 Welcome back to our lecture series, Math 1060 Trigonometry for Students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 12, I want to continue our discussion about trigonometric graphs. In previous lectures, we talked about the graphs of sine and cosine quite extensively. In this video, I want to talk about the graphs of tangent and cotangent, which naturally tangent makes sense to begin with. You can see the graph of tangent on the screen right here. Some important things to notice about tangent. First of all, unlike sine and cosine, tangent does have vertical asymptotes. There are locations, you can see here on the screen, for which tangent is undefined. Its domain is not all real numbers. The issue, of course, here is that tangent of x is the same thing as sine of x over cosine and so in particular, whenever cosine of x is equal to zero, that would imply you have a zero in the denominator of tangent, which would make tangent undefined. Now, when is cosine equal to zero? Well, as we've studied before, cosine is equal to zero at pi halves, three pi halves, five pi halves, negative pi halves, negative three pi halves. Basically, you're going to take these odd multiples of pi over two, like one pi over two, three pi over two, five pi over two, etc. That's when cosine is equal to zero and that'll make tangent undefined. At those locations, tangent will have a vertical asymptote, which you see here on the screen. Some other things to note is that tangent will equal zero exactly when sine is equal to zero, since that's the numerator of tangent. So when a sine equal to zero, that happens at zero, at pi, at two pi, at here, negative pi and negative two pi and the like. So again, integer multiples of pi, that's when tangent will equal zero. Those are some things to note. Another very important thing about tangent is that tangent's period is actually pi. Unlike cosine and sine, which are two pi periodic, tangent and cotangent are pi periodic. They repeat themselves every length of pi. So for example, if you go from this asymptote negative pi halves to this asymptote of pi halves, the length of that interval would be pi. You get pi halves plus pi halves, which is pi, and then you get this typical branch of your tangent curve. This branch is going to repeat itself over and over and over again for infinity and beyond. And that gives us the typical graph of tangent. And I'm going to switch over to Desmos for a second to show you why this is the shape of tangent. So how would we decide what the graph of tangent is going to look like? Well, we can look at specific values of tangent. Like when x equals zero, the tangent ratio at zero is itself equal to zero. So tangent is going to pass through the origin. Another important point you could consider would be pi force. At pi force, sine and cosine are actually both root two over two, so the ratio becomes one. And so we get the point right here, pi over four comma one, we get zero zero. And then by symmetry, we also get the point negative pi over four comma negative one. That's a point on tangent as well. When it came to graphing sine and cosine, I often told you you wanted to use five special points to graph one period, and then you could use that to graph the rest of the function. We want to do the same thing for tangent. Now, tangent's period is much smaller. It's only pi in length is half it half the size. So you'll notice that's of course why we're using pi force and negative pi force to help us. We didn't have to do that before when we did sine and cosine. Now, when it comes to tangent, the last two points of this collection of five won't actually be points. They'll be asymptotes. So the next one on the list would be pi halves, which there's an asymptote, and you also get negative pi halves, which gives us an asymptote as well. And so with these three points in the asymptotes, we can piece together a picture of what we would expect tangent to be. And so connecting the dots in a tangential way, we get the following picture here on the screen. Which is of course the the principal branch of a tangent curve. That is this is just one cycle. If we repeat this cycle over and over again, then we anticipate to get the picture we saw before, which is all of these tangent curves repeated. If we throw in the asymptotes, which are of course exceptions to the domain, then we get of course the graph of tangent that we saw previously. Now let's examine the graph of cotangent, which looks very similar to the graph of tangent, although it looks a little bit different. Of course, they aren't the same function. Some features to notice about cotangent. Well, since cotangent, of course, is the same thing as cosine over sine, just like tangent, we see that cotangent is undefined for various values. You see there's vertical asymptotes on its graph. This happens, of course, when sine is equal to zero. Because when sine is equal to zero makes the denominator zero, same problem as with tangent. When is sine equal to zero? Like we saw when we talked about the x-intercepts of tangent, the sine is equal to zero and multiples of pi. So you have zero pi, one pi, two pi, negative pi, negative two pi. These are going to be where the vertical asymptotes of cotangent are. In fact, every x-intercept of tangent coincides with a vertical asymptote of cotangent because these functions are reciprocal to each other. Also, where are the zeros? Where are the x-intercepts of cotangent? Well, that happens when cosine is equal to zero, which we saw earlier happens when you're at basically