Understanding Basic Trigonometric Functions

Hello everyone. Welcome to our discussion of section 5.2, where we will be talking about basic trigonometric functions. Now, if you see me wearing these heavy-duty sunglasses, it's because this room where I'm recording this video has received a brand new document camera, but the background plate for this document camera is quite bright. It's like looking at snow on a sunny day. So to protect the camera, to reduce the glare into my eyes, I'm wearing this. I hope you don't mind of this. So the emphasis in this section is on basic trigonometric functions. These are sine function, cosine function, tangent function, cotangent function, and maybe even we'll get to probably secant and cosecant function. But these are the basic ones. So, I started, as you can see, I started this, on the screen you see this unit circle, where I want to draw, suppose we have some angle, let's call it theta, in the first quadrant. Right, suppose we have this angle right here, and we want to call that theta. Now that point on the unit circle has coordinates just like the basic algebra. Remember that? It has x and y coordinates. And I want to call, but here in trigonometry we call them, or we call them, We calculate them differently, even though they're the same thing, x and y coordinates. So let's call it by its original name, point P with coordinates x, y. But that is the same thing. as, let me then expand on this, so it's equal to, instead of x in trigonometry, we have cosine of theta comma, instead of y, we have sine of theta. Sine function is on the y-axis, and cosine of theta is on the x-axis. So also, if you imagine... this, if you imagine this theta as part of a right triangle, right triangle, that means that this angle here is 90 degrees, then this kind of a presumed right triangle has a hypotenuse. Let me just for short say HYP for hypotenuse. It has an adjacent side to this angle theta. Let's call it for A, ADJ for short, and there's an opposite side to this angle, right? Let's call that OPP for short. So that side of this right triangle is opposite to theta. This side is adjacent to it, and this is the hypotenuse of this right triangle. Now, this being the terminal side of this angle theta, it has what we call... projections on the x-axis and projections on the y-axis. The projection on the y-axis is determined by the sine of theta, that's y, just compare that, y is equivalent to sine of theta, that's the projection of this terminal side of this angle on the y-axis, and cosine theta is the projection of this terminal side on the x-axis. x-axis. That's why we say x is equivalent to cosine of theta. It represents this side of this right triangle. Okay, so then given this picture or this figure, then we can write a relationship for the sine of this angle theta in the first quadrant. We can say that sine of theta is equal to the opposite side divided by the hypotenuse. Okay, that's one of the important relationships that we, basic trigonometric relationships that we want to remember. The other one is the cosine of theta is equal to the adjacent side or the projection of that terminal side along the x-axis, which would be the adjacent side, divided by the hypotenuse. That's another formula we want to remember. It's really important to, in mathematics we really don't like to memorize much, but when we do memorize, then there's a significance to that, because given these basic trigonometric functions, especially these two, sine and cosine, You can calculate all the other trigonometric functions. That's why we like to know exactly what's going on, especially with these two trigonometric functions, sine and cosine. Make sure that you fully understand these two, because like I say, you can calculate all the other trigonometric functions if you know these two. Okay, so then let's... Let me see here. So we have, then based on that, then we have, let's add to that the tangent function or the tangent of theta, like that. So for a second I thought I was off the screen, but no. So tangent of theta is equal to the opposite side divided by the adjacent side. You can look at it that way. But at the same time, more valuable is to think of it as the ratio of sine divided by the cosine. So we have sine of theta. Divided by cosine of theta is the tangent function. Like I was saying, you can look at it because the opposite side is along the y-axis, right? It's the projection of this terminal side on the y-axis. So you can also look at it as this. Or, so I'll just add, or tangent of theta. is equal to the opposite side divided by the adjacent side. Your book, I believe, mentions both, but this is more valuable to remember. But this is also useful in some occasions. So I say it's worth remembering both relationships for tangent of theta. But like you can see here, like I mentioned, if you know sine and cosine, you can calculate. the ratio and get the tangent. So that's why we like to memorize the sine and cosine. Okay, so let's then add the cotangent of theta. would be the reciprocal of this, would be ratio of cosine of theta divided by sine of theta, or by the same token, we can say that it's the adjacent side divided by the opposite side. So take the reciprocal of these two, you get the cotangent function. That is why, because cotangent function is the reciprocal of the tangent function, that is why on calculators you don't see a button for the cotangent function. Because if you know the tangent, value, you simply calculate the reciprocal of that, and you have the value for the cotangent function. Now, the other thing about these, so actually let me write that down, so you have it in your notes, that cotangent function. Or let's start with tangent. Tangent function is the reciprocal of the cotangent function. And cotangent function can be said to be the reciprocal of the tangent function. You know one, you know the other, right? Because they're reciprocal of each other. Now we also have two more basic trigonometric functions, right? So far we have shown four. So we have sine, cosine, tangent, and cotangent, right? But we also have a couple of more, and these are equivalent to these. And those are secant and cosecant. functions. So this secant function is the reciprocal of the cosine function. One over cosine function will give you the secant. Okay, I think you already see how important it is to know how to calculate sine and cosine because everything else can be calculated by based on those. Now by the same token, let me put a double-headed arrow here and actually a double-headed arrow here too because you can either go from here to there and or from here to here. These two are equivalent. The same thing we can say about cosine and secant relationship. We can say cosine of theta is equal to 1 over secant theta. It's the same thing, but this time cosine is equal to 1 over secant theta. These two are equivalent to each other. The same relationship exists between cosecant of theta and sine of theta. If you have cosecant of theta, of theta is equal to 1 over sine of theta. So the cosecant function is the reciprocal of the sine function. By the same token, we can say that the sine function is reciprocal of the cosecant function. Okay? So these are, like I say, all these formulas that you see on this page, all of these, especially these two, especially these two, that one and that one, are... mandatory memorization assignments for you. Without remembering those, it's very difficult to know what's going on with these other formulas. So let's look at each one of these trigonometric functions and their graphs and domains. The main thing is their domain, but we will look at their graphs too, because I want you guys to have a good understanding of their graphs too. That's also very, very important. Overall, this is a very, very visual math course, so you want to make sure that you connect the relationship between these trigonometric functions and their graphs. Very, very important. So we want to write this. Basic trigonometric functions, let's say they're graphs and they're domains. And the emphasis, of course, is on the domain. But the graphs are important too, so let me also highlight that. Okay, so let's start with the basic and one of the most used ones, which is the sine function. Before that, let me make a note here that all trigonometric functions are periodic functions. Periodic means like they repeat themselves infinitely many times. If you don't stop them, they keep repeating themselves in both directions on the x-axis, in the negative direction as well as the positive direction. Functions are what we call periodic functions. Okay, so let's start with the domain. Like I said, the domain is very important for us to establish right away. Let's say domain of sine of theta. So that's the first thing we will discuss. Let's draw a picture of what the assigned function looks like. So I will draw a long x-axis here, and let's say this axis represents theta, because that's like x, right? That's the independent variable we learned in basic algebra. So in this case, the independent variable is theta. It's an angle rather than any other value. And, of course, the... y-axis, and that would be for us sine of theta, right? That's our function. Or sine of theta, let's call it y. So theta being on the x-axis. So if we draw a sine function, this is how it would look. It would go through, it would start from here at the origin, or it would go through the origin, not start there, but it would... go through that. And let me actually establish that this is, so I want to stay within, I'm just going to draw some dashed lines to make sure that my graph is as accurate as possible. And so this dashed line and another one right below the theta. And I want to make sure that these are equally distant from the origin. That's necessary for my graph to look right, like that. So the sine function would look like this. It would go like this, come down here, and then go here, and then continue on. So I'm just going to put an arrow there, meaning that it's going to continue. Along the negative side of the x-axis, the same story. So it will come down here, it would curve back up. Go here and then come down and do the same thing. Repeat itself on both sides, in both directions. So because we're dealing with angles, then sign, and I'll get to this in about 15-20 minutes. It's very important that you memorize at which point these trigonometric functions are 0 or have some important value, like 1. For example, the basic sine function, which you see here, varies between 1. That's 1. That's why I drew my dashed lines. And this is negative 1. So it'll swing. It'll go up to 1, come back to negative 1, and go back up to 1, and so on and so forth. Now, if you have a number before this, like 4, it would times that, and you would get 4. 4 times 1 is 4, or 4 times... 0.5, if you're here, you would get 2. But that would have to come from a factor out here. The angle itself, with no factors, will always swing back and forth between 1 and negative 1. And look at the domain. The domain has no restriction, right? It will go from negative infinity to infinity. So the domain of sine function... sine of theta is equal to negative infinity to infinity, and we want to remember that, okay? The other thing you should, the other important part of this graph is where the sine function is equal to zero, right? When it hits on that line, that's when the sine function is zero, right? The sine function is along the y-axis, and at that point, its value would be zero. So what is that point? This would, if this is 0, so sine function is 0 at 0, or at origin, and then it's also again 0 at pi, and again 0 at 2 pi, and 0 again at 3 pi, and so on and so forth. So what would that make this peak then? What angle would that be? That would be pi over 2, because look at this, this is so symmetric, right? From 0 to pi, half of that distance is pi over 2. So if you have your calculator at your disposal, you can just take the sine of 90 degrees or pi over 2. It's going to tell you that it's equal to 1, which is true right there. Sine of pi over 2 is 1. Then you can calculate sine of pi. It's going to be 0. Sine of 2 pi, again 0. So which is this one then? This would be... Sine of that angle is negative 1, right? So what is that angle then? 3 pi over 2. Okay, I'm sorry, these are close to each other, but that's, let me change the color then. So that would be 3 pi over 2, which is that, the dashed line that you see. Okay, very, very important to very early on establish where this, what sine function does basically. Where is it zero, where is it one, where is it negative one. See, make sure that you remember what angles, at which angles sine of function is zero or one or negative one. Very, very important. The other trigonometric function that we need to know very well is the cosine function. So domain of cosine of theta, and I'm going to do the same thing, draw a long axis for theta. And then I'm going to, so that's theta, and then I'm going to draw a vertical axis for the y equals cosine of theta, right? So cosine of theta equals y. So the cosine function, and I will draw my dashed lines to make sure that I draw the graph accurately. as much as possible. So let's say that's the line y equals 1, and I'm sure you guessed what this other one will be, y equals negative 1, right? The basic cosine function, just like the sine function, will vary between y equals 1 and y equals negative 1. But unlike the sine function, the cosine function begins... at here, at 1, not as 0, not as 0 like the sine function. It starts at 1. Why? Because cosine of 0 is equal to 1. That's why I keep asking. I'm asking you guys to please memorize the value of the trigonometric functions at these special angles. Zero angle, pi over 2, pi, pi over 4. We'll get to that in 10-15 minutes. I'll show you a table that you would need to memorize basically. Or a little bit of it, if not all of it. For sine and cosine values, you should have those memorized. Very important. So we start from here. This is what the cosine function does. comes down here, goes back up, and then curves back down again when he tests the y equals 1 line. Okay, so now we know that at 0, at theta equals 0, cosine function is 1. It's at highest value, 1. Cosine of 0 angle is 1. Cosine of then, if at 0 it's 1, when is it then 0? It's pi over 2. Then when is it negative 1? If it's 1 at that, at 0, at which point is it negative 1? It's at pi. Okay, so that distance. And then again, it's 0 again at 3 pi over 2. And then finally, it's 0 at pi over 2. it is let's see 3 pi over 2, and then at 2 pi again, yes, at 2 pi, again, it hits the highest value, 1. Okay, and just like the sine function, this can go on and on on both sides. So let me actually continue that. So it can go again, do that, come back up again, do that, and so on. On this side, it will do the same thing. So it'll do this like that. So that's why we say that trigonometric functions are periodic. They do that over and over again. So take a look at this function. The domain is unrestricted, right? So just like the sine function, the domain of the cosine function is also negative infinity to infinity. Data is equal to negative infinity. infinity. Okay, and now you have a visual reference as to why that is the case. Next, let's look at tangent function. So we want to look at the domain of the tangent function, okay? Now this is why I was saying earlier to please memorize these these basic formulas because if you did that right away you know that tangent function is a ratio of two numbers. That's very important, right? You've been through algebra before, so you know when you have a fraction, you can't have a zero in the denominator. In which case, it would make the tangent in this case undefined. So if you remember this formula, then right away you want to ask, where is cosine of theta equals zero? Because those points where cosine of theta is equal to zero will be your vertical asymptotes. We talked about vertical and horizontal asymptotes a lot in Math 126. So at wherever we have cosine of theta equals zero... The tangent function has a vertical asymptote, meaning its graph cannot pass through that point. Okay, so let me again draw this long theta axis here. I will draw it here, I guess right here too. Maybe here. So we have that. This is our theta axis, and let's put the vertical axis, which would be y equals tangent of theta, right? Tangent of theta equals y. So then, given that, and I'm going to... to write that maybe here, maybe here. So tangent of theta is equal to sine of theta divided by cosine of theta. It's so important that I wrote it right away there. Now in the previous discussion, when we talked about domain of cosine of theta, notice what we found. Cosine of theta is 0 at pi over 2, at 3 pi over 2, and on and on, at 5 pi over 2, and so on and so forth. So we want to account for all of those points where cosine of theta is going to be 0. We do not want the domain of tangent theta to include those because at those points... it will be undefined. So let's draw asymptotes there. So I'm going to draw an asymptote here, and I'm going to label it as pi over 2. So this will be pi over 2. Then I'm going to draw another one, let's say here, because we can go on and on, right? But we don't want to do that. Just to make a point, a few would suffice. So these, by the way, should be equally distant. from here to here and from here to here. I don't know if I did a good job in doing that. Maybe not, but it's okay. Well, let me correct this. Before I post this on Canvas, I will delete this. But this dashed line is not in the right place. And I will tell you in a second why. Why not? So before I post these lecture notes on Canvas, I will correct this. But the correct position for this line will be here, somewhere here. So please ignore that for now. Ignore this line. Okay, I will erase it and clean this up. So the correct point would be here, like that. And that would be 3 pi over 2. Okay, so what would that be? If this is pi over 2, what would that be? That would be minus pi over 2. And then... That distance from this dashed line to that dashed line, so that would be as wide as this, roughly, right? As wide as this. I'm going to put another vertical asymptote there. And I'm going to call that minus 3 pi. over 2. So these are our asymptotes. Then the tangent function, the tangent function would graph like this. So this line would come from here, go through the origin there, and then the arrows. And then here we would have A line going like this. So do you see how these periodic functions repeat themselves, right? So from here, from this, from equivalent points is one period. From here to here. From here to here. That's one period. From here to here is one period. From here to here is one period. And then this one repeats itself on this side and on and on and on. So these are, these asymptotes are exactly where the cosine of theta is zero. Okay, I don't want to make this drawing too crowded. So I'm not going to draw the cosine graph, but if I did, you'd see that this is where the cosine graph is equal to pi. over 2 so cosine of pi over 2 would be 0 and again cosine of 3 pi over 2 would be 0 so tangent function cannot go through that. Okay so the domain looking at this graph the domain of tangent of theta is equal to all real numbers except pi over 2 plus n pi, and n being an integer. So if you plug in for n with 10, then you have 10 pi plus pi over 2 is also a vertical asymptote. So except these guys. So all real numbers except these guys. pi over 2, 3 pi over 2, negative pi over 2, and I should actually add a plus or minus, right? The minus indicates in this direction, okay? Because when this is minus pi, then you get pi over 2. 