WEBVTT 00:00:06.500 --> 00:00:13.621 Let's take a look at the tangent function. Recall that the quotient identity for tangent x is equal to 00:00:13.623 --> 00:00:24.629 sine x over cosine x which tells us that the tangent function will be undefined when the cosine of x is equal to 0. 00:00:24.631 --> 00:00:34.889 Let's see when that happens. 00:00:34.891 --> 00:00:46.899 The cosine is equal to 0 when x is equal to 2n plus 1 over 2 times pi, which is another way of saying odd multiples 00:00:46.901 --> 00:00:53.145 of pi over 2. So as you can see in my table I know that this function is going to be undefined at pi over 2. 00:00:53.149 --> 00:00:59.899 Some other values, remember it's odd multiples of pi over 2, so that would be 3 pi over 2, 5 pi over 2, 00:00:59.901 --> 00:01:10.145 negative pi over 2, and so on. Next, the tangent function will be equal to zero, or the x-intercepts, when the sine 00:01:10.147 --> 00:01:23.399 function is equal to 0. Let's see when that happens. Well the sine is equal to 0 when x is equal to n times pi, 00:01:23.401 --> 00:01:30.152 and again in both of these cases n is any integer. So that means that the zeros or x-intercepts will occur at 0, 00:01:30.154 --> 00:01:36.688 pi, 2pi, negative pi, and so on. So you can see in my table I've got the value of (0, 0). 00:01:36.690 --> 00:01:45.441 So now I've listed some values for tangent function. The first four you should know. These are special values. 00:01:45.443 --> 00:01:54.205 I've plotted those points on my graph. What we want to see is that when we choose x-values closer and closer 00:01:54.207 --> 00:02:00.455 to pi over 2 the values of the tangent function increase without bound which means that there's a vertical asymptote 00:02:00.457 --> 00:02:13.954 at pi over 2. So let's put that on our graph. And so we know that this graph from 0 to pi over 2 looks like this. 00:02:13.956 --> 00:02:22.720 Now, the tangent function happens to be odd. Let's see why. If I look at the tangent of negative x, using the quotient 00:02:22.722 --> 00:02:30.499 identity again, that's going to be the sine of negative x over the cosine of negative x. You need to recall 00:02:30.501 --> 00:02:38.253 that the sine function is odd which means that the sine of negative x is negative sine x. Cosine function 00:02:38.255 --> 00:02:44.744 remember is even which means that the cosine of negative x is the same as the cosine of x. And I can see that 00:02:44.746 --> 00:02:54.273 that's going to equal negative tangent x. So we know the tangent function is odd. What does that mean? 00:02:54.275 --> 00:03:01.283 It means that the graph is symmetric about the origin, so we can get another piece of the graph. 00:03:01.285 --> 00:03:09.287 So remember it's undefined at negative pi over 2, and if I reflect this part about the origin the graph looks like this. 00:03:09.289 --> 00:03:18.527 So here's one cycle or period for the tangent graph between negative pi over 2 and pi over 2. We call this cycle 00:03:18.529 --> 00:03:25.031 the principal cycle, and there are a couple of key special points that you need to know for this graph. 00:03:25.033 --> 00:03:40.035 Let's draw those points right now. So at pi over 2 and negative pi over 2 the tangent function is undefined. 00:03:40.037 --> 00:03:52.584 It's zero at the origin, one of the special points. The point (pi over 4, 1) is an important point as well as 00:03:52.586 --> 00:04:06.596 (negative pi over 4, -1). So this graph you should know very well. It shows where the asymptotes are. It shows 00:04:06.598 --> 00:04:12.348 these three points. We call this the center point. It's obviously in the center of the cycle. These two points, 00:04:12.350 --> 00:04:33.099 we call them halfway points, and of course this is another halfway point. And this of course is the center point. 00:04:33.101 --> 00:04:40.866 So you need to know (negative pi over 4, -1), (0,0), (pi over 4, 1) for the tangent function and the asymptotes 00:04:40.868 --> 00:04:49.392 at negative pi over 2 and pi over 2. So now that we know what one cycle looks like, let's look at some more cycles 00:04:49.394 --> 00:04:52.893 and come up with some characteristics for the tangent function. 00:04:54.899 --> 00:04:59.887 Here I've drawn three cycles for the tangent function and we want to look at some of the characteristics of 00:04:59.889 --> 00:05:07.134 the graph. The first thing we want to observe is what the domain is. The domain is all real numbers, x, such that 00:05:07.136 --> 00:05:16.170 x cannot equal 2n plus 1 over 2 times pi, where n is any integer. Again, that's saying at all odd multiples 00:05:16.172 --> 00:05:23.925 of pi over 2 it's undefined. So that's the domain. So you can see pi over 2, 3 pi over 2, those are values that are 00:05:23.927 --> 00:05:32.181 not in the domain. The range of the function, look at the y-values, is negative infinity to infinity. The period, 00:05:32.183 --> 00:05:43.199 here we have three periods, here's another one. The period is pi, and the principal cycle we will call from 00:05:43.201 --> 00:05:48.194 negative pi over 2 to pi over 2. So these are very important concepts that you need to know about the 00:05:48.196 --> 00:05:57.460 tangent function. Let's look at some more. The tangent function remember is undefined and where it's 00:05:57.462 --> 00:06:06.720 undefined there are vertical asymptotes. They occur where x is equal to 2n plus 1 over 2 times pi, again odd 00:06:06.722 --> 00:06:14.220 multiples of pi over 2. Look at the graph. I have a vertical asymptote at pi over 2, 3 pi over 2, negative pi over 2, 00:06:14.222 --> 00:06:24.484 negative 3 pi over 2. The next one would be at ... 