Trigonometric Graphs: Cycles and Amplitude

Those four have period 2 pi, and tangent and cotangent are periodic with period pi. So while on the first page of the handout, where the graphs are there, you have listed the domain and the range. On the second page at the top level, I broke them up into three, and we're going to address, in the first line, I've listed the period. Okay. Okay, how do I look at the sine and cosine graphs, or one other thing? The graph of a periodic function over an interval, and it could be any interval of period length, is called a cycle. And you can use any, any interval to refer to a cycle. I'm looking at this cycle of the graph because a cycle is the smallest portion of the graph that you need to know in order to get the graph over the entire domain, yeah? Because that's the one you can reproduce on consecutive intervals. So sine and cosine, we can look at it in two ways, and all six graphs, I like to think of them as the— and I'm only looking at it as one cycle, and all of them I'm going to start off at the origin for my cycle. Different people, for different reasons, may choose different intervals. Some people actually prefer to have the origin in the center of the period, so they go either from minus pi to pi or minus pi over 2 to pi over 2, and that's okay. I prefer to keep it at zero, and I call this the fundamental cycle or the basic cycle. Different people call it the basic cycle, something else. But you need one cycle, and you need to have that picture in your head. Is that okay? And I keep the picture in my head as a graph starting at the origin and going to the right. The sine goes up to its highest point, comes down to zero, goes down to its lowest point, and comes back to zero. That's the sine. And it goes over four equal quarters of the interval from zero to 2 pi. And because we know where this is, we have done this several times, I call these the quarter points. So this is at pi half, pi, 3 pi halves. So every cycle, I break it up into four, and the break points, I refer to them as the quarter points, as a slang maybe. I don't know. That's how I call them. Yeah. And then the basic one I know. The highest point has y-coordinate 1. The lowest point has y-coordinate negative 1. Now, if you're graphing this for me, you would label the coordinates of this point as pi/2, 1, negative pi/2, sorry, 3 pi/2, negative 1. Those are the coordinates. Now me personally, as I write 3 pi/2 and pi/2, my mind is thinking 90 degrees and 270 degrees. Yeah, because that's how I work. Is that all right? So I already have the picture of what that is in my head. Yeah. Even though these are points on the number line, in my mind they're angles, right angle going up, right angle going down. Is that okay?

Cosine, a similar thing, but because it's out of phase, and that wouldn't make more sense to you. Four quarters. But this time, unlike sine, it starts at the highest value, goes down to zero, goes down to the lowest value, comes up to zero, and then goes up to the highest value. This is how you keep in mind that sine and cosine are, we say, mutually exclusive on one and zero. Yeah, one is one, the other one is zero, guaranteed. They don't mix. And the quarter points are the same. Pi over two, 3 pi over 2, 5 pi over 2, and 2 pi. And again, you know this because we have seen right-angled triangle trigonometric standard position for quite a bit. Is that okay? Now the way in trig we state that the max and min, both relative max and relative min and absolute max and absolute min, are one and minus one, is we say that for these graphs, the amplitude, which we abbreviate as A and P often, is one. The sine function graphs are A big role in acoustics, which is, I guess, physics of sound. The goal is to identify noise and get the amplitude. The noise is a sine wave. Get the amplitude down to zero, as close to zero as possible. That's how you filter noise. Some of you, I think, wear noise-canceling headphones. Yeah, that's an application of the sine function. Yeah, you know it. So you look at several sine, several different frequencies. In fact, the number of cycles in sound, number of cycles of the graph, that's called the frequency. So noise has more cycles per unit time. You graph it on the time axis. You try to get the amplitude down. That's where the amplitude comes in. And because it's a finite number, we can identify it. Okay. So I think over there in that first column, I've written amplitude one. Yeah, is that all right? Sine can then go, sine can behave the same way. They are also out of phase. And since they're just, one is the reciprocal of sine and the other one is the reciprocal of cosine, they're really out of phase. So we first look at— Cosecant x. And cosecant x is 1 over sine x, so sine is 1 at the vertical angles, which means that the horizontal angle is going to be zero, and that's where it's undefined. So that would be at zero, pi, which is the second quarter point, and 2 pi. At those points, you have vertical asymptotes. So you have vertical asymptotes for the cosecant at multiples of pi. In the first half of the graph, you have the up cup, which has the lowest point at 1. And in the second half, you have the down cup, which has the highest point at negative 1. And when I look at the whole graph, I can't—although those points are relative extrema.