2 minus pi, you get minus pi over 2. So you get these guys. Okay, that's the domain of the tangent function. Let me box that here. Okay. I'm debating myself whether I should draw the cosine curve or not. Let me lightly draw it, because it's important for you to see that these asymptotes occur whenever the cosine curve is drawn. function is 0. So if we have, if we have say cosine function coming from here like that, then it would go like that and so on. So that's the cosine of theta. On this side, we'd have it come down here, go through that, and then come back up, go through that, and so on. This is the cosine of theta function. And look, wherever it's zero, you have the asymptotes because of this formula. Okay? I just did it once. I think it's better not to draw it because the graphs become too crowded, and I don't want to do that. Okay? Okay. All right, so that was those basic trigonometric functions. We have more, so we have the cotangent function next. So we have domain of cotangent of theta, and do you remember what the cotangent of theta was equal to? It was equal to the ratio of cosine theta over sine theta. Again we have a fraction which is important to remember because then whenever the sine of theta is zero you're gonna have a vertical asymptote. Where would those asymptotes be then? Sine of zero angle is zero, sine of pi is zero, sine of 2 pi is zero. So at theta equal zero pi 2 pi, 3 pi, 4 pi, and on and on and on. And at 0, minus pi, minus 2 pi, minus 3 pi, you're going to have vertical asymptotes for cotangent function. Okay? That's why it's so important to remember those formulas. So cosine of theta divided by sine of theta. Okay, so once again, let's draw these. We have this theta axis like that. And let's put the vertical axis. I want to make the vertical axis a little bit thicker. Maybe, well, maybe, should I do that? I think it's better if I do. Let's say, so I'll make it a little bit thicker than the x-axis because I want it to stand out here like that. So this would be y equals cotangent of theta, right, axis. Now, we just discussed that whenever the sine of theta is zero, we're going to have a vertical asymptote. So that's a vertical asymptote, right, that one. So I'm going to put a dashed. I'm going to put a dashed line right up against the y-axis. That the y-axis is an asymptote. So the graph of the cotangent function cannot cross this line. Okay? And then where will be the other asymptotes? So this is at theta equals zero. You can tell I've taught algebra a long time. So that's not x, but theta in this case. I will fix this. I will make it look better before posting this in Canvas. The next one will be theta equals pi. So let me draw it here. That will be another occasion where sine function will be 0, the denominator will be 0. Of course, we do not want that. So theta equals pi, and I'll draw another one on this side and a couple of dashed lines on that side. So we have another set of dashed lines on this side, and this will be 2 pi. Okay, and then a couple of dashed lines on this side. And these have to be equally distant from each other. From here to here and from here to here should be equal. And then now from here to here. Right, from here to here. And this would be theta equals minus pi. And one more. Theta equals minus 2 pi. Okay, so the cotangent, the graph of the cotangent function, in terms of the way it looks, it goes the opposite direction. So let me bring back the graph of the tangent function. So the graph of the tangent function was going like that, right? And also the asymptotes were different than cotangent. But the point I'm trying to make, tangent was going like this. Now cotangent is going to go like that. It's going to reverse it. itself, but it's going to stay within its own asymptotes. That's very important. And these are the asymptotes for the cotangent function. So here we will have the cotangent lines. Let me draw that. So one would go through. here and so that's one. Another one would go. I didn't draw this very symmetric. This has to be in the middle of this line, sorry about that, but I will label it so you know exactly where that should be. I will label it in a few seconds. And then we would have like that. That's better, right? So this is more in the middle of those two dashed lines. And another one, like that, like that. Sorry if my drawing isn't better than this. So you might ask, well, what are these points where it is the cotangent function is equal to 0? That's really easy to determine. because that's whenever the cosine of theta is equal to zero, right? Cosine of theta is equal to zero at pi over 2, or halfway between zero and pi, right? This is zero, this is pi. Halfway in between is pi over 2. Now this, in terms of my graph, is poorly drawn because this point should be in the middle here. But it's not. But let me label it correctly. Between pi and 2 pi, the middle point is 3 pi over 2, at which point the cotangent of theta would be equal to 0, right there. And this is the same thing, but minus. So this would be minus pi over 2. And this one here would be... I'm sure you guessed, it's minus 3 pi over 2. And of course, it'll continue in both directions because it's a periodic function. So when we write the domain for the cotangent function from this graph, you can tell that all real numbers are acceptable. except any multiples of pi, right? So 0 pi, that would be 0. It's an asymptote. So we have to exclude these asymptotes. How do we do that? This is how. We say that domain, so the domain of the cotangent of theta, is equal to all real numbers, and here's the important part, except what I've shown in dashed lines, the vertical asymptotes, except n pi, that's just n pi, 0 pi, 1 pi, 2 pi, and I should say plus or minus actually, plus or minus n pi. Okay, so because the minus... will take care of the asymptotes in this direction and positive and pi will take care of them in that direction and n is an integer any integer 0 1 2 negative 1 negative 2 and so on so this is the domain So that's why I don't want to give the domains without drawing the graphs, even though I'm not terribly good at drawing these graphs, but I'd like to give the domain after I show you the graphs, because now it makes sense. right, this except n pi are all these vertical asymptotes. And we know why, because at these points, this ratio will have a 0 denominator. And we can have that. OK, so we have a couple of more domains to look at, domain of secant function and domain of cosecant function. So let's look at those. domain of the secant of theta, which by the way, let me remind you that secant of theta is equal to 1 over, do you remember which function it was? 1 over secant is the reciprocal of cosine of theta, right? Cosine of theta. Anyway, let me put a line there, so this is different than that. Okay, so now, once again, let's draw the picture so we can make sense of the domain once we write it down. Now, from this equation, you already know that the domain is a constant. You already know where the vertical asymptotes are going to be, right? Do you need even me to tell you that? Look at this. You don't need me to tell you that, right? Whenever cosine of theta is zero, we're going to have vertical asymptotes. So this is our theta axis. Then we will have the y-axis. I'm going to draw it here somewhere. Here maybe, that's better. Like that. And then we have, of course, the y is the secant of theta in this case. Okay. All right. So first, let's draw a nice... cosine graph, right? Because we're going to need to know where this cosine is going to be 0. So here's the cosine. I've already done it for you on the previous pages. But let's do it again. So we have, I'll try to make it as accurate as I possibly can drawing these things by hand. And so on, right? On this side also. We'll come down, do this, go back up, come down again, and do that. So wherever you see the cosine, and let me write that here, that this is the cosine. cosine of theta function, okay? Wherever the cosine of theta is zero, all these points, all these points, This fact, this ratio is going to be undefined. So secant of theta will be undefined. So all those dots that you see where cosine of theta is 0 are going to be our vertical asymptotes. So let's draw them here, some of them at least. I won't draw all of them, but so it will be one here, another one here. Another one here, right? This graph didn't turn out to be too bad because the distance between these dashed lines is almost equal. Right? Not too bad. And let me also draw another one here. Another one right there. Okay, now the secant function will be in between these asymptotes. So it'll be, we will have one here. So we will have one curve here. We will have one curve here. Let me move this up a little bit. We will have one curve here like this, one curve here. And I want to draw a little bit of it. So one curve would be here like that. Like that. Okay. And on and on. So we could put another one here like that. like that but there will be more asymptotes on that side and that's on this side and that side all right so this would curve would have to curve up this way because there's there's going to be an asymptote right there as well and another asymptote here Okay, so then the question is, what are these points? Where are these? Those points are where cosine of theta is equal to zero. So this point here would have to be pi over 2. This would be 3 pi over 2. And this would be 5 pi over 2 and so on. On this side, we will have minus pi over 2, right? And then minus 3 pi over 2 and minus 5 pi over 2 and so on and so forth. Okay? So. You can see right away that the domain of the secant function is all real numbers except anything that goes through pi over 2. Pi over 2, minus pi over 2, minus 3 pi over 2, and so on. So let's write that out. So we say domain. of secant of theta is all real numbers. Here comes the important part. Except pi over 2 and anything. That is n pi. That is pi plus that. Because when you have pi over 2, for n equals 1, you'd have pi plus pi over 2. But what will you get if you do that? You'll get 3 pi over 2. What about if you put 2 pi? Well, then you get 5 pi over 2. So this piece hits all the vertical asymptotes. And if you want, we can draw one here, too, since I drew the graph for it. Let me draw one there and another one here like that. Okay. And I'm going to box that. So if you ever wonder or if you want to better understand why this domain looks the way it looks, just look at the graph. Now you have a graph for all these domains. Please refer back to the graphs. We have one more to draw for this, and that will be the cosecant function. I believe that was page, so this is page 4. So domain of cosecant theta. Do you remember what the formula was for cosecant of theta? Should I bring the first page to remind you, or do you already remember it? Do you remember that would be... fantastic. The cosecant function is the reciprocal of the sine function. So that's very important right? By now you know that because whenever sine of theta is equal to zero then we're going to have those vertical asymptotes because the cosecant of theta cannot go through those points. So that's why that those formulas are very important. Okay, alright, so just like before, before we give the domain, let's give a picture so we can make sense of the domain once we write it down. Okay, so we have something like this and the theta. I should have drawn this line a little bit lower because I have all this space, and so it's okay, it's fine. And then let me draw the vertical asymptote. Let's call this y equals the cosecant theta, right? Cosecant of theta function. And so because there's a relationship between cosecant of theta and sine theta, let's draw the sine theta graph. Let's draw this graph first, okay? Because we're going to need to know where the sine theta is equal to zero. So the sine graph does this. It does go through, do that, then come down, do that, do that, do that, and do that, and so on. Remember what this number was? That was 1. And here... That was negative 1, right? The basic sine function swings or oscillates back and forth between 1 and negative 1 on this side as well. So we can draw the graph on this side. If I can keep it nice and smooth, that would be good. Something like that. Like that. Okay? Let me put an arrow there. So you can see clearly at... At some point, points, the sine function is zero at that point, at that point, here, here, here. All those dots that I'm putting on the graph, that's where the vertical asymptotes should go, right? Because the cosecant function cannot go through those points. So I'm going to go and put a dashed line here. I'm going to put a dashed line right through that dot. where sine of theta is equal to zero, another one here, another one here, another one here. Okay, so then where are these asymptotes? We know that sine of theta is equal to zero at theta equals zero, so that's one of the angles. At theta equals zero, we have an asymptote, right? So let me actually put a dash line right next to this y-axis to indicate that the y-axis is also a vertical asymptote here, like that, okay? So let's label these. We have that's zero. What about this one? The sine of theta is also equal to zero, so Theta equals pi. What about here? 2 pi, 3 pi, and so on and so forth, right? How about this side? Minus pi, minus 2 pi, and so on and so forth. So where would the cosecant function graph be then? Well, it will be right at the top of each peak and at each valley here. So it would be like this. this, it would be like this, then like this, like this. Do you see why it's important that we draw the sine function first? Because we know exactly where these asymptotes are and where the peaks and valleys are, so we can draw our cosecant graph very nicely like this. it doesn't look too bad, like that and like that, and we can continue on either side, going in each direction. Okay, so that's that. Now the domain. Now that we have the graph, the domain is really easy to write down. I'll move this up a little bit. So the domain of the cosecant function is, again, all real numbers except these guys, right? n pi, any factor of pi, plus or minus, right? All real numbers except, I'm going to write that part with red or in red, except plus or minus n pi, where n is any integer. Let me move this up a little bit. I'm trying to keep the graph on the screen, but it looks like I need to scroll up a little bit. So integer, like that. Okay, now when you read this, you know exactly why except n pi should be there, right? Those are all the points where the sine function is equal to 0, which we cannot have because sine of theta is the denominator of this formula for the cosecant function. Okay, so now let's talk about some of the important concepts that you need to memorize in order to do well in this course. I don't know how to say it in a different way, but honestly, you need to memorize these topics or these quantities. So we're going to talk about that. So I've drawn this table, but it's blank as you can see, so we can fill it in with the right quantities. Here are the special trigonometric angles for which you need to remember all these. So, well, not all these, but at least the sine and cosine. Cosine. Okay? So here you have, on the outside of the box, I put the measures in degrees. But I know, I've already mentioned this, when I took this course as an undergraduate, first you're trying to stick with degrees. I was the same way myself, but very quickly you're going to realize that this is actually going to make things more difficult. So you need to quickly switch, even if you're comfortable with these, switch to to their equivalent measures in radians, which are these ones. So 0 degree, pi over 6 is 30 degrees, pi over 4 is 45 degrees, pi over 3 is 60 degrees, pi over 2 is 90 degrees, and so on. Make sure that you switch your mind to these measures, the radians. With these, you're going to have considerable difficulty. Now it's the beginning of the semester, but the further we go into the semester, you're going to need to know these. very well, okay, and what they represent. Pi over 6 is 30 degrees. It's easy, right? 