5 pi over 2. The y-intercept we can see is at the origin. 00:06:24.484 --> 00:06:34.237 The y-intercept is 0. For each cycle there are three important points. One of them is called the center point, 00:06:34.237 --> 00:06:44.244 so on these periods I can see I have a center point at (0, 0), center point at (pi, 0), center point at (negative pi, 0). 00:06:44.246 --> 00:06:52.492 The x-coordinates of those center points, the x-intercepts or the zeros, are of the form n pi, again where 00:06:52.494 --> 00:07:03.992 n is any integer, so again 0, pi, negative pi, 2 pi. And the y-coordinate of all those center points is 0. Next we have 00:07:03.994 --> 00:07:12.944 halfway points. For each cycle there are two halfway points. The halfway point to the left of the center point 00:07:12.946 --> 00:07:20.489 is located midway between the x-coordinate of the left vertical asymptote and the x-coordinate of the center point, 00:07:20.491 --> 00:07:28.737 and the y-coordinate of that halfway point to the left is negative 1. The halfway point to the right of the center point 00:07:28.739 --> 00:07:35.492 is located midway between the x-coordinate of the center point and the x-coordinate of the right vertical asymptote, 00:07:35.494 --> 00:07:46.065 and the y-coordinate of those halfway points is 1. Next, we see that the tangent function is odd. That means the 00:07:46.067 --> 00:07:53.313 graph is symmetric about the origin. And lastly, the graph of each cycle we can see is one-to-one. Remember your 00:07:53.315 --> 00:08:03.069 horizontal line test. If I look at one cycle, it passes the horizontal line test, so each cycle is one-to-one. 00:08:03.071 --> 00:08:10.063 You need to be very familiar with these characteristics and be able to graph one period for the tangent function. 00:08:11.089 --> 00:08:20.343 Let's take a look at sketching functions of the form y equals A tan (Bx minus C). Recall that the principal cycle 00:08:20.345 --> 00:08:28.590 for y equals tangent x was negative pi over 2 to pi over 2. When you're sketching functions in this form, the 00:08:28.592 --> 00:08:35.867 principal cycle will occur when (Bx minus C) is between negative pi over 2 and pi over 2. So in this example 00:08:35.869 --> 00:08:42.371 we have y equals 2 tan (x minus pi over 4), and we want to sketch this graph. So the first thing we're going to 00:08:42.373 --> 00:08:54.122 do is notice that B is positive. In this case B is 1. C is pi over 4. If B is negative you would have to rewrite it. 00:08:54.124 --> 00:09:00.140 Remember that the tangent function is odd to make the B positive, and then you would begin. So in this case 00:09:00.142 --> 00:09:08.893 our B is positive, so we take our (Bx minus C) which in this case is (x minus pi over 4), and we put it between 00:09:08.895 --> 00:09:15.889 negative pi over 2 to pi over 2 to determine the principal cycle for this function. So we want to solve for x. 00:09:15.891 --> 00:09:24.413 We want to add pi over 4 to each side, so that would give me negative pi over 4 less than x. Pi over 2 plus 00:09:24.415 --> 00:09:36.676 pi over 4 would be 3 pi over 4. So here's what I know. I know the principal cycle for this function is going to be 00:09:36.678 --> 00:09:48.681 negative pi over 4 to 3 pi over 4. Notice that the period, the length of that interval, is still pi. You can always 00:09:48.683 --> 00:09:56.697 find the period by taking pi and dividing by B. Remember the normal period for tangent is pi, and of course 00:09:56.699 --> 00:10:06.942 in this case, B is 1. So we can see the period is pi. So now we know the principal cycle. We can figure out where 00:10:06.946 --> 00:10:20.445 the left and the right vertical asymptotes occur for this principal cycle. The left vertical asymptote is going to be 00:10:20.447 --> 00:10:43.491 at x equals negative pi over 4, and the right vertical asymptote will occur when x is equal to 3 pi over 4. 00:10:43.493 --> 00:10:54.501 So let's go ahead and put that on our graph. So let's see, we have negative pi over 4 and 3 pi over 4, and 00:10:54.503 --> 00:11:04.506 I know those are where my vertical asymptotes are. Next we need to find those three key points. Let's start 00:11:04.508 --> 00:11:15.257 with the center point. Recall that the center point is located, the x-coordinate is going to be located midway 00:11:15.259 --> 00:11:21.280 between the x-coordinate of the left vertical asymptote and the x-coordinate of the right vertical asymptote. 00:11:21.282 --> 00:11:29.037 So that means our x-coordinate will be, you can add up these coordinates and divide by 2, which gives me 00:11:29.037 --> 00:11:38.035 2 pi over 4 divided by 2, which is pi over 4. So the x-coordinate of the center point is pi over 4. 