pi halves, three pi halves, five pi halves, negative pi halves, negative three pi halves like that. So in fact, the location of the vertical asymptotes of tangent are exactly where the x-intercepts of cotangent are. Again, this is a consequence of the two functions being reciprocals of each other. So while tangent, you have this type of shape that's increasing and is repeating itself over and over and over again, cotangent on its domain is always going to be decreasing, same basic shape, but again, it's a decreasing direction as opposed to increasing and it repeats itself every pi units. So again, the period here is going to be pi. As you go from zero to pi, that's clearly a length of pi right there. And so the period of cotangent is going to be pi. So let's again go to Desmos and take a look at why the graph of cotangent is what it is. What are some things we can say about cotangent? Well, this time, when you go at x equals zero, you don't have a point on the graph. You have a vertical asymptote. So we're going to add that to the screen. If we go to pi force, well, just like tangent, which is sine over cosine, cos tangent was just cosine over sine. At pi force, both cosine and sine are root two over two, so the ratio is going to equal one. So we get that point right there of pi force comma one. That point is shared by both of the two functions. When we go to pi halves this time, that makes cosine equal to zero and therefore cotangent will be zero as well. And then continuing on to the second quadrant there at three pi force, you'll get that cosine is negative root two over two, but sine is positive root two over two. And so you're going to get the cotangent ratio of negative one. And then continuing on to pi that will be another vertical asymptote because sine is zero at that location. So just like tangent, we're going to use these three points combined with the two vertical asymptotes to give us the basic shape of cotangent. That is, this is going to be a single cycle. And so connecting these three points and following the vertical asymptotes, you're going to get that very basic shape for cotangent. So then if we repeat this process over and over again, we get the shape of cotangent. And then of course, all of the vertical asymptotes need to be there as well. So with those observations of the graph of tangent and cotangent now addressed, let's start modifying the graphs of tangent and cotangent and go from there. So what if we want to graph the function y equals one half cotangent of negative two of x right there? Well, when it comes to trigometric functions, it's sufficient to graph one cycle. We can have one cycle, then we know we can graph as many cycles as we need to, we can extend it just by rubber stamping everything. And so we just want to graph one cycle of these things. And so what happens, what's happening here? What's the one half do? What's the negative two do? Well, like we talked about sine and cosine, whenever there is a negative sign in front, we need to deal with it. How do you deal with it? Well, cotangent is an odd function. So the negative sign does come out. So you get negative one half cotangent of two x. And if you're not sure why the negative sign comes out, remember with cotangent, if you have a cotangent of negative x, cotangent's the same thing as cosine of negative x over sine of negative x. Cosine, we should remember, is an even function. So the negative sign just disappears. Sine on the other hand is an odd function. So it spits the negative sign out. And so the net effect is you have a negative cotangent in front. You have a negative sign in front of the cotangent, I should say. So cotangent is an odd function. We're going to deal with that. If you can't remember that, just remember the symmetry of sine and cosine are going to be fine with all these other ones as well. So, okay, what's going on here then? So we have this negative sign. This negative sign suggests to us there's going to be a reflection across the x-axis. What does this one half mean? This one half means there's going to be a vertical compression, vertical compress by a factor of two. So the whole graph is going to get smaller by a factor of one half. And then what does this two here mean? This changes the period, right? And so like we've seen before, if the period, excuse me, this coefficient b is equal to two, then the period which this time is going to be pi over b. So with sine and cosine, we're talking about two pi over b. And that's because the standard period was two pi. For tangent and cotangent, the standard period is just a pi. And so we see that the period here is going to be pi halves. So we really want to label this up to pi halves. And so taking five points, I'm going to end here at five halves, pi halves, excuse me. So the middle point would be pi force. These other points we're going to have at pi eights and a three pi eights. Maybe label those as well. Three pi eights, like so. And so let's then see what then happens to the graph. So we've changed the period, which is now taken care of. We're going to graph one period on this interval right here. The vertical stretch is not going to affect the location of the vertical asymptotes. Neither will the reflection. The change of the period does, because now we're going to squish things. So instead of the standard period being from zero to pi, it's now going to be from zero to pi halves. And for cotangent, when you look at that principal branch, we have a vertical asymptote at zero. That's unaffected by this horizontal compression. But