Identifying the asymptote equations of transformed reciprocal trigonometric functions

0:00- [Instructor] We're told to identify the general equation of the vertical asymptote for the following functions. Like always, I encourage you to pause this video, have a go at it before we do it together. Now, before tackling these things that look kind of complicated, let's just do the simpler versions of these.

0:14So if we're thinking about cosecant, so cosecant of theta, notice, it's a version of this, but way simpler, that is the same thing as one over sine of theta. So you're going to have vertical asymptotes when sine of theta is equal to zero, because you would be attempting to divide by zero at those points.

0:37And so when does sine equal zero? Well, it's gonna be theta equaling zero, pi, two pi, so on and so forth. Or if you wanted to say that generally, it would be theta is equal to, we could say pi times k, where k is any integer. So k, I could write like this with math notation, k is a member of integers, if we do this kind of weird looking Z, bolded z.

1:08I could also write it, if you just wanna write it in simple terms, is where k is a member of integers either way. So k could be -1, -2, 0, 1, 2, 3, 4, 5. It could just keep going on and on and on. So let's use this information to now figure out the general equation of the vertical asymptotes for f of theta.

1:27So a few things we can do to simplify for our head. This two here scales the function up and down, but it, or not up and down, it pulls it up and down. It scales it in the vertical direction, but it doesn't change its asymptotes. Similarly, this negative one shifts the function down by one, but it does not change where the vertical asymptotes are.

1:45So we can ignore that as well. What we really just have to pay attention to is this factor right over here that is being multiplied by this expression. Actually, let me start with this expression right over here. We can replace a theta with theta plus two. If you do this, and this isn't true just of trig functions, it's true of any function, this shifts everything to the left by two.

2:10So we would shift these asymptotes to the left by two. And when you have this coefficient being multiplied by whatever your angle is, if you just took pi over four times theta, this makes you go through your periods faster. So what's going to happen is you are going to, if you are going to hit it at every pi k, you would then divide this by pi over four.

2:32So for this function over here, our vertical asymptotes are going to be at theta is equal to this expression, pi k, but we're gonna divide it by this coefficient right over here, by pi over four. And I said we're shifting everything to the left by two. Shifting everything to the left by two.

2:51So if we were to simplify this, this would, you see, a pi and a pi, you divide by one-fourth, it's the same thing as multiplying by four. You get four k minus two, where once again , k can be any integer. Those are the thetas where you're going to have a vertical asymptote.

3:07Now let's go to g of theta. And actually, let me just switch the colors, just to make this a little bit more colorful. So secant, just as a reminder, secant is, or secant of theta, I should say, secant of theta is just one over cosine of theta. So where does cosine of theta equal zero?

3:24Well, it's going to happen at pi over two, it's going to happen at three pi over two. We can keep going. Or if we just said that generally, it would be theta is equal to pi over two plus k pi, where k could be an integer. You could also have negative one times pi, negative two times pi, you could have a zero times pi right over here.

3:49So we could use this information to now look at this g of theta. Once again, this scaling factor we can ignore. Negative, it just flips it over the x-axis. The three just stretches it in the vertical direction, but it does not change where your vertical asymptotes are, so let's ignore that.

4:04So essentially if we look here, we're actually not changing the period either. We're just shifting everything to the right by three pi over four. So we'll just start with this. So it's gonna be theta is equal to pi over two plus k pi. And if we shift these vertical asymptotes to the right by three pi over four, plus three pi over four, this is going to be the same thing as, pi over two is the same thing as two pi over four.

4:32So we're gonna have k pi, k pi plus, two pi over four plus three pi over four is plus five pi over four. Once again, for, k can be any integer. This is strangely fun. All right, let's do this last one. So once again, we could ignore this thing that just shifts it up.

4:52It doesn't change where the vertical asymptotes are. Cotangent, just as a reminder, cotangent of theta is cosine of theta. It's the inverse, it's the reciprocal of tangent of theta, so it's cosine of theta over sine of theta. So you're gonna see vertical asymptotes wherever sine of theta is equal to zero, so really in the same place where you would see it for cosecant.

5:15So you would see it at theta is equal to pi k, for k being any integer. And now what do we see over here? Well, one, we have, we're gonna go through our periods five times faster or our periods are gonna be one-fifth as long. So theta is now, if we just had this five here and not this pi over seven, then it would be pi divided by this five times, or maybe I should say, well, yeah, let me write it, pi divided by five times k.

5:48But then we know if we replace our theta with theta plus pi over seven, this shifts, this shifts by, it shifts us to the left by pi over seven. So negative pi over seven. Whatever we have, we would now shift to the left. So minus pi over seven. And we're done.

6:07And in all of these cases, it's for k being any integer.