00:11:38.037 --> 00:11:49.537 The y-coordinate of the center point is 0, so the point that I have now, my center point, let's write it up here, is 00:11:49.537 --> 00:12:03.045 (pi over 4, 0). So let's go ahead and plot that point (pi over 4, 0) is my center point. Next we need to find the 00:12:03.047 --> 00:12:10.827 two halfway points. Let's find the halfway point to the left of this center point. To get the x-coordinate of the 00:12:10.829 --> 00:12:19.500 halfway point to the left, it's going to be located halfway between this left vertical asymptote and the x-coordinate 00:12:19.502 --> 00:12:28.003 of the center point. So the x-coordinate will be negative pi over 4 plus pi over 4, and again you divide that by 2. 00:12:28.005 --> 00:12:38.245 Well, the numerator is 0. 0 divided by 2 is 0. The y-coordinate, well, remember the y-coordinate for the normal 00:12:38.247 --> 00:12:44.996 tangent was going to be down at negative 1. So what you do to get the y-coordinate of this halfway point, 00:12:44.998 --> 00:12:53.991 you're going to multiply that y-coordinate of negative 1 for the basic tangent, and you're going to multiply it by A. 00:12:53.993 --> 00:13:01.288 Notice in this case, my A is 2, so I'm going to do negative 1 times 2, which gives me negative 2. So this 00:13:01.290 --> 00:13:16.571 halfway point would be (0, -2). Now we need to find the halfway point to the right of the center point. Well again, 00:13:16.573 --> 00:13:22.061 the x-coordinate will be located midway between this x-coordinate of the center point and the x-coordinate 00:13:22.063 --> 00:13:31.820 of the right vertical asymptote. So that's going to be pi over 4 plus 3 pi over 4, and again divide by 2 so that 00:13:31.822 --> 00:13:42.080 gives you pi divided by 2 is going to be the x-coordinate of this halfway point. The y-coordinate is going to be, 00:13:42.082 --> 00:13:50.090 recall the y-coordinate to the right of the normal tangent, y equals tan x, is positive 1. So you're going to multiply 00:13:50.092 --> 00:13:58.601 that positive 1 times A, which in this case is 2, so the y-coordinate of this halfway point is going to be 2. 00:13:58.601 --> 00:14:15.117 So this point will be (pi over 2, 2). So we'll plot that point. So I can see that this sketch, the graph will look like this. 00:14:15.119 --> 00:14:21.615 This is what this one cycle will look like for this particular tangent function. It's very important that you remember 00:14:21.617 --> 00:14:29.866 how to find these points and how you find the principal cycle and the left and the right vertical asymptotes. 00:14:31.383 --> 00:14:39.630 Let's look at this example. We have y equals negative tan (1/2 x plus pi), and we want to sketch this graph. 00:14:39.632 --> 00:14:46.144 The first thing we want to determine is the principal cycle. The principal cycle will occur when (Bx minus C), 00:14:46.146 --> 00:14:57.908 which in this case is (1/2 x plus pi), is between negative pi over 2 and pi over 2, which you recall negative pi over 2 00:14:57.910 --> 00:15:11.158 to pi over 2 is the principal cycle for tangent. Now we want to solve this for x. So first I want to subtract pi. 00:15:11.160 --> 00:15:26.661 So let's see. Now I need to multiply by 2. So now I know that the principal cycle for this function is negative 3 pi 00:15:26.663 --> 00:15:48.158 to negative pi. Now that I know the principal cycle I know that the left vertical asymptote is x equals negative 3 pi 00:15:48.160 --> 00:16:06.912 and the right vertical asymptote will be x equals negative pi. 00:16:06.914 --> 00:16:13.911 So again to determine the principal cycle, you take the argument (Bx minus C), you put it between negative 00:16:13.913 --> 00:16:20.489 pi over 2 and pi over 2, and you solve. That will give you the principal cycle. The left of the principal cycle tells 00:16:20.491 --> 00:16:27.495 you the left vertical asymptote, and the right gives you the right vertical asymptote. Now let's find the three 00:16:27.497 --> 00:16:40.011 special points. Let's find the center point first. Well, recall the center point is located midway between the 00:16:40.013 --> 00:16:47.771 left and the right, the x-coordinate of the center point is located midway between the left and the right 00:16:47.773 --> 00:16:59.284 vertical asymptotes. So if I take negative 3 pi plus a negative pi divided by 2, that gives me negative 2 pi 00:16:59.286 --> 00:17:09.543 for the x-coordinate of the center point. The y-coordinate of the center point is 0. So my center point let's write 00:17:09.545 --> 00:17:21.291 as (negative 2 pi, 0). So let's go over to our graph, and let's show what we have so far. Let's put in the left 00:17:21.293 --> 00:17:32.816 vertical asymptote. So let's see, that's negative pi. That's negative 3 pi, so I know I have a vertical asymptote 00:17:32.818 --> 00:17:45.330 here and here. And I know the center point is (negative 2 pi, 0). Now we need to find the halfway points. 00:17:45.332 --> 00:17:53.083 Well, the halfway point to the left of the center point, the x-coordinate is located midway between this 00:17:53.085 --> 00:18:00.835 vertical asymptote and the x-coordinate of the center point. So that will be, the x-coordinate will be negative 3 pi 00:18:00.837 --> 00:18:10.854 plus a negative 2 pi, divided by 2. Well that's negative 5 pi over 2 is the x-coordinate of this halfway point. 