Understanding the tangent function

0:00- [Lecturer] What we're going to do in this video is explore the tangent function. We clearly have a fairly complicated expression involving tangent. But what I wanna do is start with a fairly simple, let's just call it, g of theta is equal to tangent of theta.

0:20You might already be familiar with what this looks like, but let me set up some graph, a graph axes right over here. And let me... So this is y, this is theta. So we're gonna graph y as a function of theta. I guess that's one way that you could think about it.

0:40And I'm gonna put some interesting values here. So let's say that this is pi over 2. This is 2pi over 2, which is just going to be pi. This is... I wanna get equal spacing. I'm just hand-drawing it. 3pi over 2. Then this would be 2pi. And I could go the other way.

1:02This would be - pi over 2. Make sure you can see that negative. This is -pi. This is -3pi over 2. You get the idea here. And we could keep going. It's not exactly evenly spaced. Maybe something like that is -2pi. Let me put some values here. Let me say that this is 1, 2, and 3, and this is -1,

1:33- 2, and -3. - 1 right over here. So let's just first try to graph regular tangent. And if we're just graphing regular tangent, let's remind ourselves what the definition of tangent is, especially when we think about it with respect to something like the unit circle.

1:52So if that's the x-axis, that's the y-axis. And I have some angle here, theta. The tangent of this theta is going to essentially be the slope of this terminal ray, where this angle is defined by this ray and the positive x-axis, the tangent of this is the slope of this.

2:12So you can see, when our theta is zero, our tangent of that is zero, because we just have a flat line right over there, or flat ray, I guess you could say. So we would just be right over there. But then, as you increase that angle, your slope is going to be right at 1 when you get to pi over 4.

2:33That's a 45-degree angle. So halfway here, your slope is going to be at 1, so the tangent is going to be 1. And then what happens as you get closer and closer to pi over 2? Well, the slope gets higher and higher and higher, and really, it starts to approach +infinity.

2:51So it approaches +infinity. It just skyrockets up, and you essentially have a vertical asymptote right here at pi over 2. Now, what happens if we go the other way? If we go negative angles, right at -pi over 4, we're gonna have a -1 slope. So that's going to be right over there. We have a -1 slope.

3:17And as we approach -pi over 2, our slope gets more and more negative, approaching -infinity. So we have another vertical asymptote right over there. And then that cycle just keeps repeating. You could verify that. I won't go into all of it. But essentially, I could keep drawing these vertical asymptotes with a period of pi, essentially.

3:43It goes from pi to pi, where we go up to pi, and then we go up here, and then we go, and then, as our angle increases a little bit more, then our slope turns super negative again, and then it gets more and more positive as we go all the way to 3pi over 2. So it goes super negative again, and then it...

4:03So it's going to look something like that, and it's gonna keep going. And I could do it over here as well. It's gonna look something like that. So that's just a straight-up tangent. Now, that's not what we have over here. We have all of this other things that are scaling it and shifting it around.

4:27Let's say I had tangent of 3theta here, what this would do is it would make us go through the angles. Whatever the angle is, it's multiplying it by 3. So you're going to cycle through the tangent values faster. So it would actually take this period and it would divide it by whatever this coefficient is, by 3.

4:46If it's 3theta, this would be three times faster. The period, the space right over here, would be 1/3 the distance in all of those situations. We don't have anything being multiplied by the theta in our example that we're trying to do, But it's good to point out in case you were.

5:04But what we do have is we have something scaling it. If you just had a negative, it would essentially flip it around. It would flip it over the x-axis. So if you had... Maybe I'll draw it. If I had, I'm running out of letters, h of theta is equal to -tangent of theta...

5:25And this is a unique to tangent of theta. This is true of any function. We would be flipping it over the x-axis. And I'm just gonna draw one period of it, just to... So this will intersect right over there, and then we would go in and we would hit that point there.

5:42It would look something like that. And it would take time, but I would draw it over every one of these intervals. Now, if I have -3 tangent of theta... let me do that in red. So if I said, I don't know, let me make another color here, m of theta is equal to -3 tangent of theta, well, there...

6:08Let me do this, actually, another layer on my program, 'cause I think it might be fun to shift around after I do that. Well, what essentially is going to happen is it's going to... It's the -tangent of theta. We're just multiplying everything by 3. So we're going to scale it vertically.

6:23So this value, instead of being 1... Actually, lemme look at the pink one, because that was the -tan theta. Instead of this being -1, it's now going to be -3. Instead of this being 1, it's now going to be 3. And so we're going to have a graph that looks something more like this.

6:45Something more like this. So that's -3tangent of theta. Now let's go to the original thing that we're trying to do. The difference between m of theta and f of theta is we have this -pi over 4 and we have this +2. And you probably have a little bit of experience already shifting and transporting functions.

7:07If you replace your theta with a theta minus pi over 4, or if you replace your x with an x minus pi over 4, it shifts it to the right by pi over 4, and then, if you just add a 2 here, it shifts you up by 2. So what's going to happen? So if we shift to the right by pi over 4, our asymptotes are going to shift to the right by pi over 4.

7:31So it might be right about there. This asymptote is going to be right over there. This asymptote, if we shift it by pi over 4, is going to get us right over there. This isn't drawn, obviously, perfectly to scale, but hopefully you're getting the idea. This, if we go pi more, we would've had an asymptote right here at -3pi over 2, but we now shifted pi over 4 to the right, so we're going to have an asymptote now right, whoops, right over there.

8:09So now, if I go to my original -3 tangent of theta... Let me copy and paste it. That's the one here in red. So let me Control + C, Control + V. All right. So I am going to shift it up by 2. So each of these things should be 2 higher. That's close enough. It's not perfectly drawn.

8:31And I'm going to be shifting it to the right. So it will get us right over there. And now, for this one, these things I drew in blue-green are the asymptotes. And I could keep copying and pasting it. I could do another version right over here. If I shift it to the right by pi, it will look something like this.

8:56And I could keep doing it. So these ones that I've now drawn, these are our f of theta. These are the f of theta. And I haven't drawn it over all of the periods, but hopefully you get the idea of how, one, tangent works, but how we can also transform it.