually need to, in essence, reverse these two points on the \(y\) axis. So we should have a point right here at \(\pi\) negative 2, and a point right here at \(3\pi\) positive 2. So for the graph \(y = -2 \cdot \cot\left(\frac{1}{4}x\right)\), it should look something like this, where we start at the lower left, and then we go ahead and go up reach a 0 on the \(y\) axis at \(2\pi\), and then we come up here and go to the top right. So this is what our curve should look like, and because the cotangent is a repeating graph as we've learned in previous videos, this curve should continue happening between each of these asymptotes.

So, this would be the graph of our function and the solution to the problem. I hope you found this video helpful. Thanks for watching.