00:18:10.856 --> 00:18:19.610 The y-coordinate, we have to remember what the principal cycle is for tan x. Remember the halfway point to 00:18:19.612 --> 00:18:25.619 the left in y equals tan x, the y-coordinate to the left of the center point is negative 1, so we take that 00:18:25.619 --> 00:18:34.375 negative 1 value and multiply it by our A. Notice in this case our A is negative 1, so we have a negative 1 00:18:34.377 --> 00:18:51.380 times a negative 1 which of course is 1, so this halfway point is (negative 5 pi over 2, 1). So let's show that. 00:18:51.382 --> 00:18:57.400 Ok, now we have one more halfway point to find, the one of course to the right of the center point. Its 00:18:57.402 --> 00:19:03.652 x-coordinate is going to be located midway between the x-coordinate of the center point and this vertical 00:19:03.654 --> 00:19:13.001 asymptote, so that's going to be negative 2 pi plus a negative pi, over 2, that's negative 3 pi over 2 is the 00:19:13.003 --> 00:19:23.012 x-coordinate of this halfway point. And the y-coordinate you find by taking the y-coordinate for the basic tan x, 00:19:23.014 --> 00:19:32.521 which is positive 1, times A and again A is negative 1, so the y-coordinate of this halfway point is negative 1. 00:19:32.523 --> 00:19:48.038 So this point would be (negative 3 pi over 2, -1). Let's plot that point. Ok, so now we've got 00:19:48.041 --> 00:19:59.048 the center point, our halfway points, our vertical asymptotes, so here is one cycle for this tangent function. 00:20:01.000 --> 00:20:09.558 Let's take a look at the cotangent function, y equals cotangent x. First thing I want to use is, remember that 00:20:09.560 --> 00:20:18.560 the cotangent function using the quotient identity is the same thing as cosine x over sine x. So this means 00:20:18.562 --> 00:20:31.082 that the cotangent function will be undefined when sine x is equal to 0. And when does that happen? 00:20:31.084 --> 00:20:50.349 That occurs when x is equal to n times pi, where n is an integer. So it's undefined for example at 0, pi, 2 pi, 00:20:50.351 --> 00:21:03.104 negative pi. I've listed those in my table. The next thing is it's equal to 0, or the x-intercepts occur, when 00:21:03.106 --> 00:21:17.874 cosine x is equal to 0, and that occurs when x is equal to 2n plus 1 over 2 times pi, or odd multiples of pi over 2. 00:21:17.876 --> 00:21:27.137 So at pi over 2 it's 0. The next one would be 3 pi over 2, 5 pi over 2, and so on. So I've got that point. 00:21:27.139 --> 00:21:34.146 Pi over 4 and 3 pi over 4, those are some special values you should know. The cotangent of pi over 4 is 1. 00:21:34.148 --> 00:21:41.647 The cotangent of 3 pi over 4 is negative 1. So I've plotted those three points. Where the cotangent function is 00:21:41.649 --> 00:21:47.404 undefined, that's where your vertical asymptotes will be. So there's one here on the y-axis, and there's one 00:21:47.406 --> 00:22:02.405 here at x equals pi. So one cycle for this cotangent will look like this. We can see the length of that interval is pi, 00:22:02.407 --> 00:22:13.432 so the period is pi. And we call this cycle from 0 to pi, we call this the principal cycle for cotangent. 00:22:13.432 --> 00:22:19.936 You need to know where those asymptotes are, and then let's look at these points. There are three special points 00:22:19.938 --> 00:22:27.695 for the cotangent. We have the center point located at (pi over 2, 0). We have a halfway point to the left of that, 00:22:27.697 --> 00:22:36.704 (pi over 4, 1), and our other halfway point to the right of the center point is (3 pi over 4, -1). It's very important 00:22:36.706 --> 00:22:44.207 that you learn this graph, that you can draw one cycle including the asymptotes and the three important points. 00:22:45.955 --> 00:22:52.199 Here I've drawn three cycles for the cotangent function, and we want to look at the characteristics of the 00:22:52.201 --> 00:23:00.707 cotangent function. The first is the domain, that's the values of x. x can be any real number except n pi, 00:23:00.709 --> 00:23:10.758 where n is any integer. So that means for example it's undefined at 0, pi, 2 pi, negative pi, and all multiples of pi. 00:23:10.760 --> 00:23:19.027 So that's the domain. Next let's look at the range, the y-values. Well we can look at a cycle, and we can see that 00:23:19.029 --> 00:23:28.789 the y-values go from negative infinity to infinity. The period for the cotangent function we can see is pi. This is 00:23:28.789 --> 00:23:35.292 what we call the principal cycle. We can see another period over here, so you need to know that the period 00:23:35.294 --> 00:23:45.050 for cotangent is pi, and you need to know that the principal cycle for the cotangent function is 0 to pi. 00:23:45.052 --> 00:23:55.051 From here to here is the principal cycle. Now, let's look at some other characteristics. Because it's undefined, 00:23:55.053 --> 00:24:01.570 there are vertical asymptotes at all values x equals n pi, so on this graph you can see I have a vertical asymptote 00:24:01.572 --> 00:24:10.569 at x equals 0, x equals pi, x equals 2 pi, and x equals negative pi. All multiples of pi there will be vertical 00:24:10.571 --> 00:24:17.072 asymptotes, and you need to know where they occur. We can see from the graph that there's no y-intercept 00:24:17.074 --> 00:24:25.568 since x cannot equal 0. Now for each cycle there are three special points, and you need to learn these 00:24:25.570 --> 00:24:38.070 points and be able to plot them. Let's talk about the center point first. The center point is located halfway 00:24:38.072 --> 00:24:44.569 between the left vertical asymptote and the right vertical asymptote. So