180 divided by 6 is 30 degrees. So then at the top, we have these trigonometric functions that we've been talking about so far. Sine of theta and cosine of theta, and then we have the tangent, and not theta, t, this time t. So let me put that in parentheses. So tangent of t t, t being the angle, right, and then cotangent of t, and then secant of t, and then, of course, the last one, cosecant of t. So sine of 0 is 0. And let me actually switch to a different color here, 0. Sine of pi over 6 is 1 half. Sine of pi over 4 is square root of 2 over 2. Sine of pi over 3 is square root of 3 over 2. Sine of pi over 2 is 1. Okay? It's very important that you remember these quantities for sine and cosine. And then cosine of 0 is 1. Cosine of 30 degrees is square root of 3 over 2. Cosine of pi over 4 is square root of 2 over 2. Cosine of pi over 3 is 1 half. Cosine of 90 degrees is, of course, 0. So if you look, there is a pattern, right? The pi over 4 or 45 degrees has the same sine and cosine. The sine of 30 degrees is equal to the cosine of 60 degrees. And the cosine of 30 degrees is equal to the sine of 60 degrees. Okay? There are relationships here. There's a pattern here. Okay? You've got to remember this because if you do, then you can calculate the rest of this table based on your knowledge of these two functions. Okay? So tangent is sine. over cosine, so that'll be 0. 0 divided by 1, 0, that's what I'm saying. Once you know these two columns, this, this, you can calculate all of this. So this will be undefined, right? Do you know why? Why is this undefined? Because cotangent is cosine divided by sine. So 1 divided by 0 is undefined. So these two columns are very important. Let me... emphasize that by putting, highlighting that and that. Okay, these two columns are very, very important. And then let me fill in the rest of this, so 1. But you can verify. If you know this, you can verify all these quantities based on this and the formulas I gave you on the first page of this session. And cosecant would be undefined. And this would be tangent of t would be square root of 3 over 3. Then this would be square root of 3. And then 2 square root of 3 over 3, 2. And then here would be 1, 1 again. And then square root of 2, square root of 2. And then we have square root of 3, square root of 3 over 3, and 2 and 2 square root of 3 over 3. And then finally at the bottom we have undefined, 0, undefined again. So test yourself. Memorize these and see if you can calculate these, any of these. Okay, tangent of 60 degrees is square root of 3. Secant of 45 degrees is square root of 2, and so on. Then the other thing that you should remember and have this in your memory at all times is the sign of the algebraic functions in each quadrant. So this is... quadrant 1, let me use my thicker marker here, this quadrant is quadrant 1, in here this is quadrant 2, this is quadrant 3, and this of course is quadrant 4, right? So make sure that you know the algebraic sign for these trigonometric functions. When we started this this session, right, I mentioned that cosine is equivalent to x, sine is equivalent to y. So whenever cosine is, x is negative, cosine is negative. X is positive, cosine is positive. Whenever the y-axis is positive, sine of theta is positive. Whenever the y-axis is negative, sine of theta is negative. is negative. And the other trigonometric functions are the ratios of those. So we can easily then obtain those two. So we have here, in the first quadrant, everything is easy, right? Everything is positive because the x and the y-axis are positive. So positive, positive, positive, all of these are positive. Now here, things are getting a little bit more interesting, right? Because in the second quadrant, which... axis is positive? The y-axis. Y-axis is tantamount to the sine of a function, right? So then I should have said sine theta, but you get my point what I'm talking about here. So this would still be positive, but look at the x-axis. It's negative, so cosine would be negative. Tangent is a ratio of sine divided by cosine. Positive divided by negative, negative. Cotangent is cosine divided by sine. So negative divided by positive, negative. Secant is the reciprocal of the cosine function, so negative again. Cosecant is the reciprocal of the sine function, so one over a positive number, it's going to be a positive number, right? So in the second quadrant, look at that. The only positive trigonometric functions are sine and cosecant, okay? I'm sure you're seeing the benefit. of this, to remember this. It'll come in handy a lot, as you'll see in the examples that are coming up. Okay, so what about the third quadrant? Look at that. Is x, what is x? Is it negative or positive? It is negative. What about the y axis? Is it positive or negative? Negative. So the sine function, which has to do with the, which is equivalent of the y axis, is negative. Cosine function, which is equivalent of the x axis, negative. negative. What about the tangent? What about the tangent? So tangent is the ratio of the sine divided by cosine, right? So negative divided by negative, it is positive. Cotangent is the ratio of cosine divided by sine. So again, negative divided by negative, positive. How about the secant function? Secant is the reciprocal of which function? Reciprocal of? Cosine function. So if cosine is negative, take the reciprocal of a negative number, you get a negative number. So negative. How about cosecant? Cosecant is the reciprocal of the sine function. And sine function is negative in the third quadrant. So take the reciprocal of a negative number. What do you get? You get a negative number. So in the third quadrant, the only positive trigonometric functions are tangent and cotangent. But do you have to memorize? this? No. You just have to memorize that sign representing trigonometry. Cosine represents the y-axis. Cosine represents the x-axis. And then go from there. Then you can easily translate wherever the axis is negative or positive, then the trigonometric function associated with it will take the same sign. And how about the fourth quadrant, right? So in the fourth quadrant, the x-axis is positive. What does that mean? Then which one of these should be positive? The x-axis is associated with cosine, so that's going to be positive. What else? What else should be positive here? Well, secant function is the reciprocal of cosine function, so the secant function also should be positive, right? Now sine for sure should be negative, right? Because that's sine function is associated with the y-axis, so that's negative. In the fourth quadrant, y-axis is negative. Tangent function is the ratio of sine divided by cosine and negative number divided by a positive number we get a negative number. Cotangent is the ratio of cosine divided by sine. Positive divided by negative we get a negative number. Coscant function is the reciprocal of which function? Do you remember? Coscant function is the reciprocal of the sine function. So if you take the reciprocal of a negative number what do you get? You get a negative number. number. So in the fourth quadrant, the only positive ones are cosine and secant, which are reciprocals of each other, by the way. Okay, so in addition to this table that I asked you to at least memorize these two first columns for sine and cosine, you also need to remember the algebraic sine of trigonometric functions in each quadrant. Of course, as I said, only you need to know, remember that's what sine and cosine are. and then you can calculate the algebraic sign for all the other trigonometric functions based on those two, sine and cosine. So before we go over some examples, there are some things that we need to know. Some of it from, actually one or two of it is from the precalculus one. So it's nice that that stuff that we learned in the previous class... useful in this class as well. So I'm going to say recall, recall from pre-calc 1 or math 126, you remember that course? Okay, that odd functions are functions such that if you replace the x with minus x, it will give the result is minus f of x. And even functions... would be if you replace the x with minus x, that'll be equal to f of x, right? Minus x and x, for an even function, they're equal, okay? This, we remember... two let me box them separately very important in this course as well as the pre-calculus one we learned before right the other thing that I want to make a note of now we want to connect this to this stuff to trigonometric stuff right which functions are odd which functions are even so let's connect them to trigonometry So the cosine function, cosine of theta, and secant function, secant of theta. So this argument can be anything, right? T, theta, x, whatever you want it to be. I just chose theta. These two functions are the only even trigonometric functions. Functions are even, right? So just like any other even function, if you replace theta with its equivalent negative equivalent, you still get theta. So let's write them down. So therefore... cosine of minus theta is equal to, just like that, just like the even functions in algebra, is equal to cosine of theta. Okay, that's one thing we should know. And the secant function, which is a reciprocal of the cosine function and vice versa. So secant of negative theta is equal to secant of theta. Like that. These are the only two even trigonometric functions. The other ones are odd. So let's write them. Let's make sure that you have this in your notes. So note that sine of theta. Then what is tangent of theta? What else? We have cotangent of theta. And lastly, cosecant of theta, right? Cosecant of theta is the reciprocal of sine theta. Are all odd functions. All odd functions. Okay, so what does that mean? This is what it means. Just like what the odd functions meant in basic algebra, this minus in the argument translates in the minus in front of the function. Those two are equivalent. So then we can say that... Sine of, we start with that, sine of negative theta is equal to minus sine of theta. You know, trust me, this simplifies when we get to where we have to solve trigonometric functions. functions, which is several or many sections later. These help us simplify expressions or simplify our calculations so much. So please make sure that you remember this. Then tangential functions are the ones that are important to us. So tangential functions are the ones that are important to us. So tangential functions are the ones that are important to us. tangent of theta, tangent of, sorry, minus theta is equal to minus tangent of theta. So you can take that negative out and put it in front for sine, tangent, cotangent, and cosecant. Okay, and then so cotangent of negative theta is equal to minus cotangent of theta. And finally, of course, cosecant of negative theta is equal to minus cosecant of theta. I guess I can box all of these at once. These are very, very important to remember so let me just emphasize here and then also here. The other things that your book talks about in this section, and these are very important, again they become very very useful when we start solving trigonometric equations, are some formulas. So let's, your book mentions them in this section, so I'm going to mention them in this section. and they are called the Pythagorean identities. That's what they're called. So we have the sine squared t, some angle t, plus notice that I've used theta, now I use t. You have to get used to these various forms of the argument being represented. Cosine squared t, so sine squared t plus cosine squared t equals 1. Very nice and easy formula for us. us to remember sine squared plus cosine squared is equal to 1 then the other form the other two there's there's three of them three Pythagorean identities 1 plus tangent squared of T is equal to secant squared T 1 plus cotangent squared t. is equal to cosecant squared t. Very useful formula, so you'll find this out. So the sooner you memorize these, the better. Okay, finally, we get to some examples. So all of this was just introducing these trigonometric functions, their relationships with each other, their domains, their graphs. All of this is very important for us to connect with each other, the graphs with the functions, with formulas, domains, and everything. Okay, so let's start. The first example, we want to find... the exact value of the trigonometric function at the given real number. So do you know when we say the exact, see this is very important, the exact value, do you know what that means? Let me bring this table back. See, these are exact values. That means you don't want to calculate the decimal values. These are all exact values, all of them. So you don't want to calculate this with your calculator and say, oh gee, this is point. Let's say 6, 4, something. You don't want to do that. So you leave them in exact form. These are all exact forms. In trigonometry, most of the time, we want exact values. Sometimes we also want to have exact values. also calculate the decimals. But look at all of this table you have. All you have are exact values. Very important. Okay, so that's what they mean. So the first one, we're going to do a number of these examples. The first one, let's call it A. is we want to calculate tangent of 5 pi over 6. Okay? So you remember in one of the previous sections, or section 5.1, we talked about reference number, right? So now that becomes very, very handy here to remember to connect that topic with this. So in these, I'm going to draw graphs, but you don't have to. I'm just drawing graphs for all of them. these for this for sure you don't have to draw graphs i'm sure there are some of you who will answer this knowing knowing the table knowing that this table and remembering what we said about reference numbers in section 5.1 i'm sure some of you will be able to answer this without drawing any graphs very easily too but i draw graphs just in case some people want to have a visual reference I myself would like visual references in topics like these. So let me draw a little graph here before answering this. And so let's call this, I don't know, if you wanna call it theta or something, any, it could be x2. And then the y-axis here. This is our function. And then so here the function is tangent of 5 pi over 6. So where is this angle? So if this was 6 pi over 6, let's translate it, right? So we can say 5 pi over 6 is equal to 6 pi over 6 minus pi over 6, right? You can break it up into this. 6 pi over 6 is just pi over 6. pi minus pi over 6 is going to give you 5 pi over 6. Right? So where is this angle? Let's see. Pi minus pi over 6. So now you know that 0 to pi is from here to here. 0 to pi. Going in the counterclockwise direction because pi is positive. It's not minus pi. Otherwise I would go in the clockwise direction. If this was minus pi, I would go this way, clockwise. But it's positive, so I'm going counterclockwise from 0 to pi. And this is pi, and that's 0. Theta equals 0, and theta equals... equals pi, right? So 0 to pi is that, but is that, that angle is pi minus pi over 6. So not only we go to pi, but then we have to come back, we have to subtract pi over 6 from it. So where do we end up? We end up in the second quadrant. Okay, so 5 pi over 6 is this angle. Well, let me draw it with my red ink. So that is 5 pi over 6, or pi minus pi over 6. Okay? So in section 5.1, we learned that if you find the equivalent of this angle in reference numbers, you can easily calculate what that number is, right? So let's call this theta would be pi over 6, right? This angle, let me just say pi over 6. Now the reference number is pi over 6, so the tangent of that angle of this big angle is equal to tangent of this reference number but you have to remember please that you're in the second quadrant so you calculate tangent of pi over six but in the second quadrant so let's do that so tangent of Tangent of 5 pi over 6 is equal to tangent of pi over 6 in quadrant 2. Okay? In quadrant 2. Okay, so that would be tangent of pi over 6 is equal to sine of pi over 6 divided by cosine of pi over 6. This is where remembering those relationships that I mentioned will be very, very useful to you. And also that table, right? Sine of pi over 6 is 1 over 2. That's in the table that I asked you to memorize. Cosine of pi over 6 is equal to square root of 3 over 2. Okay? So this is a complex fraction. 2 and 2 cancel out. We get 1 over square root of 3. But then this is, let's rationalize this. So we're going to multiply it by... by square root of 3, because we don't want a square root in the denominator, right? Multiply it by square root of 3 over square root of 3, because what is the square root of 3 over square root of 3? It's 1, right? 