Graphs of Secant and Cosecant Functions - Video Tutorials & Practice Problems
On a tight schedule?
Get a 10 bullets summary of the topic
1
concept
Graphs of Secant and Cosecant Functions
Video duration:
6m
Play a video:
Hey, everyone. So up to this point, we spent a lot of time dealing with the sine and cosine functions and we've discussed their graphs in different ways that the graphs can be transformed like being stretched or shifted in some kind of way. Well, in this video, we're going to learn that we can graph more trigonometric functions such as the sequin 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 sin and cosine to graph the co and seek it. So without further ado, let's get right into this. Now, we're called that the co and seek are reciprocals, they're reciprocal identities for the sin and cosine. So cos it is one over the sign and seek. It is one over the cosine. And you can use these relations chips to figure out what their graphs look like. So I'm just going to jump right into the co sequent graph. And if I go ahead and take the reciprocal for all these outputs that I see the reciprocal of one is just one, the reciprocal of negative one is negative one because one divided by negative one would be negative one. And then the reciprocal of one is just one there. So that means at an X value of pi over two, we'll have one for three pi over two, we're going to have negative one and then for five pi over two, we're going to have one 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 gonna be one over 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 sign value, our co 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 pi and we'll have another Asymptote at an X value of two 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 and smaller as we go to the left. And as we go to the right, and what happens take the reciprocal of a number that gets smaller and smaller and smaller. Well, the whole fraction is going to get bigger and bigger. And because of this, our co 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, 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 frowning faces where we have the peaks and valleys respectively. Now let's also take a look and see how the C function behaves. Well, I can use the same logic by taking the reciprocal of all these values for the cosine. So the reciprocal of one is one and then we have the reciprocal of negative one and the reciprocal of one. So going to our graph, our points are going to be at one at the peak there at negative one for the valley at next value pi and then at another peak, we're going to have one as well. Now because see is a reciprocal of the cosine, all the places that we see zero for the cosine are going to be undefined for the second. So when doing this, we can actually draw asymptotes at all these undefined value. So at an X value of pi over two, we're going to have an Asymptote at three pi over two, we're going to have another Asymptote. And then at five pi over two, 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 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 pi to pi basically just in your multiples of pi. And for the sequent, we ended up with asymptotes at pi over 23 pi over 25 pi over two or basically just odd multiples of pi over two. Now, something that really nice about the co and seeking 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 co and seeking 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 cos of two X. 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 and then build off of that. Now, one of the things I know about the co is I know the cosecant is a reciprocal of the sign. And so what I'm actually going to do is first find the graph of Y equals the sine of two X. Now this is not the graph that we're looking for. But this will give me an idea as to what the coin is going to look like since I know their relationship. So recall that for the sine of two X. Well, we first need to figure out what the period is and the period for a sin or cosine graph is two pi divided by B. Now, the B value that we have in front is two. So we're going to have two pi divided by two 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 that we're going to cross through pi over two 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 three pi over two and then goes here to two pi and then we're going to reach another peak at five pi over two. And that's what our s of two X graph is going to look like. And to figure out what the co is going to look like. Well, we can just recognize that for Y equals the cosecant of two X. 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 what we're going to have is asymptotes for every place where this graph reaches zero or crosses the X axis. So we'll have an Asymptote here at zero. We'll have an Asymptote at pi over two. We'll have another Asymptote at pi we'll have an Asymptote at three pi over two and at two 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 frowning 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 peek 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 ska and coin. Let me know if you have any questions. Thanks for watching.
2
Problem
Problem
Below is a graph of the function y=sec(bx−π). Determine the value of b.
A
b=2
B
b=4
C
b=2π
D
b=π
3
Problem
Problem
Below is a graph of the function y=csc(bx). Determine the value of b.