the x-coordinate is right in the middle 00:24:44.571 --> 00:24:52.829 between the left and the right vertical asymptote. And the y-coordinate of the center point is 0. So all of these, 00:24:52.831 --> 00:24:59.379 you can see three center points. They always occur, the x-coordinate is midway between the left vertical 00:24:59.381 --> 00:25:07.876 asymptote and the right vertical asymptote and the y-coordinate is always 0. Now let's look at the halfway points 00:25:07.878 --> 00:25:16.388 for each cycle. Let's look at the halfway point to the left of the center point. Let's look at this one. 00:25:16.390 --> 00:25:24.386 The x-coordinate of this halfway point is located midway between the left vertical asymptote and the x-coordinate 00:25:24.388 --> 00:25:33.892 of the center point, and the y-coordinate of the halfway point is 1. Next, let's look at the halfway point to the right 00:25:33.894 --> 00:25:41.665 of the center point. Let's look at this one. Its x-coordinate is located midway between the x-coordinate of the 00:25:41.667 --> 00:25:49.428 center point and the right vertical asymptote, and the y-coordinate of this halfway point is negative 1. 00:25:49.430 --> 00:25:55.937 So you need to know how to plot and find those halfway points and center points. 00:25:55.939 --> 00:26:03.193 Next we can see that the graph of the cotangent is odd. I can see that the graph is symmetric about the origin. 00:26:03.195 --> 00:26:12.200 Remember, because it's odd that means that the cotangent of negative x is the opposite of the cotangent of x. 00:26:12.202 --> 00:26:18.454 And lastly, we can see that for each cycle for the cotangent function that it passes the horizontal line test 00:26:18.456 --> 00:26:28.460 which means that for each cycle the cotangent function is one-to-one on each cycle. 00:26:28.462 --> 00:26:36.204 You need to learn all these characteristics of the cotangent function and be able to graph at least one period 00:26:36.206 --> 00:26:41.955 showing the asymptotes and showing the center point and the halfway points. 00:26:43.200 --> 00:26:50.703 Let's take a look at sketching functions of the form y equals A cotangent of (Bx minus C). In this example, 00:26:50.705 --> 00:26:56.993 we want to sketch the graph of 3 cotangent (x plus pi over 3). When you have functions in this form the 00:26:56.995 --> 00:27:04.495 first thing you want to note is if B is positive. If B is not positive, then you would rewrite this function using the 00:27:04.497 --> 00:27:12.510 fact that the cotangent is odd. In this case, I can see my B is 1, so we don't need to rewrite it. Recall that 00:27:12.512 --> 00:27:19.756 the cotangent function the principal cycle for y equals cotangent x was between 0 and pi. So in order to find 00:27:19.758 --> 00:27:26.511 the principal cycle for functions in this form, we're going to let (Bx minus C), in this case (x plus pi over 3), 00:27:26.513 --> 00:27:35.261 we're going to put it between 0 and pi. This will give us the principal cycle for this function. Solving for x, 00:27:35.263 --> 00:27:56.055 subtracting the pi over 3, we can see that the principal cycle for this function is negative pi over 3 to 2 pi over 3. 00:27:56.057 --> 00:28:08.056 Now the length of that interval would be the period, and the length of that interval is pi. We can always find the 00:28:08.058 --> 00:28:17.075 period by taking pi, the normal period for cotangent x, and dividing by B. In this case B of course is 1, so I can see 00:28:17.077 --> 00:28:31.339 that that's pi. Now that we have the principal cycle we can find the left vertical asymptote and the right. 00:28:31.341 --> 00:28:46.356 The left vertical asymptote will occur when x is equal to negative pi over 3, and the right vertical asymptote 00:28:46.358 --> 00:29:00.700 would be when x is equal to 2 pi over 3. Now we want to find the center point. Well, the center point is 00:29:00.702 --> 00:29:06.622 located midway between the left and the right vertical asymptotes. Let's go ahead and put our asymptotes 00:29:06.624 --> 00:29:21.133 on the graph now. So we have negative pi over 3. Let's see, if this is 2 pi over 3, then about right here. So I have 00:29:21.135 --> 00:29:31.387 an asymptote here, and I have an asymptote here. So the center point is located midway between the left 00:29:31.389 --> 00:29:37.407 and the right vertical asymptote. Let's find the x-coordinate. Well, I don't need a right coordinate. Let's just write 00:29:37.409 --> 00:29:52.413 x equals. So the x-coordinate will be negative pi over 3 plus 2 pi over 3, divided by 2. Pi over 3 divided by 00:29:52.415 --> 00:30:07.663 2 is pi over 6, and the y-coordinate of this center point is 0. So the center point is (pi over 6, 0). So remember 00:30:07.665 --> 00:30:19.370 it's midway between so let's try to eyeball. That looks about right, so (pi over 6, 0) is my center point. 00:30:19.372 --> 00:30:28.213 Now we need to find the halfway points. Well, let's find the halfway point to the left of the center point. 