5 over... 5 is 1, negative 2 over negative 2 is 1, so we're multiplying by 1, really, which is important because that means we're not changing this. Anything multiplied by 1 is not going to change, but... What we're achieving with this is that we're going to rationalize the denominator, which we learned to do in pre-calculus 1. So this will be square root of 3 over 3. Am I done? No. That's why it's important for you to remember that what you're calculating here is the reference number, not this, not tangent of 5 pi over 6. So what difference is that? that make? The sign, the algebraic sign in quadrant 2, in quadrant 2, the tangent is negative. So because of that, we have to put a negative here. So the answer is negative square root of 3 over 3. Okay? That, which quadrant we are, very important because that determines the sign. If we were in the first quadrant, that would be positive. But we're in the second quadrant, where tangent is negative. Okay? And that's why I showed you guys that here. That's why you have page 5. Remember, page 5, when you download the lecture notes, page 5 has the table and this circle with all the algebraic signs. In the second quadrant, the tangent is negative. is negative. Okay, do you see that? Very important. Alright, so another, this would be part B, but we're going to do the same thing, tangent of, we're going to find the exact value of this. of this quantity, tangent of 7 pi over 6, right? That's the question. Well, just like before, break this up into terms that are easily calculated. So this will be equal to... Let me... So this would be the solution. So that is equal to, let's see, 6 pi over 6. Let's break it up. 6 pi over 6 plus pi over 6, right? So 7 pi over 6, I'm just breaking up the angle, and then we're going to get involved with the tangent later. 7 pi over 6, you can write it as 6 pi over 6 plus pi over 6, which is 6 and x. 6 cancel out, you get pi plus pi over 6. So can you see which quadrant this angle is? I'm sure you can, most, if not all of you, most of you are seeing where this is. Well, if you're not sure, let's draw a graph. So this is our theta. And we have the y-axis. So this angle, which is 7 pi over 6, is pi plus pi over 6. Note that both angles are positive. So the direction of movement is going to be in the counterclockwise direction. So from here to here is pi, then plus another pi over 6, so we're in the third quadrant. So let's draw the angle right there. Somewhere here, we are somewhere here. Here, that's the terminal side, and the angle itself, how did I draw it over there? With red ink, I'm going to do the same thing here. So that is the angle. This angle is 7 pi over 6. but what about the reference number? The reference number is the angle this is something we learned in section 5.1 the angle between the terminal side of the triangle this angle and the x-axis so that's our reference angle theta okay it's pi over six because we know to get there we we did pi plus pi over six so we know that that angle is pi over six so remember we got to be very careful because it's really easy to calculate tangent of pi over six but then you got to remember to apply the appropriate sign for the third quadrant. Okay, just remember that. Alright, so now we're going to involve the tangent function. The reference number is pi over 6. So we say tangent of 7 pi over 6 is equal to tangent of our reference number pi over 6 in quadrant 3. Okay? So, that determines the sine, right? In the third quadrant, the sine and cosine are both negative. So, the tangent is the ratio of sine divided by cosine. So, negative divided by negative, tangent is positive in the third quadrant. And if you have any doubts... Let me bring this back, this sheet, this circle. In the third quadrant, the tangent function is positive because tangent is the ratio of sine divided by cosine. So negative divided by another negative, we get positive. Okay, so now tangent of pi over 6 is equal to sine of pi over 6 divided by cosine of pi over 6. Once again, sine of pi over 6 is... 1 over 2, cosine of pi over 6 is square root of 3 over 2. This is from the table, right? Sine of pi over 6 is 1 over 2. Cosine of pi over 6 is the square root of 3 over 2. That's why I ask you memorize these first two columns because then you can calculate the tangent based on that. So we are on our way to calculating that. tangent pi over 6. So we're getting there, but it's in the table. If you can remember it, you can write it right away. So this complex fraction, we get just like before, this is very similar to part a. We get 1 over square root of 3. Again, we don't like, we learned in pre-calculus 1 that we don't like to see radicals in the denominator. So let's rationalize it by multiplying it by square root of 3 over square root of 3. And we get square root of 3 over 3. Do we put a minus? No. We just established that in the third quadrant, the tangent function is positive. So we leave it like that. So this is our final answer for that. Square root of 3 over 3. We have more examples, so let's go over those. This one is interesting. So this is part c. We want to calculate tangent of 11 pi over 6, right? So, I mean, I forgot to mention, but please pause the video. On all of these examples, pause the video, do your own calculations on a piece of scratch paper, and then come back to the video. Please do that because I get worried if you just copy my stuff down, you won't progress in this class fast enough. But if you do, if you do pause and you use your own brain power, even if you're... solutions may not be correct right away, you will get to full accuracy very quickly. But you've got to resist the temptation to look at the solution before you do your own work. So I hope you've been doing that, but I forgot to mention it. Okay, so just like before, let's break this up, right? So because this is a big angle, let's break it up to something that we can, to angles that we can understand. smaller angles like the ones in the table. So if we break this up, we can say that 11 pi over 6 is equal to 12 pi over 6 minus pi over 6, isn't it? Right? Because 12 pi minus pi, that's 11 pi over 6. So we break it down. broke it up into these two. But why did we do that? Well, now look at this. 12 pi over 6, we get 2 pi minus pi over 6. So before I draw the graph, ask yourself, in which quadrant is this angle? This 2 pi minus pi over 6. In which quadrant is it? Well, pause the video and think about it. Okay, now let's draw your graph. So hold your thought. I'm sure you have an answer in mind, but if you're sure, perfect. But if not sure, let's see what the graph tells us. Okay, that's theta, some angle. So according to this, which is the broken up pieces of this, it says go to positive 2 pi. This 2 pi, look at it, is positive. So it indicates movement in the counterclockwise direction, right? Go 2 pi in the counterclockwise. If I do that, I'm going to be back here where I started, 360 degrees, right? But then it says subtract. minus pi over 6 minus pi over 6 minus pi over 6 indicates movement in the clockwise direction so if I do that where will I end up in which quadrant in the fourth quadrant so this angle is in the fourth quadrant here okay So this is, so 11 pi over 6 is this angle. That's 11 pi over 6, right? But the reference angle for this 11 pi over 6, sorry, the reference number is this one. Theta is, of course, is pi over 6. That's the reference angle. So that makes it really easy to calculate this again, right? We've done tangent pi over 6. So let me write this just like what I did in the previous examples. Tangent of, what we're saying is tangent of 11 pi over 6 is equal to tangent of the reference number pi over 6 in quadrant... Four, okay? We're in the fourth quadrant. Very important in terms of the algebraic sign of what we need to calculate. So let's, well, what is the sine? the tangent function in the fourth quadrant? Well the x-axis or the cosine is positive in the fourth quadrant, the y-axis or the sine is negative and And tangent is sine divided by cosine. Negative divided by positive. We get negative. So let's calculate this. We get, again, tangent of pi over 6. I know I've done this in the previous examples, but I'll do it again. Sorry, this is folding over there. So let me do it again for this example. So we have sine of pi over 6 divided by... cosine of pi over 6, right? Tangent is equal to sine over cosine. And then this is equal to 1 over 2 over a square root of 3 over 2. Once again, the 2s cancel out. We get 1 over square root of 3. We need to rationalize this. Square root of 3 over square root of 3. We end up with square root of 3 over 3 over 3. Are we done? No. We just said that the tangent function in the fourth quadrant has a negative sign. And if I bring this back, this circle that I drew for you guys a few minutes ago, tangent in the fourth quadrant is negative. Of course it makes sense because it's sine divided by cosine. And sine, which is the y-axis, is negative, divided by the cosine, which is the x-axis. x is negative divided by positive, we get negative. So we have to put a negative here. And I will then box that. So that is the correct answer. And this, in which quadrant we are, is so... important in calculating these that I want to highlight it like that. Here we were in quadrant three, and here we were in quadrant in the fourth quadrant here. Okay, very, very important. Okay, I think we have more examples, so let's go over those. So here again we have similar but this time we have other functions like sine, cosine, and again tangent. So we have this example where we want to find The exact value of the, again we want exact, right? So we want to leave it as, say, a square root of 2 over 2, or 1 over 2, or the measures that you just saw us give as the final answer, right? So we want to give our answers in exact value form of the trigonometric functions. Okay, and we have several parts, so A, part A. We want to calculate the... Exact value of sine of 2 pi over 3. Okay, so that's what we want to calculate. Well, please pause the video and answer this. Answer this on your own. You know plenty by now. to answer this. So I'm assuming you've done that and now you're back. So this first thing we want to establish is in which quadrant is this angle? So we can break it up just like before. We can break it up into two easy pieces. So let's say that that is equal, 2 pi over 3 is equal to 3 pi over 3 minus pi over 3, right? We can do that. 3 pi minus pi is 2 pi over 3. So this is a good, these are good pieces for that because they're both very easy to calculate. 3 pi over 3 is pi minus... pi over 3. So which quadrant do you think this is located then? So we go pi in the counterclockwise direction, because this is positive, and then we subtract, or go in the clockwise direction, pi over 3. So I know by now at least 90% of you, if not all of you, know in which quadrant this is. But if you have any doubts, then you can go to the next slide. Let me draw a little graph for this. Right? This is theta. So this is saying go pi in the counterclockwise direction because it's positive. And then minus pi over 3. So they can go back in the clockwise direction pi over 3. So we are in the second quadrant. Right? We are in here. Okay, so I'm just going to draw a line in that quadrant like that. So 2 pi over 3 is this angle. This is 2 pi over 3. What is the reference number then? The reference number is this, pi over 3. We already know that, right? Because that's what we have. We went... and then we went back pi over 3. So that's our reference number. So we can easily calculate this. We can say that sine of 2 pi over 3 is equal to sine of pi over 3 in quadrant 2. Okay, great. So now we're going to, from the table, we know sine of pi over 3 is the square root of 3 over 2. Here, let's bring it back. This is why I asked you guys to memorize this, sine of pi over 3. That's 60 degrees, pi over 3, sine of pi over 3 is square root of 3 over 2. But you've got to be careful because you're in the second quadrant. The sine function represents the... the y-axis, right? What is the sine function in the, what is the algebraic sine of the sine function in the second quadrant? It is, is it negative or positive? It's positive, right? So we have, therefore, we have positive square root of 3 over 2. Okay, so this is our final answer for that. Part B is the same angle, so we don't have to draw the graph again. We have it right there. It's 2 pi over 3, but... Sorry, let me move this up so you can see it. I don't know if I can keep the graph on the screen. I'll try. Okay, like that. So this time we have to calculate cosine of 2 pi over 3. It's the same angle. angle, but this time instead of, and the reference number will be the same, it's still pi over 3, and we're going to be in the second quadrant, but instead of sine will be cosine. So let's write that down. So cosine of 2 pi over 3 is equal to cosine of that reference number pi over 3 in quadrant 2. Very important, right? Let me highlight that one and this one. Very important. So cosine of pi over 3 is 1 over 2, but in the second quadrant, cosine is associated with the x-axis. In the second quadrant, that x-axis is negative. So if I write 1 over 2, it's incorrect, because in the second quadrant, the x-axis or the cosine is negative. So we should write minus 1 half. Okay, we have that. Sorry, if I was off the screen, sorry. This computer in front of me, the screen doesn't show. I'll have to turn around and look at the screen to make sure that I am on the screen. I think I've been on the screen most of the time. I think so. So next, part c, and let me move up so everything is visible like that. We have now tangent of 2 pi over 3. So once again, the angle is... stays the same, the reference number stays the same, so we can write, and we're still in the second quadrant, so we can write that tangent of 2 pi over 3 is equal to tangent of pi over 3 in quadrant 2. Like that. And then, so, we got to be careful because in the second quadrant, tangent, remember, is the ratio of sine divided by cosine, right? In the second quadrant, sine, which is... associated with the y-axis is positive, but cosine, which is associated with the x-axis, is negative. So positive divided by negative, because sine divided by cosine is tangent, so positive divided by negative is negative. So that will be equal to sine of pi over 3 divided by cosine of pi over 3. Sine of pi over 3 is square root of 3 over 2. divided by cosine of pi over 3, which is 1 half. Remember, the final answer should be negative, okay? Negative 4. So we have square root of 3 over 1 is a square root of 3, but then we have, we're in the second quadrant, so we have to be, tangent has to be negative in that quadrant, okay? Like that. I hope this is making sense. If not, please join the WebEx sessions and ask me directly. I'll be happy to go over these with you if you have any further questions. Let's do a little more examples. This is going to be a little bit different example than these. So we want to, it says given. The terminal point, we know about terminal point, right? We learned about these in section 5.1. Given the terminal point square root of 5 over 4 comma square root of 11 divided by 4, given that, we want to find sine of, sine of t, cosine of t, and tangent of t. Okay. Interesting, isn't it? Here's why. So here's the solution. Some of these questions are too easy, I think. So they give us the terminal point, right? So let's write it down here. We have square root of 5 over 4, comma, square root of negative 11 over 4. So we already know that in trigonometry, the x component is associated with... Well, let me first write this in algebra. So x, y. So this is our y component. This is our x. component, right? And in trigonometry, x is represented along the cosine, or the cosine is represented along the x-axis. So we have cosine of t, comma, y-axis is represented by the, or the sine is represented by the y-axis. So we have sine of t here, right? These are the terminal point. Terminal point is just a point, right? So it has some coordinate. But we're in trigonometry, so we know that those x and y coordinates translate into x translates into cosine of t and y sine of t. So when they say, given this terminal point, find sine cosine, it's already been given. This is our cosine. This is our sine, okay? So then that's too easy. So cosine of t then, cosine of t is equal to, let me bring it into the screen right there, cosine of t is equal to square root of 5 over 4. Notice we didn't have to do any calculations. Right? And sine of t, these two are immediately known, right? We don't have to do any calculations for these. They're right there in front of us. So, and then sine of t is equal to minus square root of 11 over 4. like that. Then we of course have to calculate tangent of t. Alright, so tangent is sine divided by cosine. So sine is this, minus square root of 11, divided by 4, divided by cosine of t, which is square root of 5, divided by 4. Okay? So the 4 and 4 cancel each other out. What do we get? We get... minus square root of 11 over square root of 5. As before, we don't want to see square roots or the radical expressions in the denominator, right? So let's rationalize this by multiplying the numerator and the denominator by square root of 5. Right? So then we have minus square root of 55 divided by 5. Okay? So that is the value of tangent of that terminal point. So we have that. Okay, and we have that as shown. So these beginning sections in this course are very, very important because they're laying down the foundation. These basic trigonometric functions are easy functions, but remembering them is very important. in the more complicated calculations we'll do later. So please take the time to memorize that table and the circle and the algebraic signs for these trigonometric functions. Of course, those relations... that I mentioned at the beginning on page one, all those relationships, please remember those formulas. And I thank you for watching this video until the end. And please remember, even if I forget, you don't forget to pause the video repeatedly and write down your own solution before you look at what I have done in these examples. So for now, have a nice day. see you again next time.