I can't call them absolute extrema because I have points above them and below them when I look at the entire graph. So period is 2 pi, once again, because we have done this in standard position a lot. I know where these quarter points happen: pi half, pi, 3 pi halves, and 2 pi. Okay? So I refer to these two points as points. I can't say amplitude here because that doesn't make sense here. I refer to these points as points with y-coordinate is 1, yeah, or in magnitude, so 1 is 1. So I refer to them as 1 points. And I say, in, I guess I call it my slang, I say the 1 points are plus or minus 1 over here. And the 1 points happen for cosecant at the first quarter and at the third quarter, yeah. On the handout, I've listed them as coordinates so that, because you're told for official documents you can't use slang, but it's proper math. That make sense? Cosecant essentially pulls this graph back by pi over 2, so it'll really start at the bottom point of the up top, and then you've got to get a whole period then. I want to squeeze it in here, this bottom here, so I get all six in. So cosecant has the vertical asymptotes at all multiples of pi halves, because cosecant is the flat trig function, which means it's zero at 90 degrees, 270 degrees, yeah, as the straight angles, upright angles. Zero pi halves, pi, three pi halves, and two pi. And there are vertical asymptotes at these odd multiples. So you start off with the left half of the up cup, and then in the middle you have the down cup completely, and then the left half of the up cup over here. Are you seeing that on the picture that you're looking at your handout? Yeah. And notice that these quarter points that are off the axes, which matter, are always at y coordinate one. Sine and cosine, I can refer to them. I could call them one point, but we have the word amplitude. Over here, we don't have the word amplitude, so I just call them one points to refer to them as those are the points whose y-coordinate is one, because that one gets affected when you throw in some constants. Okay? Tangent and cotangent are of the same rate, right? The only difference now is that the period is pi. So let's do tangent first. I still start out with zero, and I'm still looking at this graph as being made up of four of them. Tangent is sine over cosine, and cosine is zero at the vertical angles. I know that the period is pi, so the endpoint is pi, which means the halfway point is pi halves, which means over here tangent will be undefined, and so there's a vertical asymptote at that point. What is the first quarter point going to be? What's in the middle of zero and ninety? Forty-five, right? Forty-five.

So this is pi over 4, and this would be 2 pi over 4, so that one better be 3 pi over 4. That's how I think about it. And then 4 pi over 4 is just pi, yeah. As soon as I realize that my quarter points are, the denominator 4, I think of all four of them with denominator 4, and that's how I count them. And mentally I'm good with the fractions, so I know how to reduce. 2 pi over 4, I can write as pi over 2. 4 pi over 4, I can write as pi. All right? And we all know that pi over 4 tangent is 1, 3 pi over 4 tangent is negative 1. And we know that from 0 to pi, we can't use quadrants here. 0 to pi/2 is the first quadrant, and then pi/2 to pi is the second quadrant where tangent is negative. Is that okay? Are you okay, Mandy? I have a question. Yeah, a question. No, I got it now. I was just wondering where 40... Yeah. Or you can say it's half of 90, right? I think that's what I was trying to get. It's one half of pi/2, or one half of 90, which is 45, or however you get it. Getting the hang of that as I do it? Cotangent is the same thing. There's going to be a shift. But there'll also be a flip over because they are the reciprocals of each other. But this can cotangent is cosine over sine, and sine is zero on the flat angle, so at zero, pi, 2 pi, that's where you're going to have issues. And so I've still drawn the quarter points, which means my asymptote come at the end. The origin—I'm sorry, the y-axis is also an asymptote for cotangent, just like it's an asymptote for cosecant. Both of them have sine in the denominator. And the midway would be pi half, and this would be pi over 4, and this would be 3 pi over 4. And again, you can think of it as 0 pi over 4, 1 pi over 4, 2 pi over 4, which is pi over 2, 3 pi over 4, and then 4 pi over 4, which is pi. And this time here, here's what's going to happen. It's going to look like that. The graph is pulled and then flipped over on the line 1, a horizontal line 1. So this is at 1, and this is at negative 1. And those are my one points. So all six graphs I have one points. They correspond to one of those four quarter points. It just varies depending on the shape of the graph, which depends on the function. And I call them one points because the y-coordinate is 1 in all six of them. These corner points are the most important. That's really what you always need to get a hold of. A calculus, when you're going looking for it, you're not going to be graphing. You're really trying to get, rather quickly, these key points that, how shall I put it, determine the actual shape of the graph, because those are the ones you're going to try to manipulate. Yeah? I told you a little bit about sound acoustics. They need that amplitude to kill the sound and not the conversation. Yeah? When they transmit conversation across the airwaves, a lot of sound corrupts it. Then when they try to de-sound, convert those waves into sound, they only want to convert the actual conversation and not the noise. Sometimes if you use some old-fashioned phones, you get a lot of static, and that's the noise. That's where they have to kill those frequencies of sound. Yeah? And so, in physics and even in electronics and a lot of things, those points play a big role, and you have to get a hold of those. You have to be able to recognize those, yes, and then you can manipulate your AI and your computer to get it. But if you don't have a basic understanding of that.