00:30:28.215 --> 00:30:35.963 The x-coordinate to the left of the center point is located midway between the left vertical asymptote and this 00:30:35.965 --> 00:30:43.215 x-coordinate of the center point. Well, let's see what we get. That's negative 2 pi over 6. So negative pi over 6 00:30:43.217 --> 00:30:57.230 divided by 2 is negative pi over 12. Now we need to find the y-coordinate. Well in order to find the y-coordinate 00:30:57.232 --> 00:31:03.991 we have to recall when you look at the graph of y equals cotangent of x the halfway point to the left has a 00:31:03.993 --> 00:31:12.993 y-coordinate of positive 1, so we want to take that y-coordinate of 1 and we want to multiply it by A. In this case 00:31:12.995 --> 00:31:25.015 my A is 3, so I want to multiply 1 times 3 which of course is 3, so this halfway point is (negative pi over 12, 3). 00:31:25.017 --> 00:31:37.520 So, looks about right here, negative pi over 12, and we want to be up at 3. Now we need to find the halfway 00:31:37.522 --> 00:31:43.518 point to the right of the center point. Well the x-coordinate for this halfway point is going to be located midway 00:31:43.520 --> 00:31:53.266 between the x-coordinate of the center point and the right vertical asymptote, so let's see what that gives us. 00:31:53.268 --> 00:32:03.324 Pi over 6 plus 2 pi over 3 is the same thing as 4 pi over 6. I have pi over 6 plus 4 pi over 6 is 5 pi over 6. 00:32:03.326 --> 00:32:14.081 Well that gives me 5 pi over 12 is the x-coordinate. And then to find the y-coordinate we need to remember 00:32:14.083 --> 00:32:21.339 again the y-coordinate for the halfway point for the basic cotangent is negative 1, so we need to do 00:32:21.341 --> 00:32:33.588 negative 1 times our A, and again my A is 3, so this will be negative 3. So this halfway point is (5 pi over 12, -3), 00:32:33.590 --> 00:32:45.309 and it's located halfway between there. So this would be 5 pi over 12, and then I need negative 3. So this is the 00:32:45.311 --> 00:32:58.875 sketch of this particular cotangent function. Once again for cotangent it's very important that you remember that 00:32:58.877 --> 00:33:06.630 your (Bx minus C) has to be between 0 and pi because that's the principal cycle for the normal cotangent. 00:33:06.632 --> 00:33:12.131 Also make sure you know how to find the center point and the halfway points. 00:33:13.887 --> 00:33:20.633 Let's take a look at the cosecant function. Recall that the cosecant function is the reciprocal of the sine 00:33:20.635 --> 00:33:29.388 function, so it's equal to 1 over the sine of x. In order to get the graph of the cosecant, what we're going to do 00:33:29.390 --> 00:33:36.635 is first look at the sine function and then use the reciprocal identity. So let's do this first. Let's lightly sketch 00:33:36.637 --> 00:33:43.423 the sine function over here. You should remember the sine function very well. Its period is 2 pi. 00:33:43.425 --> 00:34:08.948 It's (0, 0), (pi over 2, 1), (pi, 0), (3 pi over 2, -1), and then (2 pi, back to 0). Of course we can draw another period. 00:34:08.950 --> 00:34:18.197 Ok, so we've sketched sine x, two periods for the sine x. If the cosecant, which it is, is the reciprocal of the sine 00:34:18.199 --> 00:34:31.960 function. First of all, notice that the cosecant is going to be undefined when the sine of x is 0, 00:34:31.962 --> 00:34:46.983 and the sine of x is equal to 0, we can see, when x is equal to n times pi, where n is an integer. 00:34:46.985 --> 00:34:54.732 Now when the sine of x is 0, that's going to give you your vertical asymptotes. So we know at all multiples of n pi 00:34:54.734 --> 00:35:00.951 there are going to be vertical asymptotes. So I know there's going to be an asymptote here on the y-axis, here 00:35:00.953 --> 00:35:19.014 at pi, here at 2 pi, negative pi, and then negative 2 pi. Now, notice that the maximum point for the sine function is 00:35:19.016 --> 00:35:28.776 (pi over 2, 1). The reciprocal of 1 is 1, so this point will also be a point on the cosecant function. Look at all 00:35:28.778 --> 00:35:36.032 these values for sine x between 0 and pi, the sine function, the values are between 0 and 1. If you take the 00:35:36.034 --> 00:35:41.785 reciprocal of those values you get numbers bigger than 1, and it has to approach those asymptotes. So this portion 00:35:41.787 --> 00:35:49.535 of the graph looks like this. Similarly, over here at 3 pi over 2, the sine function is negative 1. The reciprocal 00:35:49.537 --> 00:35:57.043 of negative 1 is also negative 1. So this is going to be a point on the cosecant function. And all these values of sine 00:35:57.045 --> 00:36:03.060 between negative 1 and 0, the reciprocal of those values are numbers less than negative 1, and it of course 00:36:03.062 --> 00:36:10.016 approaches the asymptote, so this portion of the cosecant graph looks like this. So here's one cycle. 00:36:10.018 --> 00:36:23.328 Let's draw another one. Now let's look at some characteristics of the cosecant function. 00:36:25.030 --> 00:36:32.290 Let's look at the characteristics of the cosecant function. First of all, remember that when the sine of x is 0 00:36:32.293 --> 00:36:41.549 it's undefined, so the domain is all real numbers except x equals n pi, where n is an integer. The range... 00:36:41.551 --> 00:36:51.807 The domain is all real numbers except n pi. The range, we need to look at the y-values. So let's look at the y-values. 00:36:51.809 --> 00:37:00.069 Goes down to negative infinity, goes up to negative 1. And then it picks up at 1 and goes to infinity. 