So to protect the camera, to reduce the glare into my eyes, I'm wearing this. I hope you don't mind of this. So the emphasis in this section is on basic trigonometric functions. These are sine function, cosine function, tangent function, cotangent function, and maybe even we'll get to probably secant and cosecant function. But these are the basic ones.

So, I started, as you can see, I started this, on the screen you see this unit circle, where I want to draw, suppose we have some angle, let's call it theta, in the first quadrant. Right, suppose we have this angle right here, and we want to call that theta. Now that point on the unit circle has coordinates just like the basic algebra. Remember that? It has x and y coordinates.

And I want to call, but here in trigonometry we call them, or we call them, We calculate them differently, even though they're the same thing, x and y coordinates. So let's call it by its original name, point P with coordinates x, y. But that is the same thing. as, let me then expand on this, so it's equal to, instead of x in trigonometry, we have cosine of theta comma, instead of y, we have sine of theta.

Sine function is on the y-axis, and cosine of theta is on the x-axis. So also, if you imagine... this, if you imagine this theta as part of a right triangle, right triangle, that means that this angle here is 90 degrees, then this kind of a presumed right triangle has a hypotenuse.

Let me just for short say HYP for hypotenuse. It has an adjacent side to this angle theta. Let's call it for A, ADJ for short, and there's an opposite side to this angle, right? Let's call that OPP for short. So that side of this right triangle is opposite to theta.

This side is adjacent to it, and this is the hypotenuse of this right triangle. Now, this being the terminal side of this angle theta, it has what we call... projections on the x-axis and projections on the y-axis. The projection on the y-axis is determined by the sine of theta, that's y, just compare that, y is equivalent to sine of theta, that's the projection of this terminal side of this angle on the y-axis, and cosine theta is the projection of this terminal side on the x-axis.

x-axis. That's why we say x is equivalent to cosine of theta. It represents this side of this right triangle.

Okay, so then given this picture or this figure, then we can write a relationship for the sine of this angle theta in the first quadrant. We can say that sine of theta is equal to the opposite side divided by the hypotenuse. Okay, that's one of the important relationships that we, basic trigonometric relationships that we want to remember. The other one is the cosine of theta is equal to the adjacent side or the projection of that terminal side along the x-axis, which would be the adjacent side, divided by the hypotenuse.

That's another formula we want to remember. It's really important to, in mathematics we really don't like to memorize much, but when we do memorize, then there's a significance to that, because given these basic trigonometric functions, especially these two, sine and cosine, You can calculate all the other trigonometric functions. That's why we like to know exactly what's going on, especially with these two trigonometric functions, sine and cosine. Make sure that you fully understand these two, because like I say, you can calculate all the other trigonometric functions if you know these two. Okay, so then let's...

Let me see here. So we have, then based on that, then we have, let's add to that the tangent function or the tangent of theta, like that. So for a second I thought I was off the screen, but no. So tangent of theta is equal to the opposite side divided by the adjacent side.

You can look at it that way. But at the same time, more valuable is to think of it as the ratio of sine divided by the cosine. So we have sine of theta.

Divided by cosine of theta is the tangent function. Like I was saying, you can look at it because the opposite side is along the y-axis, right? It's the projection of this terminal side on the y-axis. So you can also look at it as this. Or, so I'll just add, or tangent of theta.

is equal to the opposite side divided by the adjacent side. Your book, I believe, mentions both, but this is more valuable to remember. But this is also useful in some occasions.

So I say it's worth remembering both relationships for tangent of theta. But like you can see here, like I mentioned, if you know sine and cosine, you can calculate. the ratio and get the tangent. So that's why we like to memorize the sine and cosine. Okay, so let's then add the cotangent of theta.

would be the reciprocal of this, would be ratio of cosine of theta divided by sine of theta, or by the same token, we can say that it's the adjacent side divided by the opposite side. So take the reciprocal of these two, you get the cotangent function. That is why, because cotangent function is the reciprocal of the tangent function, that is why on calculators you don't see a button for the cotangent function.

Because if you know the tangent, value, you simply calculate the reciprocal of that, and you have the value for the cotangent function. Now, the other thing about these, so actually let me write that down, so you have it in your notes, that cotangent function. Or let's start with tangent.

Tangent function is the reciprocal of the cotangent function. And cotangent function can be said to be the reciprocal of the tangent function. You know one, you know the other, right?

Because they're reciprocal of each other. Now we also have two more basic trigonometric functions, right? So far we have shown four. So we have sine, cosine, tangent, and cotangent, right? But we also have a couple of more, and these are equivalent to these.

And those are secant and cosecant. functions. So this secant function is the reciprocal of the cosine function. One over cosine function will give you the secant.

Okay, I think you already see how important it is to know how to calculate sine and cosine because everything else can be calculated by based on those. Now by the same token, let me put a double-headed arrow here and actually a double-headed arrow here too because you can either go from here to there and or from here to here. These two are equivalent.

The same thing we can say about cosine and secant relationship. We can say cosine of theta is equal to 1 over secant theta. It's the same thing, but this time cosine is equal to 1 over secant theta.

These two are equivalent to each other. The same relationship exists between cosecant of theta and sine of theta. If you have cosecant of theta, of theta is equal to 1 over sine of theta. So the cosecant function is the reciprocal of the sine function. By the same token, we can say that the sine function is reciprocal of the cosecant function.

Okay? So these are, like I say, all these formulas that you see on this page, all of these, especially these two, especially these two, that one and that one, are... mandatory memorization assignments for you.

Without remembering those, it's very difficult to know what's going on with these other formulas. So let's look at each one of these trigonometric functions and their graphs and domains. The main thing is their domain, but we will look at their graphs too, because I want you guys to have a good understanding of their graphs too.

That's also very, very important. Overall, this is a very, very visual math course, so you want to make sure that you connect the relationship between these trigonometric functions and their graphs. Very, very important. So we want to write this. Basic trigonometric functions, let's say they're graphs and they're domains.

And the emphasis, of course, is on the domain. But the graphs are important too, so let me also highlight that. Okay, so let's start with the basic and one of the most used ones, which is the sine function. Before that, let me make a note here that all trigonometric functions are periodic functions. Periodic means like they repeat themselves infinitely many times.

If you don't stop them, they keep repeating themselves in both directions on the x-axis, in the negative direction as well as the positive direction. Functions are what we call periodic functions. Okay, so let's start with the domain. Like I said, the domain is very important for us to establish right away.

Let's say domain of sine of theta. So that's the first thing we will discuss. Let's draw a picture of what the assigned function looks like.

So I will draw a long x-axis here, and let's say this axis represents theta, because that's like x, right? That's the independent variable we learned in basic algebra. So in this case, the independent variable is theta. It's an angle rather than any other value.

And, of course, the... y-axis, and that would be for us sine of theta, right? That's our function. Or sine of theta, let's call it y. So theta being on the x-axis.

So if we draw a sine function, this is how it would look. It would go through, it would start from here at the origin, or it would go through the origin, not start there, but it would... go through that.

And let me actually establish that this is, so I want to stay within, I'm just going to draw some dashed lines to make sure that my graph is as accurate as possible. And so this dashed line and another one right below the theta. And I want to make sure that these are equally distant from the origin.

That's necessary for my graph to look right, like that. So the sine function would look like this. It would go like this, come down here, and then go here, and then continue on.

So I'm just going to put an arrow there, meaning that it's going to continue. Along the negative side of the x-axis, the same story. So it will come down here, it would curve back up.

Go here and then come down and do the same thing. Repeat itself on both sides, in both directions. So because we're dealing with angles, then sign, and I'll get to this in about 15-20 minutes.

It's very important that you memorize at which point these trigonometric functions are 0 or have some important value, like 1. For example, the basic sine function, which you see here, varies between 1. That's 1. That's why I drew my dashed lines. And this is negative 1. So it'll swing. It'll go up to 1, come back to negative 1, and go back up to 1, and so on and so forth.

Now, if you have a number before this, like 4, it would times that, and you would get 4. 4 times 1 is 4, or 4 times... 0.5, if you're here, you would get 2. But that would have to come from a factor out here. The angle itself, with no factors, will always swing back and forth between 1 and negative 1. And look at the domain.

The domain has no restriction, right? It will go from negative infinity to infinity. So the domain of sine function...

sine of theta is equal to negative infinity to infinity, and we want to remember that, okay? The other thing you should, the other important part of this graph is where the sine function is equal to zero, right? When it hits on that line, that's when the sine function is zero, right?

The sine function is along the y-axis, and at that point, its value would be zero. So what is that point? This would, if this is 0, so sine function is 0 at 0, or at origin, and then it's also again 0 at pi, and again 0 at 2 pi, and 0 again at 3 pi, and so on and so forth.

So what would that make this peak then? What angle would that be? That would be pi over 2, because look at this, this is so symmetric, right?

From 0 to pi, half of that distance is pi over 2. So if you have your calculator at your disposal, you can just take the sine of 90 degrees or pi over 2. It's going to tell you that it's equal to 1, which is true right there. Sine of pi over 2 is 1. Then you can calculate sine of pi. It's going to be 0. Sine of 2 pi, again 0. So which is this one then?

This would be... Sine of that angle is negative 1, right? So what is that angle then?

3 pi over 2. Okay, I'm sorry, these are close to each other, but that's, let me change the color then. So that would be 3 pi over 2, which is that, the dashed line that you see. Okay, very, very important to very early on establish where this, what sine function does basically. Where is it zero, where is it one, where is it negative one. See, make sure that you remember what angles, at which angles sine of function is zero or one or negative one.

Very, very important. The other trigonometric function that we need to know very well is the cosine function. So domain of cosine of theta, and I'm going to do the same thing, draw a long axis for theta.

And then I'm going to, so that's theta, and then I'm going to draw a vertical axis for the y equals cosine of theta, right? So cosine of theta equals y. So the cosine function, and I will draw my dashed lines to make sure that I draw the graph accurately. as much as possible. So let's say that's the line y equals 1, and I'm sure you guessed what this other one will be, y equals negative 1, right?

The basic cosine function, just like the sine function, will vary between y equals 1 and y equals negative 1. But unlike the sine function, the cosine function begins... at here, at 1, not as 0, not as 0 like the sine function. It starts at 1. Why?

Because cosine of 0 is equal to 1. That's why I keep asking. I'm asking you guys to please memorize the value of the trigonometric functions at these special angles. Zero angle, pi over 2, pi, pi over 4. We'll get to that in 10-15 minutes.

I'll show you a table that you would need to memorize basically. Or a little bit of it, if not all of it. For sine and cosine values, you should have those memorized. Very important. So we start from here.

This is what the cosine function does. comes down here, goes back up, and then curves back down again when he tests the y equals 1 line. Okay, so now we know that at 0, at theta equals 0, cosine function is 1. It's at highest value, 1. Cosine of 0 angle is 1. Cosine of then, if at 0 it's 1, when is it then 0? It's pi over 2. Then when is it negative 1?

If it's 1 at that, at 0, at which point is it negative 1? It's at pi. Okay, so that distance.

And then again, it's 0 again at 3 pi over 2. And then finally, it's 0 at pi over 2. it is let's see 3 pi over 2, and then at 2 pi again, yes, at 2 pi, again, it hits the highest value, 1. Okay, and just like the sine function, this can go on and on on both sides. So let me actually continue that. So it can go again, do that, come back up again, do that, and so on. On this side, it will do the same thing. So it'll do this like that.

So that's why we say that trigonometric functions are periodic. They do that over and over again. So take a look at this function. The domain is unrestricted, right?

So just like the sine function, the domain of the cosine function is also negative infinity to infinity. Data is equal to negative infinity. infinity. Okay, and now you have a visual reference as to why that is the case. Next, let's look at tangent function.

So we want to look at the domain of the tangent function, okay? Now this is why I was saying earlier to please memorize these these basic formulas because if you did that right away you know that tangent function is a ratio of two numbers. That's very important, right? You've been through algebra before, so you know when you have a fraction, you can't have a zero in the denominator.

In which case, it would make the tangent in this case undefined. So if you remember this formula, then right away you want to ask, where is cosine of theta equals zero? Because those points where cosine of theta is equal to zero will be your vertical asymptotes.

We talked about vertical and horizontal asymptotes a lot in Math 126. So at wherever we have cosine of theta equals zero... The tangent function has a vertical asymptote, meaning its graph cannot pass through that point. Okay, so let me again draw this long theta axis here. I will draw it here, I guess right here too. Maybe here.

So we have that. This is our theta axis, and let's put the vertical axis, which would be y equals tangent of theta, right? Tangent of theta equals y.

So then, given that, and I'm going to... to write that maybe here, maybe here. So tangent of theta is equal to sine of theta divided by cosine of theta. It's so important that I wrote it right away there. Now in the previous discussion, when we talked about domain of cosine of theta, notice what we found.

Cosine of theta is 0 at pi over 2, at 3 pi over 2, and on and on, at 5 pi over 2, and so on and so forth. So we want to account for all of those points where cosine of theta is going to be 0. We do not want the domain of tangent theta to include those because at those points... it will be undefined.

So let's draw asymptotes there. So I'm going to draw an asymptote here, and I'm going to label it as pi over 2. So this will be pi over 2. Then I'm going to draw another one, let's say here, because we can go on and on, right? But we don't want to do that. Just to make a point, a few would suffice. So these, by the way, should be equally distant.

from here to here and from here to here. I don't know if I did a good job in doing that. Maybe not, but it's okay. Well, let me correct this.

Before I post this on Canvas, I will delete this. But this dashed line is not in the right place. And I will tell you in a second why. Why not?

So before I post these lecture notes on Canvas, I will correct this. But the correct position for this line will be here, somewhere here. So please ignore that for now.

Ignore this line. Okay, I will erase it and clean this up. So the correct point would be here, like that.

And that would be 3 pi over 2. Okay, so what would that be? If this is pi over 2, what would that be? That would be minus pi over 2. And then...

That distance from this dashed line to that dashed line, so that would be as wide as this, roughly, right? As wide as this. I'm going to put another vertical asymptote there. And I'm going to call that minus 3 pi. over 2. So these are our asymptotes.

Then the tangent function, the tangent function would graph like this. So this line would come from here, go through the origin there, and then the arrows. And then here we would have A line going like this. So do you see how these periodic functions repeat themselves, right? So from here, from this, from equivalent points is one period.

From here to here. From here to here. That's one period. From here to here is one period. From here to here is one period.

