Graphs of Other Trigonometric Functions: Videos & Practice Problems

Understanding the graphs of trigonometric functions is essential, particularly for cosecant, secant, tangent, and cotangent. The cosecant and secant are reciprocals of sine and cosine, respectively, with asymptotes at points where sine or cosine equals zero. The tangent function, defined as sine over cosine, has asymptotes at odd multiples of π/2 and a period of π. The cotangent graph mirrors the tangent but is flipped, with asymptotes at integer multiples of π. Both functions share transformation rules, enhancing their graphing process.

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, discussing their graphs and different ways that the graphs can be transformed, such as 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. This might sound a bit scary at first, but don't worry 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. Cosecant is ⅛ divided by the sine and secant is ⅛ divided by the cosine. 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.

If I go ahead and take the reciprocal for all these outputs that I see, the reciprocal of ⅛ is just ⅛. The reciprocal of negative ⅛ is negative ⅛ because one divided by negative ⅛ would be negative ⅛, and then the reciprocal of ⅛ is just ⅛ there. So, that means at an x value of π/2, we'll have ⅛. For 3π/2, we're going to have negative ⅛, and then for 5π/2, we're going to have ⅛ again. Now I'm also going to need to take the reciprocal of these values.

But notice that all of these values are zero. And if I take the reciprocal of zero, that's just going to be ⅛ divided by zero. But this is actually a problem, because recall that in math, it's a fundamental rule that we cannot divide by zero. So what does this mean? Well, that means that every place we see zero 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 zero. So we're going to have an asymptote at an x value of zero. We'll have another asymptote at an x value of π.

And we'll have another asymptote at an x value of 2π. 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 increase 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 ⅛ is ⅛, and then we have the reciprocal of negative ⅛ and the reciprocal of ⅛. So going to our graph, our points are going to be at ⅛ at the peak there, at negative ⅛ for the valley at x value of π, and then at another peak, we're going to have ⅛ as well. Now because secant is a reciprocal of the cosine, all the places that we see zero 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 x value of π/2, we're going to have an asymptote.

At 3π/2, we're going to have another asymptote. And then at 5π/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 zero, π, 2π, basically just integer multiples of π. And for the secant, we ended up with asymptotes at π/2, 3π/2, 5π/2, or basically just odd multiples of π/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 wh

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. $b=\pi$

Answer: (A)

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. $b=\frac{3\pi}{4}$

Answer: (C)

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 divided by 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 zero. 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 zero, notice that we're going to have at an x value of zero that our tangent is sine over cosine. Well, that's just going to be zero divided by one, which is zero. 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 π4. This is going to give you 22 divided by 22, which is just a value of one. So π4, we have an output of one for our tangent graph. Now if I go to the left, we're going to have negative 22 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 π2 and negative π2 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 zero. We're going to have negative one over zero and one over zero. We learn in math that you cannot divide by zero, so that means that these values are undefined.

And because we get undefined values at negative π2 and at π2, 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 then 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 32π, negative 12π, 12π, and 32π.

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 zero, which are going to be at odd multiples of π2. So because of this, on our tangent graph, we're going to see another asymptote at negative 32π, and an asymptote at positive 32π. 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.

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 two π units because the distance from here to there on the x axis would be two 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 two π 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 π2 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 two π over b. And the b value is what we have in front of the x, so when we have π divided by π2.

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 two π divided by π, because this will stay in the denominator. And then those π's are going to cancel giving us a period of two. So if our period is two, that means that every two 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 one, two, three on the x axis, and then we have negative one, negative two, negative three. Now we've learned that how the tangent graph starts right here at the middle, and then it curves up to the right and down to the left. And we also know that this tangent graph is going to have a period of two. 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 two 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 two, if I start here where this asymptote is and I go two units to the right, we're going to have another asymptote at positive three. And if I go to negative one, we can go two units to the left and get another asymptote at negative three. 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. $b=\frac12$

Answer: (D)

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 one half times the tangent of \( x \) minus \( \frac{\pi}{2} \). Now, to solve this problem, what I'm going to do is I'm first going to graph the function \( y \) is equal to one half times the tangent of \( x \). So I'm going to ignore this \( \frac{\pi}{2} \) for now, but we will get to this later. So 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 \( -\frac{\pi}{2} \) and at \( \frac{\pi}{2} \). Now we do have more asymptotes on this graph, we're going to have other asymptotes at odd multiples of \( \frac{\pi}{2} \), but for now, I'm just going to draw these two. Now as for the curve, I know the curve starts here at the middle, and then it goes up and to the right, and then down and to the left. Now another thing that I recall for this tangent is that we have points at one, or at \( \frac{\pi}{4} \), \((1)\) and then at \( -\frac{\pi}{4} \), \( (-1) \).

So 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 one and negative one on the \( y \) axis, these are going to be reduced. So we're then going to have a point instead at one half, and then at negative one half for these two values of \( \frac{\pi}{4} \) and \( -\frac{\pi}{4} \) respectively.