00:37:00.071 --> 00:37:09.579 So the range is this. The period is the same as the period for sine which is 2 pi, and because it's undefined it has 00:37:09.581 --> 00:37:17.579 infinitely many vertical asymptotes at x equals n pi, again where n is an integer. You can see these on the graph. 00:37:17.581 --> 00:37:26.598 So asymptotes here, I've shown at negative 2 pi, negative pi, 0, pi, and 2 pi. Now there are a couple of other points. 00:37:26.600 --> 00:37:34.359 For each cycle there are two important points for the cosecant. Let's look at this one first. This point is a 00:37:34.361 --> 00:37:41.862 relative maximum. This value of negative 1 is a relative max, because in this little neighborhood that is the highest 00:37:41.864 --> 00:37:51.369 y-value. Where do they occur? Well it occurs at 3 pi over 2, plus of course 2 pi times n, where n is any integer. 00:37:51.371 --> 00:38:00.626 Adding the period gets me to the next one. The maximum value is of course the y-value, which is negative 1. 00:38:00.628 --> 00:38:09.137 Now let's look at this point. This point is where the relative min occurs. The x-coordinate where it occurs is at 00:38:09.139 --> 00:38:16.390 pi over 2, and of course to get to the next one you add 2 pi, the period, times n, again where n is an integer. 00:38:16.392 --> 00:38:27.657 The minimum value is 1. So those are two important points to know about. So let's see. Let's underline them, 00:38:27.659 --> 00:38:39.156 the relative max and the relative min, and it's important what the y-value is. And the function is odd, 00:38:39.158 --> 00:38:44.906 which means that the cosecant of negative x is negative cosecant of x, which means the graph is symmetric 00:38:44.908 --> 00:38:52.406 about the origin. You need to learn this graph and be able to graph it with showing the two points on each cycle 00:38:52.408 --> 00:38:54.406 and the asymptotes. 00:38:55.906 --> 00:39:03.942 Let's look at sketching functions of the form y equals A cosecant of (Bx minus C). Here my example is y equals 00:39:03.944 --> 00:39:11.198 negative 3 cosecant of (x minus pi). To graph functions in this form, we're going to graph its reciprocal function, 00:39:11.200 --> 00:39:19.951 A sine (Bx minus C), that you should be very good at graphing. So our first step is going to be to sketch the graph 00:39:19.953 --> 00:39:32.471 of negative 3 sine of (x minus pi). For this graph we know the period is 2 pi because B is 1. We know there's a 00:39:32.473 --> 00:39:46.724 phase shift of pi to the right, and we know the amplitude is the absolute value of negative 3 which is 3. 00:39:46.726 --> 00:39:55.728 So let's sketch that graph lightly first. So the shift is pi. So remember, let me make some tick marks here. 00:39:55.730 --> 00:40:15.974 So let's see, pi and we need 3. Ok, so since the phase shift is pi, that means (pi, 0) is going to be a point 00:40:15.976 --> 00:40:22.522 for the sine. And then because A is negative, remember it's going to go down first, so at negative 3 pi over 2 00:40:22.524 --> 00:40:36.766 it's down at negative 3, 0, up to 3, back down to 0. Alright, so there's one cycle for the sine. Let's go ahead and 00:40:36.768 --> 00:40:57.523 sketch another one. So this is negative pi, so it's going to go up to 3, back to 0, down to negative 3, back to 0. 00:40:57.525 --> 00:41:09.060 Ok, now we have two periods for the reciprocal function negative 3 sine of (x minus pi). We want to get the 00:41:09.062 --> 00:41:17.570 graph of the cosecant. If the sine is 0, then we know we're going to have vertical asymptotes. So in this case, 00:41:17.572 --> 00:41:27.833 every time the sine is 0, here at negative pi, 0, pi, 2 pi, 3 pi, that's where your asymptotes are going to be. So 00:41:27.835 --> 00:41:43.605 we have an asymptote here, we have one on the y, we have one at pi, 2 pi, and 3 pi. Ok, so now that we know 00:41:43.607 --> 00:41:55.364 where the asymptotes are, we want to figure out the two points on each cycle. This point right here, this 00:41:55.366 --> 00:42:04.617 (pi over 2, 3), we can see is a relative max for the sine function. That's going to be a relative min for the cosecant. 00:42:04.619 --> 00:42:20.894 And this point, which is, let's go ahead and label it. It was 3 pi over 2, it's down at negative 3. That minimum 00:42:20.896 --> 00:42:36.404 point for the sine becomes a maximum point for the cosecant. Let's draw a graph. Ok. So we've got our points 00:42:36.406 --> 00:42:58.679 labeled. The next thing we want to know is what are the relative maximum points for the cosecant? Well, let's see 00:42:58.681 --> 00:43:05.678 where they are. This point is a relative maximum point and this is a relative maximum point. What point is that? 00:43:05.680 --> 00:43:29.177 It occurs at (negative pi over 2, -3), and then this one is at (3 pi over 2, -3). Ok and for these two periods those are 00:43:29.179 --> 00:43:35.672 the only two relative maximum points. Now what are the relative minimum points? Well for these two cycles 00:43:35.674 --> 00:43:56.423 there are two relative minimum points. Let's go ahead and list those. This one is at (pi over 2, 3) and this one 00:43:56.425 --> 00:44:13.176 is at (5 pi over 2, 3). So we have determined the easiest way to graph the cosecant is always to graph its 00:44:13.178 --> 00:44:18.995 reciprocal of sine which you should be good at. And then once you get that, it's very easy. 00:44:18.997 --> 00:44:25.992 Every time the sine is 0, that's where your vertical asymptotes are. Label, find your two points for each cycle, 00:44:25.994 --> 00:44:32.995 and then sketch your graph. Remember you approach the asymptotes. Once you have that, you can easily find 00:44:32.997 --> 00:44:38.025 your relative maximum and relative minimum points. 