And then this one repeats itself on this side and on and on and on. So these are, these asymptotes are exactly where the cosine of theta is zero. Okay, I don't want to make this drawing too crowded. So I'm not going to draw the cosine graph, but if I did, you'd see that this is where the cosine graph is equal to pi.

over 2 so cosine of pi over 2 would be 0 and again cosine of 3 pi over 2 would be 0 so tangent function cannot go through that. Okay so the domain looking at this graph the domain of tangent of theta is equal to all real numbers except pi over 2 plus n pi, and n being an integer. So if you plug in for n with 10, then you have 10 pi plus pi over 2 is also a vertical asymptote.

So except these guys. So all real numbers except these guys. pi over 2, 3 pi over 2, negative pi over 2, and I should actually add a plus or minus, right?

The minus indicates in this direction, okay? Because when this is minus pi, then you get pi over 2. 2 minus pi, you get minus pi over 2. So you get these guys. Okay, that's the domain of the tangent function. Let me box that here.

Okay. I'm debating myself whether I should draw the cosine curve or not. Let me lightly draw it, because it's important for you to see that these asymptotes occur whenever the cosine curve is drawn.

function is 0. So if we have, if we have say cosine function coming from here like that, then it would go like that and so on. So that's the cosine of theta. On this side, we'd have it come down here, go through that, and then come back up, go through that, and so on.

This is the cosine of theta function. And look, wherever it's zero, you have the asymptotes because of this formula. Okay? I just did it once.

I think it's better not to draw it because the graphs become too crowded, and I don't want to do that. Okay? Okay. All right, so that was those basic trigonometric functions. We have more, so we have the cotangent function next.

So we have domain of cotangent of theta, and do you remember what the cotangent of theta was equal to? It was equal to the ratio of cosine theta over sine theta. Again we have a fraction which is important to remember because then whenever the sine of theta is zero you're gonna have a vertical asymptote. Where would those asymptotes be then?

Sine of zero angle is zero, sine of pi is zero, sine of 2 pi is zero. So at theta equal zero pi 2 pi, 3 pi, 4 pi, and on and on and on. And at 0, minus pi, minus 2 pi, minus 3 pi, you're going to have vertical asymptotes for cotangent function.

Okay? That's why it's so important to remember those formulas. So cosine of theta divided by sine of theta. Okay, so once again, let's draw these. We have this theta axis like that.

And let's put the vertical axis. I want to make the vertical axis a little bit thicker. Maybe, well, maybe, should I do that?

I think it's better if I do. Let's say, so I'll make it a little bit thicker than the x-axis because I want it to stand out here like that. So this would be y equals cotangent of theta, right, axis.

Now, we just discussed that whenever the sine of theta is zero, we're going to have a vertical asymptote. So that's a vertical asymptote, right, that one. So I'm going to put a dashed.

I'm going to put a dashed line right up against the y-axis. That the y-axis is an asymptote. So the graph of the cotangent function cannot cross this line. Okay? And then where will be the other asymptotes?

So this is at theta equals zero. You can tell I've taught algebra a long time. So that's not x, but theta in this case. I will fix this. I will make it look better before posting this in Canvas.

The next one will be theta equals pi. So let me draw it here. That will be another occasion where sine function will be 0, the denominator will be 0. Of course, we do not want that. So theta equals pi, and I'll draw another one on this side and a couple of dashed lines on that side. So we have another set of dashed lines on this side, and this will be 2 pi.

Okay, and then a couple of dashed lines on this side. And these have to be equally distant from each other. From here to here and from here to here should be equal. And then now from here to here.

Right, from here to here. And this would be theta equals minus pi. And one more.

Theta equals minus 2 pi. Okay, so the cotangent, the graph of the cotangent function, in terms of the way it looks, it goes the opposite direction. So let me bring back the graph of the tangent function. So the graph of the tangent function was going like that, right? And also the asymptotes were different than cotangent.

But the point I'm trying to make, tangent was going like this. Now cotangent is going to go like that. It's going to reverse it. itself, but it's going to stay within its own asymptotes. That's very important.

And these are the asymptotes for the cotangent function. So here we will have the cotangent lines. Let me draw that. So one would go through.

here and so that's one. Another one would go. I didn't draw this very symmetric.

This has to be in the middle of this line, sorry about that, but I will label it so you know exactly where that should be. I will label it in a few seconds. And then we would have like that.

That's better, right? So this is more in the middle of those two dashed lines. And another one, like that, like that.

Sorry if my drawing isn't better than this. So you might ask, well, what are these points where it is the cotangent function is equal to 0? That's really easy to determine.

because that's whenever the cosine of theta is equal to zero, right? Cosine of theta is equal to zero at pi over 2, or halfway between zero and pi, right? This is zero, this is pi.

Halfway in between is pi over 2. Now this, in terms of my graph, is poorly drawn because this point should be in the middle here. But it's not. But let me label it correctly.

Between pi and 2 pi, the middle point is 3 pi over 2, at which point the cotangent of theta would be equal to 0, right there. And this is the same thing, but minus. So this would be minus pi over 2. And this one here would be...

I'm sure you guessed, it's minus 3 pi over 2. And of course, it'll continue in both directions because it's a periodic function. So when we write the domain for the cotangent function from this graph, you can tell that all real numbers are acceptable. except any multiples of pi, right?

So 0 pi, that would be 0. It's an asymptote. So we have to exclude these asymptotes. How do we do that?

This is how. We say that domain, so the domain of the cotangent of theta, is equal to all real numbers, and here's the important part, except what I've shown in dashed lines, the vertical asymptotes, except n pi, that's just n pi, 0 pi, 1 pi, 2 pi, and I should say plus or minus actually, plus or minus n pi. Okay, so because the minus... will take care of the asymptotes in this direction and positive and pi will take care of them in that direction and n is an integer any integer 0 1 2 negative 1 negative 2 and so on so this is the domain So that's why I don't want to give the domains without drawing the graphs, even though I'm not terribly good at drawing these graphs, but I'd like to give the domain after I show you the graphs, because now it makes sense.

right, this except n pi are all these vertical asymptotes. And we know why, because at these points, this ratio will have a 0 denominator. And we can have that.

OK, so we have a couple of more domains to look at, domain of secant function and domain of cosecant function. So let's look at those. domain of the secant of theta, which by the way, let me remind you that secant of theta is equal to 1 over, do you remember which function it was?

1 over secant is the reciprocal of cosine of theta, right? Cosine of theta. Anyway, let me put a line there, so this is different than that. Okay, so now, once again, let's draw the picture so we can make sense of the domain once we write it down.

Now, from this equation, you already know that the domain is a constant. You already know where the vertical asymptotes are going to be, right? Do you need even me to tell you that? Look at this.

You don't need me to tell you that, right? Whenever cosine of theta is zero, we're going to have vertical asymptotes. So this is our theta axis. Then we will have the y-axis. I'm going to draw it here somewhere.

Here maybe, that's better. Like that. And then we have, of course, the y is the secant of theta in this case.

Okay. All right. So first, let's draw a nice... cosine graph, right? Because we're going to need to know where this cosine is going to be 0. So here's the cosine.

I've already done it for you on the previous pages. But let's do it again. So we have, I'll try to make it as accurate as I possibly can drawing these things by hand. And so on, right? On this side also.

We'll come down, do this, go back up, come down again, and do that. So wherever you see the cosine, and let me write that here, that this is the cosine. cosine of theta function, okay?

Wherever the cosine of theta is zero, all these points, all these points, This fact, this ratio is going to be undefined. So secant of theta will be undefined. So all those dots that you see where cosine of theta is 0 are going to be our vertical asymptotes. So let's draw them here, some of them at least. I won't draw all of them, but so it will be one here, another one here.

Another one here, right? This graph didn't turn out to be too bad because the distance between these dashed lines is almost equal. Right? Not too bad.

And let me also draw another one here. Another one right there. Okay, now the secant function will be in between these asymptotes. So it'll be, we will have one here. So we will have one curve here.

We will have one curve here. Let me move this up a little bit. We will have one curve here like this, one curve here.

And I want to draw a little bit of it. So one curve would be here like that. Like that. Okay.

And on and on. So we could put another one here like that. like that but there will be more asymptotes on that side and that's on this side and that side all right so this would curve would have to curve up this way because there's there's going to be an asymptote right there as well and another asymptote here Okay, so then the question is, what are these points?

Where are these? Those points are where cosine of theta is equal to zero. So this point here would have to be pi over 2. This would be 3 pi over 2. And this would be 5 pi over 2 and so on. On this side, we will have minus pi over 2, right?

And then minus 3 pi over 2 and minus 5 pi over 2 and so on and so forth. Okay? So.

You can see right away that the domain of the secant function is all real numbers except anything that goes through pi over 2. Pi over 2, minus pi over 2, minus 3 pi over 2, and so on. So let's write that out. So we say domain. of secant of theta is all real numbers.

Here comes the important part. Except pi over 2 and anything. That is n pi. That is pi plus that. Because when you have pi over 2, for n equals 1, you'd have pi plus pi over 2. But what will you get if you do that?

You'll get 3 pi over 2. What about if you put 2 pi? Well, then you get 5 pi over 2. So this piece hits all the vertical asymptotes. And if you want, we can draw one here, too, since I drew the graph for it. Let me draw one there and another one here like that. Okay.

And I'm going to box that. So if you ever wonder or if you want to better understand why this domain looks the way it looks, just look at the graph. Now you have a graph for all these domains. Please refer back to the graphs.

We have one more to draw for this, and that will be the cosecant function. I believe that was page, so this is page 4. So domain of cosecant theta. Do you remember what the formula was for cosecant of theta?

Should I bring the first page to remind you, or do you already remember it? Do you remember that would be... fantastic. The cosecant function is the reciprocal of the sine function.

So that's very important right? By now you know that because whenever sine of theta is equal to zero then we're going to have those vertical asymptotes because the cosecant of theta cannot go through those points. So that's why that those formulas are very important. Okay, alright, so just like before, before we give the domain, let's give a picture so we can make sense of the domain once we write it down. Okay, so we have something like this and the theta.

I should have drawn this line a little bit lower because I have all this space, and so it's okay, it's fine. And then let me draw the vertical asymptote. Let's call this y equals the cosecant theta, right?

Cosecant of theta function. And so because there's a relationship between cosecant of theta and sine theta, let's draw the sine theta graph. Let's draw this graph first, okay? Because we're going to need to know where the sine theta is equal to zero. So the sine graph does this.

It does go through, do that, then come down, do that, do that, do that, and do that, and so on. Remember what this number was? That was 1. And here... That was negative 1, right? The basic sine function swings or oscillates back and forth between 1 and negative 1 on this side as well.

So we can draw the graph on this side. If I can keep it nice and smooth, that would be good. Something like that. Like that. Okay?

Let me put an arrow there. So you can see clearly at... At some point, points, the sine function is zero at that point, at that point, here, here, here.

All those dots that I'm putting on the graph, that's where the vertical asymptotes should go, right? Because the cosecant function cannot go through those points. So I'm going to go and put a dashed line here. I'm going to put a dashed line right through that dot. where sine of theta is equal to zero, another one here, another one here, another one here.

Okay, so then where are these asymptotes? We know that sine of theta is equal to zero at theta equals zero, so that's one of the angles. At theta equals zero, we have an asymptote, right? So let me actually put a dash line right next to this y-axis to indicate that the y-axis is also a vertical asymptote here, like that, okay? So let's label these.

We have that's zero. What about this one? The sine of theta is also equal to zero, so Theta equals pi.

What about here? 2 pi, 3 pi, and so on and so forth, right? How about this side? Minus pi, minus 2 pi, and so on and so forth.

So where would the cosecant function graph be then? Well, it will be right at the top of each peak and at each valley here. So it would be like this. this, it would be like this, then like this, like this.

Do you see why it's important that we draw the sine function first? Because we know exactly where these asymptotes are and where the peaks and valleys are, so we can draw our cosecant graph very nicely like this. it doesn't look too bad, like that and like that, and we can continue on either side, going in each direction. Okay, so that's that. Now the domain.

Now that we have the graph, the domain is really easy to write down. I'll move this up a little bit. So the domain of the cosecant function is, again, all real numbers except these guys, right?

n pi, any factor of pi, plus or minus, right? All real numbers except, I'm going to write that part with red or in red, except plus or minus n pi, where n is any integer. Let me move this up a little bit. I'm trying to keep the graph on the screen, but it looks like I need to scroll up a little bit. So integer, like that.

Okay, now when you read this, you know exactly why except n pi should be there, right? Those are all the points where the sine function is equal to 0, which we cannot have because sine of theta is the denominator of this formula for the cosecant function. Okay, so now let's talk about some of the important concepts that you need to memorize in order to do well in this course. I don't know how to say it in a different way, but honestly, you need to memorize these topics or these quantities.

So we're going to talk about that. So I've drawn this table, but it's blank as you can see, so we can fill it in with the right quantities. Here are the special trigonometric angles for which you need to remember all these. So, well, not all these, but at least the sine and cosine. Cosine.

Okay? So here you have, on the outside of the box, I put the measures in degrees. But I know, I've already mentioned this, when I took this course as an undergraduate, first you're trying to stick with degrees.

I was the same way myself, but very quickly you're going to realize that this is actually going to make things more difficult. So you need to quickly switch, even if you're comfortable with these, switch to to their equivalent measures in radians, which are these ones. So 0 degree, pi over 6 is 30 degrees, pi over 4 is 45 degrees, pi over 3 is 60 degrees, pi over 2 is 90 degrees, and so on.

Make sure that you switch your mind to these measures, the radians. With these, you're going to have considerable difficulty. Now it's the beginning of the semester, but the further we go into the semester, you're going to need to know these.

very well, okay, and what they represent. Pi over 6 is 30 degrees. It's easy, right? 180 divided by 6 is 30 degrees.

So then at the top, we have these trigonometric functions that we've been talking about so far. Sine of theta and cosine of theta, and then we have the tangent, and not theta, t, this time t. So let me put that in parentheses.