So that means the graph is going to look a little bit more like this, where we can see that we're 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 \( \frac{\pi}{2} \). So we need to figure out what \( y \) equals one half times the tangent of \( x \) minus \( \frac{\pi}{2} \) looks like. Well, something that I can see is that we have a \( \frac{\pi}{2} \) 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 \( h/b \) 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 \( \frac{\pi}{2} \) divided by \( b \), which in this case is any number in front of the \( x \), but since there's nothing there, we just write one. So we're going to have \( h/b \) is equal to just \( \frac{\pi}{2} \), 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 \( \frac{\pi}{2} \) units to the right, because we got a positive \( h/b \).

So what I can do is take this point right here, and I can shift it \( \frac{\pi}{2} \) units to the right. In fact, I can take this asymptote and shift it \( \frac{\pi}{2} \) units to the right, which would put us at \( \pi \). Then I can take this asymptote and shift it \( \frac{\pi}{2} \) units to the right, which would put us at zero. So this is what the graph is going to look like. And because we have asymptotes at \( \pi \) and then at zero, we're going to have another asymptote at negative \( \pi \) on the \( x \) axis.

And what I can do from here is continue drawing these curves since they're going to repeat every \( \pi \) amount of units. So we'd have another curve which is right back here. It's going to 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 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. 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.

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. 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. The first thing 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.

If I look at the tangent graph, I can see that we reach zero at negative π, zero, and π. So we reach zero on the y-axis at all these points. And it is a fundamental math rule that you cannot divide by zero. So any place that we see zero 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. There's going to be an asymptote right here at negative π.

There's also going to be an asymptote here at zero on the x-axis, and there's going to be an asymptote over here at positive π. Now from here, we need to draw the curves. 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. 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 zero. 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 three π over two. And on our tangent graph, our graph gets really, really small when we get to negative three π over two in the negative direction. So basically, we go to negative infinity.

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 zero approximately, or it's going to approach zero. So that is why we have zero at negative three π over two, at negative π over two, at π over two, and at three π over two. 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. One of the main similarities between the tangent and cotangent graph is the distance on our x-axis where the curves repeat. Notice for our tangent graph that we repeat from negative π over two to π over two, which is just a distance of π.

It is a distance of π for each of these curves on this graph. That is one of the nice things about the tangent graph as we repeat every π units, and it turns out that the cotangent graph is the exact same thing. If you look, we can see here our graph goes from negative π to zero, and that's just the distance of π. And we also go from zero to π, which is also a distance of π. So it turns out that this graph repeats every π amount of units as well.

Since these graphs have the same repeating behavior, their periods are also going to be the same. So just like the period of the tangent was π/b, the period of the cotangent is also π/b. The way these graphs differ from each other is the values of their asymptotes. 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 π over two. Notice we have an asymptote at negative three π over two, negative π over two.

We have an asymptote at π over two, three π over two. Then if we kept extending this graph, we'd also have asymptotes at five π over two and seven π over two. Now, notice for our cotangent graph, our asymptotes are at negative π, they're at zero, they're at π. If we kept extending this graph, we'd see asymptotes at two π, three π, four π. So it turns out for the cotangent graph, the asymptotes are always going to be at integer multiples of π.

The location of the asymptotes on the x-axis, as well as the fact that these curves are flipped upside down, are the two main differences between the cotangent and tangent graph. Now to make sure we understand this concept well, let's try an example where we have to deal with the cotangent. Recall that the period of the cotangent is just π/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 that b value is equal to π. So we're going to have π/π, and whenever you divide one number by itself, it's just going to be equal to one because they're both going to cancel and give you one. So that means that our period is equal to one. So what I can do on our graph is I can draw points every one unit on our x-axis.

So I'll go one, two, three to the right, and I'll go negative one, negative two, negative three to the left. Now the period of this graph is one, meaning I can draw these asymptotes for every one unit. So we're going to have an asymptote here at zero, we're going to have it at one, and then at two, and then at three. And these asymptotes are also going to repeat behind the graph. So I'll have an asymptote at negative one, negative two, negative three, 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. 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.

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. $b=\frac12$

Answer: (A)

Introduction to Cotangent Graph Example 2

Hey everyone. So let's see how we can go about solving this example. Here we are asked to graph the function y = -2∙cotx4. Now what I noticed 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 two here, which is going to change the amplitude or tallness of our graph. And we have this one fourth inside 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 going to ignore this amplitude and negative sign out front and just focus on y = cotx4.

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 π / b, where b is the number that is in front of the x. Since I can see that that's one-fourth, I get that we have π/14. You can flip this fraction and bring it to the top, which is going to give you 4π. So, the period is 4π.