00:44:38.027 --> 00:44:43.528 You need to practice working these examples until you have mastered the skills and concepts involved. 00:44:46.030 --> 00:44:51.789 Let's take a look at the secant function, y equals secant x. Recall that the secant function is the reciprocal 00:44:51.791 --> 00:44:59.289 of the cosine. I'm going to write this as 1 over the cosine of x, which means the secant will be undefined 00:44:59.291 --> 00:45:14.811 when the cosine is 0. And when is that? Let's see, undefined. The cosine is 0 at odd multiples of pi over 2, 00:45:14.813 --> 00:45:27.576 which is a way of saying 2n plus 1 over 2 times pi, where n is an integer. What's going to happen for the 00:45:27.578 --> 00:45:34.080 secant is that's where the vertical asymptotes will be. First let's do this. Let's sketch our basic function 00:45:34.082 --> 00:45:43.831 y equals cosine x that you should be very familiar with. So let's see, remember cosine (0, 1), (pi over 2, 0), 00:45:43.833 --> 00:46:00.360 (pi, -1), (3 pi over 2, 0), (2 pi, back up to 1). And then let's draw another period to the left, so (negative pi over 2, 0), 00:46:00.362 --> 00:46:12.608 (negative pi, -1), negative 3 pi over 2, back up. Alright, so here's our basic cosine function. 00:46:12.610 --> 00:46:19.122 So since the secant is the reciprocal of the cosine, we're going to be able to get some points. First of all, remember 00:46:19.124 --> 00:46:27.636 we said that wherever the cosine is 0 the secant will be undefined. That's where the vertical asymptotes will be. 00:46:27.636 --> 00:46:39.389 So I can see that there's going to be a vertical asymptote at pi over 2, 3 pi over 2, negative pi over 2, 00:46:39.391 --> 00:46:52.910 negative 3 pi over 2. Next, this point here on the cosine (0, 1), well the reciprocal of 1 is 1, so this point will also 00:46:52.912 --> 00:46:59.660 be a point on the secant function. And all these values of cosine between 0 and 1, the reciprocal of those values 00:46:59.662 --> 00:47:07.420 are numbers bigger than one, and it approaches the asymptotes. So this is what that portion of the graph looks like. 00:47:07.422 --> 00:47:16.687 Here we have a point for cosine (pi, -1). Well, the reciprocal of negative 1 is of course negative 1, so that will be a 00:47:16.687 --> 00:47:24.194 point on the secant. And all these values for the cosine between negative 1 and 0, the reciprocal of those values are 00:47:24.194 --> 00:47:32.702 numbers less than negative 1, and it approaches the vertical asymptotes as well. Here's one period for the secant. 00:47:32.704 --> 00:47:41.711 Let's draw another. Oh I forgot this portion. Of course reciprocal of 1 is 1, and then it approaches the asymptote. 00:47:41.723 --> 00:47:52.207 So here are two cycles for the secant functions. You need to learn this graph and learn where the asymptotes are 00:47:52.209 --> 00:47:56.714 and those two points for each cycle. 00:47:58.212 --> 00:48:07.464 Let's look at characteristics of the secant function. First the domain, values of x are all real numbers except 00:48:07.466 --> 00:48:16.241 when x is equal to 2n plus 1 over 2 times pi, where n is an integer. That's another way of saying odd multiples 00:48:16.243 --> 00:48:25.003 of pi over 2 are not in the domain. The range would be the y-values. If we look at the graph we can see the 00:48:25.005 --> 00:48:34.253 y-values go from negative infinity up to and including negative 1. And then it jumps up. It goes from 1 to infinity. 00:48:34.255 --> 00:48:43.003 So that would be the range. The period is the same as the period for cosine which is 2 pi. It has infinitely many 00:48:43.005 --> 00:48:51.527 vertical asymptotes, and they occur where it's undefined which is x equals 2n plus 1 over 2 times pi. Again 00:48:51.529 --> 00:48:57.283 same thing as odd multiples of pi over 2. So we can see our asymptotes are at negative 3 pi over 2, 00:48:57.285 --> 00:49:09.298 negative pi over 2, pi over 2, and 3 pi over 2. The next one would be at 5 pi over 2. For each cycle for the 00:49:09.300 --> 00:49:19.303 secant function, there are two important points. One is where a relative max occurs, and it occurs at the x-value pi. 00:49:19.305 --> 00:49:29.301 The next one would occur 2 pi away times n, again n is an integer, and the maximum value is negative 1. 00:49:29.303 --> 00:49:38.826 Again, in this region this is the highest y-coordinate. Here we have a relative minimum point. It occurs 00:49:38.828 --> 00:49:47.584 at x equals all multiples of 2 pi n. So one is at 0. The next would be at 2 pi, negative 2 pi. Again, n is an integer, 00:49:47.586 --> 00:49:58.594 and the minimum value is 1. The function is even, same as the cosine, which means the secant of negative x 00:49:58.596 --> 00:50:08.607 is the secant of x. That's important. Let's underline relative max and relative min. And because it's even, 00:50:08.609 --> 00:50:15.112 that means that the graph is symmetric about the y-axis. It's important that you know how to graph the secant, 00:50:15.114 --> 00:50:21.110 and know where the asymptotes are and where those two points are.