So tangent of t t, t being the angle, right, and then cotangent of t, and then secant of t, and then, of course, the last one, cosecant of t. So sine of 0 is 0. And let me actually switch to a different color here, 0. Sine of pi over 6 is 1 half. Sine of pi over 4 is square root of 2 over 2. Sine of pi over 3 is square root of 3 over 2. Sine of pi over 2 is 1. Okay? It's very important that you remember these quantities for sine and cosine. And then cosine of 0 is 1. Cosine of 30 degrees is square root of 3 over 2. Cosine of pi over 4 is square root of 2 over 2. Cosine of pi over 3 is 1 half.

Cosine of 90 degrees is, of course, 0. So if you look, there is a pattern, right? The pi over 4 or 45 degrees has the same sine and cosine. The sine of 30 degrees is equal to the cosine of 60 degrees. And the cosine of 30 degrees is equal to the sine of 60 degrees. Okay?

There are relationships here. There's a pattern here. Okay?

You've got to remember this because if you do, then you can calculate the rest of this table based on your knowledge of these two functions. Okay? So tangent is sine. over cosine, so that'll be 0. 0 divided by 1, 0, that's what I'm saying.

Once you know these two columns, this, this, you can calculate all of this. So this will be undefined, right? Do you know why?

Why is this undefined? Because cotangent is cosine divided by sine. So 1 divided by 0 is undefined. So these two columns are very important.

Let me... emphasize that by putting, highlighting that and that. Okay, these two columns are very, very important.

And then let me fill in the rest of this, so 1. But you can verify. If you know this, you can verify all these quantities based on this and the formulas I gave you on the first page of this session. And cosecant would be undefined.

And this would be tangent of t would be square root of 3 over 3. Then this would be square root of 3. And then 2 square root of 3 over 3, 2. And then here would be 1, 1 again. And then square root of 2, square root of 2. And then we have square root of 3, square root of 3 over 3, and 2 and 2 square root of 3 over 3. And then finally at the bottom we have undefined, 0, undefined again. So test yourself.

Memorize these and see if you can calculate these, any of these. Okay, tangent of 60 degrees is square root of 3. Secant of 45 degrees is square root of 2, and so on. Then the other thing that you should remember and have this in your memory at all times is the sign of the algebraic functions in each quadrant.

So this is... quadrant 1, let me use my thicker marker here, this quadrant is quadrant 1, in here this is quadrant 2, this is quadrant 3, and this of course is quadrant 4, right? So make sure that you know the algebraic sign for these trigonometric functions. When we started this this session, right, I mentioned that cosine is equivalent to x, sine is equivalent to y.

So whenever cosine is, x is negative, cosine is negative. X is positive, cosine is positive. Whenever the y-axis is positive, sine of theta is positive. Whenever the y-axis is negative, sine of theta is negative. is negative.

And the other trigonometric functions are the ratios of those. So we can easily then obtain those two. So we have here, in the first quadrant, everything is easy, right? Everything is positive because the x and the y-axis are positive.

So positive, positive, positive, all of these are positive. Now here, things are getting a little bit more interesting, right? Because in the second quadrant, which...

axis is positive? The y-axis. Y-axis is tantamount to the sine of a function, right?

So then I should have said sine theta, but you get my point what I'm talking about here. So this would still be positive, but look at the x-axis. It's negative, so cosine would be negative.

Tangent is a ratio of sine divided by cosine. Positive divided by negative, negative. Cotangent is cosine divided by sine.

So negative divided by positive, negative. Secant is the reciprocal of the cosine function, so negative again. Cosecant is the reciprocal of the sine function, so one over a positive number, it's going to be a positive number, right? So in the second quadrant, look at that.

The only positive trigonometric functions are sine and cosecant, okay? I'm sure you're seeing the benefit. of this, to remember this.

It'll come in handy a lot, as you'll see in the examples that are coming up. Okay, so what about the third quadrant? Look at that. Is x, what is x? Is it negative or positive?

It is negative. What about the y axis? Is it positive or negative? Negative.

So the sine function, which has to do with the, which is equivalent of the y axis, is negative. Cosine function, which is equivalent of the x axis, negative. negative. What about the tangent?

What about the tangent? So tangent is the ratio of the sine divided by cosine, right? So negative divided by negative, it is positive.

Cotangent is the ratio of cosine divided by sine. So again, negative divided by negative, positive. How about the secant function? Secant is the reciprocal of which function?

Reciprocal of? Cosine function. So if cosine is negative, take the reciprocal of a negative number, you get a negative number.

So negative. How about cosecant? Cosecant is the reciprocal of the sine function. And sine function is negative in the third quadrant.

So take the reciprocal of a negative number. What do you get? You get a negative number. So in the third quadrant, the only positive trigonometric functions are tangent and cotangent. But do you have to memorize?

this? No. You just have to memorize that sign representing trigonometry.

Cosine represents the y-axis. Cosine represents the x-axis. And then go from there. Then you can easily translate wherever the axis is negative or positive, then the trigonometric function associated with it will take the same sign.

And how about the fourth quadrant, right? So in the fourth quadrant, the x-axis is positive. What does that mean?

Then which one of these should be positive? The x-axis is associated with cosine, so that's going to be positive. What else?

What else should be positive here? Well, secant function is the reciprocal of cosine function, so the secant function also should be positive, right? Now sine for sure should be negative, right?

Because that's sine function is associated with the y-axis, so that's negative. In the fourth quadrant, y-axis is negative. Tangent function is the ratio of sine divided by cosine and negative number divided by a positive number we get a negative number.

Cotangent is the ratio of cosine divided by sine. Positive divided by negative we get a negative number. Coscant function is the reciprocal of which function? Do you remember? Coscant function is the reciprocal of the sine function.

So if you take the reciprocal of a negative number what do you get? You get a negative number. number. So in the fourth quadrant, the only positive ones are cosine and secant, which are reciprocals of each other, by the way.

Okay, so in addition to this table that I asked you to at least memorize these two first columns for sine and cosine, you also need to remember the algebraic sine of trigonometric functions in each quadrant. Of course, as I said, only you need to know, remember that's what sine and cosine are. and then you can calculate the algebraic sign for all the other trigonometric functions based on those two, sine and cosine. So before we go over some examples, there are some things that we need to know. Some of it from, actually one or two of it is from the precalculus one.

So it's nice that that stuff that we learned in the previous class... useful in this class as well. So I'm going to say recall, recall from pre-calc 1 or math 126, you remember that course? Okay, that odd functions are functions such that if you replace the x with minus x, it will give the result is minus f of x. And even functions...

would be if you replace the x with minus x, that'll be equal to f of x, right? Minus x and x, for an even function, they're equal, okay? This, we remember... two let me box them separately very important in this course as well as the pre-calculus one we learned before right the other thing that I want to make a note of now we want to connect this to this stuff to trigonometric stuff right which functions are odd which functions are even so let's connect them to trigonometry So the cosine function, cosine of theta, and secant function, secant of theta.

So this argument can be anything, right? T, theta, x, whatever you want it to be. I just chose theta. These two functions are the only even trigonometric functions. Functions are even, right?

So just like any other even function, if you replace theta with its equivalent negative equivalent, you still get theta. So let's write them down. So therefore... cosine of minus theta is equal to, just like that, just like the even functions in algebra, is equal to cosine of theta.

Okay, that's one thing we should know. And the secant function, which is a reciprocal of the cosine function and vice versa. So secant of negative theta is equal to secant of theta. Like that. These are the only two even trigonometric functions.

The other ones are odd. So let's write them. Let's make sure that you have this in your notes.

So note that sine of theta. Then what is tangent of theta? What else?

We have cotangent of theta. And lastly, cosecant of theta, right? Cosecant of theta is the reciprocal of sine theta.

Are all odd functions. All odd functions. Okay, so what does that mean? This is what it means. Just like what the odd functions meant in basic algebra, this minus in the argument translates in the minus in front of the function.

Those two are equivalent. So then we can say that... Sine of, we start with that, sine of negative theta is equal to minus sine of theta. You know, trust me, this simplifies when we get to where we have to solve trigonometric functions.

functions, which is several or many sections later. These help us simplify expressions or simplify our calculations so much. So please make sure that you remember this. Then tangential functions are the ones that are important to us.

So tangential functions are the ones that are important to us. So tangential functions are the ones that are important to us. tangent of theta, tangent of, sorry, minus theta is equal to minus tangent of theta.

So you can take that negative out and put it in front for sine, tangent, cotangent, and cosecant. Okay, and then so cotangent of negative theta is equal to minus cotangent of theta. And finally, of course, cosecant of negative theta is equal to minus cosecant of theta. I guess I can box all of these at once. These are very, very important to remember so let me just emphasize here and then also here.

The other things that your book talks about in this section, and these are very important, again they become very very useful when we start solving trigonometric equations, are some formulas. So let's, your book mentions them in this section, so I'm going to mention them in this section. and they are called the Pythagorean identities. That's what they're called. So we have the sine squared t, some angle t, plus notice that I've used theta, now I use t.

You have to get used to these various forms of the argument being represented. Cosine squared t, so sine squared t plus cosine squared t equals 1. Very nice and easy formula for us. us to remember sine squared plus cosine squared is equal to 1 then the other form the other two there's there's three of them three Pythagorean identities 1 plus tangent squared of T is equal to secant squared T 1 plus cotangent squared t.

is equal to cosecant squared t. Very useful formula, so you'll find this out. So the sooner you memorize these, the better.

Okay, finally, we get to some examples. So all of this was just introducing these trigonometric functions, their relationships with each other, their domains, their graphs. All of this is very important for us to connect with each other, the graphs with the functions, with formulas, domains, and everything.

Okay, so let's start. The first example, we want to find... the exact value of the trigonometric function at the given real number.

So do you know when we say the exact, see this is very important, the exact value, do you know what that means? Let me bring this table back. See, these are exact values.

That means you don't want to calculate the decimal values. These are all exact values, all of them. So you don't want to calculate this with your calculator and say, oh gee, this is point. Let's say 6, 4, something. You don't want to do that.

So you leave them in exact form. These are all exact forms. In trigonometry, most of the time, we want exact values. Sometimes we also want to have exact values.

also calculate the decimals. But look at all of this table you have. All you have are exact values. Very important. Okay, so that's what they mean.

So the first one, we're going to do a number of these examples. The first one, let's call it A. is we want to calculate tangent of 5 pi over 6. Okay?

So you remember in one of the previous sections, or section 5.1, we talked about reference number, right? So now that becomes very, very handy here to remember to connect that topic with this. So in these, I'm going to draw graphs, but you don't have to. I'm just drawing graphs for all of them. these for this for sure you don't have to draw graphs i'm sure there are some of you who will answer this knowing knowing the table knowing that this table and remembering what we said about reference numbers in section 5.1 i'm sure some of you will be able to answer this without drawing any graphs very easily too but i draw graphs just in case some people want to have a visual reference I myself would like visual references in topics like these.

So let me draw a little graph here before answering this. And so let's call this, I don't know, if you wanna call it theta or something, any, it could be x2. And then the y-axis here. This is our function.

And then so here the function is tangent of 5 pi over 6. So where is this angle? So if this was 6 pi over 6, let's translate it, right? So we can say 5 pi over 6 is equal to 6 pi over 6 minus pi over 6, right? You can break it up into this. 6 pi over 6 is just pi over 6. pi minus pi over 6 is going to give you 5 pi over 6. Right?

So where is this angle? Let's see. Pi minus pi over 6. So now you know that 0 to pi is from here to here.

0 to pi. Going in the counterclockwise direction because pi is positive. It's not minus pi. Otherwise I would go in the clockwise direction.

If this was minus pi, I would go this way, clockwise. But it's positive, so I'm going counterclockwise from 0 to pi. And this is pi, and that's 0. Theta equals 0, and theta equals...

equals pi, right? So 0 to pi is that, but is that, that angle is pi minus pi over 6. So not only we go to pi, but then we have to come back, we have to subtract pi over 6 from it. So where do we end up? We end up in the second quadrant. Okay, so 5 pi over 6 is this angle.

Well, let me draw it with my red ink. So that is 5 pi over 6, or pi minus pi over 6. Okay? So in section 5.1, we learned that if you find the equivalent of this angle in reference numbers, you can easily calculate what that number is, right? So let's call this theta would be pi over 6, right?

This angle, let me just say pi over 6. Now the reference number is pi over 6, so the tangent of that angle of this big angle is equal to tangent of this reference number but you have to remember please that you're in the second quadrant so you calculate tangent of pi over six but in the second quadrant so let's do that so tangent of Tangent of 5 pi over 6 is equal to tangent of pi over 6 in quadrant 2. Okay? In quadrant 2. Okay, so that would be tangent of pi over 6 is equal to sine of pi over 6 divided by cosine of pi over 6. This is where remembering those relationships that I mentioned will be very, very useful to you. And also that table, right?

Sine of pi over 6 is 1 over 2. That's in the table that I asked you to memorize. Cosine of pi over 6 is equal to square root of 3 over 2. Okay? So this is a complex fraction. 2 and 2 cancel out. We get 1 over square root of 3. But then this is, let's rationalize this.

So we're going to multiply it by... by square root of 3, because we don't want a square root in the denominator, right? Multiply it by square root of 3 over square root of 3, because what is the square root of 3 over square root of 3? It's 1, right? 5 over...

5 is 1, negative 2 over negative 2 is 1, so we're multiplying by 1, really, which is important because that means we're not changing this. Anything multiplied by 1 is not going to change, but... What we're achieving with this is that we're going to rationalize the denominator, which we learned to do in pre-calculus 1. So this will be square root of 3 over 3. Am I done?

No. That's why it's important for you to remember that what you're calculating here is the reference number, not this, not tangent of 5 pi over 6. So what difference is that? that make? The sign, the algebraic sign in quadrant 2, in quadrant 2, the tangent is negative. So because of that, we have to put a negative here.

So the answer is negative square root of 3 over 3. Okay? That, which quadrant we are, very important because that determines the sign. If we were in the first quadrant, that would be positive.

But we're in the second quadrant, where tangent is negative. Okay? And that's why I showed you guys that here.

That's why you have page 5. Remember, page 5, when you download the lecture notes, page 5 has the table and this circle with all the algebraic signs. In the second quadrant, the tangent is negative. is negative. Okay, do you see that?