then the asymptote that would normally go through pi will now go through pi halves instead. So we're going to add these vertical asymptotes to our graph. Okay. Some other things to note here is that cotangent would normally have a x-intercept at pi halves, but because we squished the graph horizontally, that vertical asymptote, excuse me, that x-intercept is now going to be at pi force, like so. The standard cotangent would then go like this, but we've reflected this across the x-axis. So it actually needs to go the other way. Okay. If these values here are listed as increments of one-half, one-half, one, three halves. So we get negative one-half, negative one, negative three halves, like so. Then we see that the point that's normally at pi force one, it'll then have its x-coordinate be pi eight, and its y-coordinate will be one, but it's actually negative one. So we get a point right here. And then this one's going to have a y-coordinate of one right here. And so connecting together these points, fitting between the vertical asymptotes in the usual tangential way, we then get the graph of y equals one-half cotangent of negative 2x. Let's look at a graph here of tangent. This one has a lot of stuff going on here, but we can dissect all of it. We get y equals 3 plus 2 tangent of x over 2 plus pi over 8. Whenever it comes to these graphs, it's important that you factor the inside of the function, the so called horizontal zone. So we're going to factor away the coefficient that's in front of the x, so we're going to factor out a one-half that leaves behind x plus pi force because we took away the one-half there. And so now we can diagnose all these things. So the plus 3 right there is going to be a vertical shift, a shift up by 3. A, the 2 there, that's going to be a vertical stretch by a factor of 2. This one-half right here, what does it do? It's going to affect the period, right? So if B is equal to this one-half, we see the period is going to equal pi over one-half, that is to say 2 pi. So we actually elongated the graph. A one-half in there is actually doing a horizontal stretch. And then what is this, what is this pi force right here? This is going to be a shift to the left by pi force, like so. And so let's keep track of these things. So for tangent, when it comes to graphing tangent, we typically like to graph from negative pi halves all the way up to pi halves or something like that. That's typically what we want to do. And so we're going to start off by drawing the vertical asymptotes, which would be at pi halves and negative pi halves. But of course, things have changed, right? These shouldn't be pi halves anymore. This should be negative pi, and this should be pi. Because we've elongated the period, the period is now 2 pi, this graph will repeat itself from negative pi to pi. But since we're shifting everything to the left by pi force, it also makes sense that we move everything over by that little amount as well. And so with that in mind, let me label my x-coordinates here to show you what I want to be graphing. So I'm going to put this one to be pi halves. And this is going to be pi force. That would put this as 3 pi force. And then continuing on, we get a negative pi force right here, a negative pi halves right here, negative 3 pi force. The next one would be a negative pi. And then the one after that would be negative pi force, like so. So without the horizontal shift, if you ignore that for a second, because of the period change, we should have asymptotes at pi and negative pi. But because of the horizontal shift, we're going to move all of those to the left by pi force. So instead of having one at pi, the vertical asymptote is going to take place at 3 pi force, like so. And also, instead of having one at negative pi, it should be taking place at negative pi force. It just bumped everything to the left by one, like so. So we have our asymptotes. Then let's look for these middle points as well. Well, for a tangent function, it should have an x intercept at 0. But we changed things. Everything got moved up by 3. So in fact, our midline, our midline is now going to be y equals 3. So that's worth putting on the screen right there, our midline. So we're not going to worry about x intercepts anymore. We're going to see where do we intersect this midline. And so normally, for tangent, that it should be hitting its midline at x equals 0. But we would never think to the left by pi force. So we actually then get the point negative pi force, three, that should be then be on the graph. So how do we find the next point right here? Well, for a typical tangent, you go from the x intercept, you go one step, which is equal to pi force, and you go one up, which is equal to one right there. But we've stretched things, right? So there's a vertical stretch of two, and there's a horizontal stretch of two. So we really need to go two steps to the left, to the right, excuse me, and then two steps up. So we end up with this point right here of pi force, comma five. We then of course have negative pi force, comma three. We want to do the same thing, take two steps to the left and two steps down. So we end up with the point right here of negative three pi force, comma one. And as this is a tangent function, no reflection happens. If you connect these dots in the typical tangential way, you then get one cycle of the graph, y equals three plus two tangent x over two plus pi over eight. And so this gives us some examples on how we could graph tangent or cotangent functions with transformations.