Something that I know about the cotangent graph is that the cotangent graph has an asymptote that starts at an x-value of zero. Since we start at an x-value of zero and our period is 4π when we have this b value in front of x, that means that starting at zero, we're going to repeat every 4π units. I can draw another asymptote here at 4π, and I can draw another asymptote over there at 8π. 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 zero 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.

So this over here should be π and this should be 3π, and that makes sense. Do we have π, 2π, 3π, 4π? And at these values, you should see an output of one. So what we should have is an output right about there, which would be at one, and we should see an output right here, which would typically be at negative one. Adjusting our curve a little, this is what the function would look like for the cotangent of one fourth x, 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 cotangent of one-fourth x. We're trying to find -2∙cotx4. Now, this number out in front here, the two, is going to change the amplitude or basically vertically stretch our graph. So, rather than having an output of one when you're at π, you should actually have an output of two at this value.

And then, rather than having an output at negative one for 3π, you should actually have an output at negative two. Adjusting our curve, the graph should look a bit more stretched vertically when we incorporate this two. 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.

We should have a point right here at π − 2, and a point right here at 3π + 2. So for the graph y = -2∙cotx4, it should look something like this, where we start at the lower left, and then we go ahead and go up, reach a zero on the y-axis at 2π, and then we come up here and go to the top right. 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. This would be the graph of our function and the solution to the problem. I hope you found this video helpful.

Thanks for watching.

Here’s what students ask on this topic:

What are the key differences between the graphs of secant and cosecant functions?

The key differences between the graphs of secant and cosecant functions lie in the placement of their asymptotes and their relationship to sine and cosine. The secant function is the reciprocal of cosine, so its asymptotes occur where the cosine function equals zero, which are at odd multiples of π/2 (e.g., π/2, 3π/2). On the other hand, the cosecant function is the reciprocal of sine, so its asymptotes occur where the sine function equals zero, which are at integer multiples of π (e.g., 0, π, 2π). Both graphs exhibit repeating "smiley" and "frowny" shapes between their asymptotes, but their asymptote locations are distinct due to their reciprocal relationships with sine and cosine.

How do you determine the period of a tangent or cotangent function?

The period of a tangent or cotangent function is determined using the formula $\frac{\pi }{b}$, where $b$ is the coefficient of $x$ in the function. For example, in $y=\tan \left(\frac{\pi }{2}x\right)$, $b$ is $\frac{\pi }{2}$. The period is calculated as $\frac{\pi }{\frac{\pi }{2}}$, which simplifies to 2. This means the graph repeats every 2 units. For cotangent, the same formula applies, but the graph is flipped compared to tangent.

Why do secant and cosecant graphs have asymptotes?

Secant and cosecant graphs have asymptotes because they are the reciprocals of cosine and sine, respectively. Asymptotes occur at points where the sine or cosine function equals zero, as division by zero is undefined. For the secant function, asymptotes appear at odd multiples of π/2 (e.g., π/2, 3π/2), where cosine equals zero. For the cosecant function, asymptotes occur at integer multiples of π (e.g., 0, π, 2π), where sine equals zero. These asymptotes represent points where the secant or cosecant function approaches infinity or negative infinity, creating vertical lines on the graph.

How do transformations affect the graphs of tangent and cotangent functions?

Transformations affect the graphs of tangent and cotangent functions in similar ways to other trigonometric functions. Vertical stretches or compressions occur when the function is multiplied by a constant. Horizontal stretches or compressions are determined by the coefficient $b$ in the function, which changes the period using the formula $\frac{\pi }{b}$. Phase shifts occur when a constant is added or subtracted inside the function, shifting the graph left or right. Vertical shifts occur when a constant is added or subtracted outside the function, moving the graph up or down. These transformations allow for precise adjustments to the graph's shape and position.

What is the relationship between the tangent and cotangent graphs?

The tangent and cotangent graphs are closely related but have key differences. Both have a period of $\pi$ and repeat every $\pi$ units. However, the tangent graph has asymptotes at odd multiples of $\frac{\pi }{2}$ (e.g., $\frac{\pi }{2}$, $\frac{3}{\pi }$), while the cotangent graph has asymptotes at integer multiples of $\pi$ (e.g., 0, $\pi$, $2\pi$). Additionally, the tangent graph increases from bottom left to top right, while the cotangent graph decreases from top left to bottom right, making them mirror images of each other.

How do you graph the cosecant function using the sine function?

To graph the cosecant function using the sine function, follow these steps: 1) Start by graphing the sine function, as the cosecant is its reciprocal. 2) Identify the points where the sine function equals zero (e.g., 0, π, 2π). These are the locations of the vertical asymptotes for the cosecant graph, as division by zero is undefined. 3) At the peaks and valleys of the sine graph, plot the corresponding points for the cosecant graph, as the reciprocal of 1 is 1 and the reciprocal of -1 is -1. 4) Draw the "smiley" and "frowny" curves of the cosecant graph between the asymptotes, ensuring they approach infinity near the asymptotes. This method leverages the relationship between sine and cosecant for accurate graphing.