Very important. Alright, so another, this would be part B, but we're going to do the same thing, tangent of, we're going to find the exact value of this. of this quantity, tangent of 7 pi over 6, right? That's the question.

Well, just like before, break this up into terms that are easily calculated. So this will be equal to... Let me...

So this would be the solution. So that is equal to, let's see, 6 pi over 6. Let's break it up. 6 pi over 6 plus pi over 6, right? So 7 pi over 6, I'm just breaking up the angle, and then we're going to get involved with the tangent later.

7 pi over 6, you can write it as 6 pi over 6 plus pi over 6, which is 6 and x. 6 cancel out, you get pi plus pi over 6. So can you see which quadrant this angle is? I'm sure you can, most, if not all of you, most of you are seeing where this is. Well, if you're not sure, let's draw a graph. So this is our theta.

And we have the y-axis. So this angle, which is 7 pi over 6, is pi plus pi over 6. Note that both angles are positive. So the direction of movement is going to be in the counterclockwise direction. So from here to here is pi, then plus another pi over 6, so we're in the third quadrant.

So let's draw the angle right there. Somewhere here, we are somewhere here. Here, that's the terminal side, and the angle itself, how did I draw it over there? With red ink, I'm going to do the same thing here. So that is the angle.

This angle is 7 pi over 6. but what about the reference number? The reference number is the angle this is something we learned in section 5.1 the angle between the terminal side of the triangle this angle and the x-axis so that's our reference angle theta okay it's pi over six because we know to get there we we did pi plus pi over six so we know that that angle is pi over six so remember we got to be very careful because it's really easy to calculate tangent of pi over six but then you got to remember to apply the appropriate sign for the third quadrant. Okay, just remember that.

Alright, so now we're going to involve the tangent function. The reference number is pi over 6. So we say tangent of 7 pi over 6 is equal to tangent of our reference number pi over 6 in quadrant 3. Okay? So, that determines the sine, right?

In the third quadrant, the sine and cosine are both negative. So, the tangent is the ratio of sine divided by cosine. So, negative divided by negative, tangent is positive in the third quadrant.

And if you have any doubts... Let me bring this back, this sheet, this circle. In the third quadrant, the tangent function is positive because tangent is the ratio of sine divided by cosine.

So negative divided by another negative, we get positive. Okay, so now tangent of pi over 6 is equal to sine of pi over 6 divided by cosine of pi over 6. Once again, sine of pi over 6 is... 1 over 2, cosine of pi over 6 is square root of 3 over 2. This is from the table, right? Sine of pi over 6 is 1 over 2. Cosine of pi over 6 is the square root of 3 over 2. That's why I ask you memorize these first two columns because then you can calculate the tangent based on that. So we are on our way to calculating that.

tangent pi over 6. So we're getting there, but it's in the table. If you can remember it, you can write it right away. So this complex fraction, we get just like before, this is very similar to part a.

We get 1 over square root of 3. Again, we don't like, we learned in pre-calculus 1 that we don't like to see radicals in the denominator. So let's rationalize it by multiplying it by square root of 3 over square root of 3. And we get square root of 3 over 3. Do we put a minus? No. We just established that in the third quadrant, the tangent function is positive.

So we leave it like that. So this is our final answer for that. Square root of 3 over 3. We have more examples, so let's go over those. This one is interesting.

So this is part c. We want to calculate tangent of 11 pi over 6, right? So, I mean, I forgot to mention, but please pause the video.

On all of these examples, pause the video, do your own calculations on a piece of scratch paper, and then come back to the video. Please do that because I get worried if you just copy my stuff down, you won't progress in this class fast enough. But if you do, if you do pause and you use your own brain power, even if you're... solutions may not be correct right away, you will get to full accuracy very quickly.

But you've got to resist the temptation to look at the solution before you do your own work. So I hope you've been doing that, but I forgot to mention it. Okay, so just like before, let's break this up, right?

So because this is a big angle, let's break it up to something that we can, to angles that we can understand. smaller angles like the ones in the table. So if we break this up, we can say that 11 pi over 6 is equal to 12 pi over 6 minus pi over 6, isn't it?

Right? Because 12 pi minus pi, that's 11 pi over 6. So we break it down. broke it up into these two. But why did we do that?

Well, now look at this. 12 pi over 6, we get 2 pi minus pi over 6. So before I draw the graph, ask yourself, in which quadrant is this angle? This 2 pi minus pi over 6. In which quadrant is it? Well, pause the video and think about it.

Okay, now let's draw your graph. So hold your thought. I'm sure you have an answer in mind, but if you're sure, perfect. But if not sure, let's see what the graph tells us. Okay, that's theta, some angle.

So according to this, which is the broken up pieces of this, it says go to positive 2 pi. This 2 pi, look at it, is positive. So it indicates movement in the counterclockwise direction, right?

Go 2 pi in the counterclockwise. If I do that, I'm going to be back here where I started, 360 degrees, right? But then it says subtract.

minus pi over 6 minus pi over 6 minus pi over 6 indicates movement in the clockwise direction so if I do that where will I end up in which quadrant in the fourth quadrant so this angle is in the fourth quadrant here okay So this is, so 11 pi over 6 is this angle. That's 11 pi over 6, right? But the reference angle for this 11 pi over 6, sorry, the reference number is this one. Theta is, of course, is pi over 6. That's the reference angle.

So that makes it really easy to calculate this again, right? We've done tangent pi over 6. So let me write this just like what I did in the previous examples. Tangent of, what we're saying is tangent of 11 pi over 6 is equal to tangent of the reference number pi over 6 in quadrant...

Four, okay? We're in the fourth quadrant. Very important in terms of the algebraic sign of what we need to calculate. So let's, well, what is the sine?

the tangent function in the fourth quadrant? Well the x-axis or the cosine is positive in the fourth quadrant, the y-axis or the sine is negative and And tangent is sine divided by cosine. Negative divided by positive. We get negative. So let's calculate this.

We get, again, tangent of pi over 6. I know I've done this in the previous examples, but I'll do it again. Sorry, this is folding over there. So let me do it again for this example. So we have sine of pi over 6 divided by...

cosine of pi over 6, right? Tangent is equal to sine over cosine. And then this is equal to 1 over 2 over a square root of 3 over 2. Once again, the 2s cancel out. We get 1 over square root of 3. We need to rationalize this. Square root of 3 over square root of 3. We end up with square root of 3 over 3 over 3. Are we done?

No. We just said that the tangent function in the fourth quadrant has a negative sign. And if I bring this back, this circle that I drew for you guys a few minutes ago, tangent in the fourth quadrant is negative.

Of course it makes sense because it's sine divided by cosine. And sine, which is the y-axis, is negative, divided by the cosine, which is the x-axis. x is negative divided by positive, we get negative. So we have to put a negative here. And I will then box that.

So that is the correct answer. And this, in which quadrant we are, is so... important in calculating these that I want to highlight it like that. Here we were in quadrant three, and here we were in quadrant in the fourth quadrant here. Okay, very, very important.

Okay, I think we have more examples, so let's go over those. So here again we have similar but this time we have other functions like sine, cosine, and again tangent. So we have this example where we want to find The exact value of the, again we want exact, right?

So we want to leave it as, say, a square root of 2 over 2, or 1 over 2, or the measures that you just saw us give as the final answer, right? So we want to give our answers in exact value form of the trigonometric functions. Okay, and we have several parts, so A, part A.

We want to calculate the... Exact value of sine of 2 pi over 3. Okay, so that's what we want to calculate. Well, please pause the video and answer this. Answer this on your own. You know plenty by now.

to answer this. So I'm assuming you've done that and now you're back. So this first thing we want to establish is in which quadrant is this angle?

So we can break it up just like before. We can break it up into two easy pieces. So let's say that that is equal, 2 pi over 3 is equal to 3 pi over 3 minus pi over 3, right? We can do that.

3 pi minus pi is 2 pi over 3. So this is a good, these are good pieces for that because they're both very easy to calculate. 3 pi over 3 is pi minus... pi over 3. So which quadrant do you think this is located then? So we go pi in the counterclockwise direction, because this is positive, and then we subtract, or go in the clockwise direction, pi over 3. So I know by now at least 90% of you, if not all of you, know in which quadrant this is. But if you have any doubts, then you can go to the next slide.

Let me draw a little graph for this. Right? This is theta. So this is saying go pi in the counterclockwise direction because it's positive. And then minus pi over 3. So they can go back in the clockwise direction pi over 3. So we are in the second quadrant.

Right? We are in here. Okay, so I'm just going to draw a line in that quadrant like that. So 2 pi over 3 is this angle.

This is 2 pi over 3. What is the reference number then? The reference number is this, pi over 3. We already know that, right? Because that's what we have. We went...

and then we went back pi over 3. So that's our reference number. So we can easily calculate this. We can say that sine of 2 pi over 3 is equal to sine of pi over 3 in quadrant 2. Okay, great. So now we're going to, from the table, we know sine of pi over 3 is the square root of 3 over 2. Here, let's bring it back. This is why I asked you guys to memorize this, sine of pi over 3. That's 60 degrees, pi over 3, sine of pi over 3 is square root of 3 over 2. But you've got to be careful because you're in the second quadrant.

The sine function represents the... the y-axis, right? What is the sine function in the, what is the algebraic sine of the sine function in the second quadrant?

It is, is it negative or positive? It's positive, right? So we have, therefore, we have positive square root of 3 over 2. Okay, so this is our final answer for that.

Part B is the same angle, so we don't have to draw the graph again. We have it right there. It's 2 pi over 3, but...

Sorry, let me move this up so you can see it. I don't know if I can keep the graph on the screen. I'll try. Okay, like that. So this time we have to calculate cosine of 2 pi over 3. It's the same angle.

angle, but this time instead of, and the reference number will be the same, it's still pi over 3, and we're going to be in the second quadrant, but instead of sine will be cosine. So let's write that down. So cosine of 2 pi over 3 is equal to cosine of that reference number pi over 3 in quadrant 2. Very important, right? Let me highlight that one and this one.

Very important. So cosine of pi over 3 is 1 over 2, but in the second quadrant, cosine is associated with the x-axis. In the second quadrant, that x-axis is negative. So if I write 1 over 2, it's incorrect, because in the second quadrant, the x-axis or the cosine is negative.

So we should write minus 1 half. Okay, we have that. Sorry, if I was off the screen, sorry.

This computer in front of me, the screen doesn't show. I'll have to turn around and look at the screen to make sure that I am on the screen. I think I've been on the screen most of the time.

I think so. So next, part c, and let me move up so everything is visible like that. We have now tangent of 2 pi over 3. So once again, the angle is... stays the same, the reference number stays the same, so we can write, and we're still in the second quadrant, so we can write that tangent of 2 pi over 3 is equal to tangent of pi over 3 in quadrant 2. Like that.

And then, so, we got to be careful because in the second quadrant, tangent, remember, is the ratio of sine divided by cosine, right? In the second quadrant, sine, which is... associated with the y-axis is positive, but cosine, which is associated with the x-axis, is negative. So positive divided by negative, because sine divided by cosine is tangent, so positive divided by negative is negative.

So that will be equal to sine of pi over 3 divided by cosine of pi over 3. Sine of pi over 3 is square root of 3 over 2. divided by cosine of pi over 3, which is 1 half. Remember, the final answer should be negative, okay? Negative 4. So we have square root of 3 over 1 is a square root of 3, but then we have, we're in the second quadrant, so we have to be, tangent has to be negative in that quadrant, okay? Like that. I hope this is making sense.

If not, please join the WebEx sessions and ask me directly. I'll be happy to go over these with you if you have any further questions. Let's do a little more examples. This is going to be a little bit different example than these. So we want to, it says given.

The terminal point, we know about terminal point, right? We learned about these in section 5.1. Given the terminal point square root of 5 over 4 comma square root of 11 divided by 4, given that, we want to find sine of, sine of t, cosine of t, and tangent of t.

Okay. Interesting, isn't it? Here's why. So here's the solution. Some of these questions are too easy, I think.

So they give us the terminal point, right? So let's write it down here. We have square root of 5 over 4, comma, square root of negative 11 over 4. So we already know that in trigonometry, the x component is associated with...

Well, let me first write this in algebra. So x, y. So this is our y component.

This is our x. component, right? And in trigonometry, x is represented along the cosine, or the cosine is represented along the x-axis. So we have cosine of t, comma, y-axis is represented by the, or the sine is represented by the y-axis. So we have sine of t here, right?

These are the terminal point. Terminal point is just a point, right? So it has some coordinate.

But we're in trigonometry, so we know that those x and y coordinates translate into x translates into cosine of t and y sine of t. So when they say, given this terminal point, find sine cosine, it's already been given. This is our cosine. This is our sine, okay?

So then that's too easy. So cosine of t then, cosine of t is equal to, let me bring it into the screen right there, cosine of t is equal to square root of 5 over 4. Notice we didn't have to do any calculations. Right? And sine of t, these two are immediately known, right?

We don't have to do any calculations for these. They're right there in front of us. So, and then sine of t is equal to minus square root of 11 over 4. like that. Then we of course have to calculate tangent of t. Alright, so tangent is sine divided by cosine.

So sine is this, minus square root of 11, divided by 4, divided by cosine of t, which is square root of 5, divided by 4. Okay? So the 4 and 4 cancel each other out. What do we get?

We get... minus square root of 11 over square root of 5. As before, we don't want to see square roots or the radical expressions in the denominator, right? So let's rationalize this by multiplying the numerator and the denominator by square root of 5. Right?

So then we have minus square root of 55 divided by 5. Okay? So that is the value of tangent of that terminal point. So we have that.

Okay, and we have that as shown. So these beginning sections in this course are very, very important because they're laying down the foundation. These basic trigonometric functions are easy functions, but remembering them is very important. in the more complicated calculations we'll do later. So please take the time to memorize that table and the circle and the algebraic signs for these trigonometric functions.

Of course, those relations... that I mentioned at the beginning on page one, all those relationships, please remember those formulas. And I thank you for watching this video until the end.

And please remember, even if I forget, you don't forget to pause the video repeatedly and write down your own solution before you look at what I have done in these examples. So for now, have a nice day. see you again next time.

Transcript for:Understanding Basic Trigonometric Functions

Transcript for:
Understanding Basic Trigonometric Functions