It becomes difficult, yeah. So by actually spending time doing these graphs, you will learn to rather quickly get these points without having to struggle for them and know what you're doing, yeah. Is that all right? Okay. So now, so all that is put in the ... here. So I now am going to go to the second part, and I'm going to write down what I want to write down, and then play a little bit on Desmos, and then we get to the exercise and try to do as many problems as we can in class. And then the ones I haven't done, hopefully you'll try to do on your own without Desmos, and then check your answer to see if you're doing it correctly. The graph of, we are only going to look at y is equal to trig function, sorry, a times trig function bx plus c. So we are only going to look at graphs of this kind of trig functions. There are three constants, a, b, and c. a is in front, b is the coefficient of x, and c is the constant that's part of the angle. Is that all right? I should have mentioned that, written it out there. I was in ... actually this handout is the old one. I wasn't paying attention when I created it that time. When I used it the first time, I realized, oh, I should have put this in. Something distracted me, and I never did. And I used it several times. I think this is more than 10 years old. To this day, I haven't made that correction. So this time I said, well, I'm going to make the correction. You remember I printed out the handout and put it in, and because I made the correction and uploaded the corrected version, just to not have problems, we require b to be positive. The other two can be any, as long as they are numbers. If b is not positive, then we are dealing with—if b is zero, for example, then we're just dealing with a number, yeah, or something that's undefined. If b is negative, we run into trouble, because then how do we define period in c without throwing abstract value? Period is always positive number, yeah? So that's why b should be positive. I didn't include that, so we want to add that there. This is similar to the graph of y is equal to trig function of x. So the shape is the same. It's just modified with the following changes. But first point in A affects the points whose coordinates are one. Yeah, so the first point A, first of all, if A is negative, you have a reflection with respect to the x-axis. And essentially what that means is the graph just flips over the x-axis. So you just draw it upside down, or downside up, however you like to look at it. But here, if I use my slang, the one points, I just say for slot will be a plus or minus A. And it doesn't matter whether A is positive or negative, plus or minus A, one will be positive, one will be negative. Yeah, so those points with y-coordinates one and minus one, they change over to A and minus A. Yeah, that's what it does.

The period, there's a change. And it changes to the original period over b. So this is the original period over b. So in the case of cosine, sine, secant, and cosecant, it'll be 2 pi divided by whatever that number is. Is that okay? There is a phase shift, and I'll tell you what the phase shift on the graph. And the phase shift is always minus c over b. And again, there you need b to be a non-zero number. That's why we make sure that is the case and positive, and we can use it in the period. Is that all right? So if c was a negative number to start out with, then the phase shift will be a positive phase shift. If c is a negative number, yeah. What this means is, and we take this mean, if you want to write it in your own lingo, is x equals zero, that is the starting point of these four graphs, and that's why I look at these six graphs as being the fundamental cycle, because that's my starting point. It moves. To x is equal to minus c over b. Wherever minus c over b is, that's where it's going to move. Is that all right? So let me go through a series of very basic examples, and before we get to the exercise, let me look at them in Desmos for contrast. And we'll work with sine, and maybe in Desmos I'll do maybe one or two with tangent, and then talk through the exercise. I won't do reflection over here, but we do it in Desmos. If I had y is equal to 2 sine x, with just that constant, yeah, that constant only affects the amplitude. And since we are drawing sketches, I know the shape of sine. I just reproduced; my starting point is not affected. Yeah, because there's no coefficient in front of x, my period is not affected. So on the x-axis, I don't change anything. My quarter points are going to be exactly the same. So let's just, so we have different numbers. Let's make that three rather than two. We don't have to deal with two complicated fractions. That'll be working, yeah. The only thing that is changed by this number is the amplitude, which means the points with coordinates one and minus one will go to minus three and three. Yeah, so the amplitude changes to three and minus three. That's the only change. Is that okay? And so in physics, you learn, how do I get that down to zero, or as close to zero as possible on the noise? How do I keep the ones that are not noise as high as possible, you know? I don't know how they do it now, but in the old days when they created all these transistor radios and all that, the noise that came out was very, very soft. So they put amplifiers that boosted. Yeah. In my time, we used to buy...

Specifically, the amplifier that slides that regularly distorted noise and frequency. But that's really what you're doing over there. You're taking the sine wave and killing the amplitude, yeah. Now we keep the same y is equal to 3 sine x, but now I'm going to put that 2 in there, yeah. I know that my starting point is not shifted at all. It's going to start at the origin because I don't have that constant term. There's no phase shift, we say. There's no shift in phase, yeah. So once again, as long as I draw the right shape, if it was the cosine, I would take this shape. If it was the secant, I would take that shape. If it was the cotangent, I would take that shape. So the shape is going to be the same, yeah. And all six shapes have four pieces. Is that all right? We already dealt with the amplitude, so I know that this is going to be 3 and minus 3. I know my starting point is going to be 0. I need to figure this out. In this case, my period, I use p for period, is going to change. It's going to change from 2 pi to 2 pi divided by 2, which is the constant in front of x. And 2 pi divided by 2 is pi, yeah. So I know my period ends from 0 because I started at 0. I end at five. Am I making sense? Now watch the board as I do this. Well, I need to get those four quarter points, and I know they have to have the same length, so I take my period and divide it by four. If I take pi and divide it by four, and this is in my head, pi divided by four is one pi over four. And then I keep adding. I start at zero, which is zero plus one is pi over four. One plus one is two pi over four, but because we're good in math, two pi over four is pi over two. Yeah. Two plus one is three pi over four, which I can't simplify, and three plus one is four pi over four, and I have it. Is that all right? Actually, in my head, I cheat. I don't say pi over four. I say 45, 90, 135, 180. Yeah? In my mind, I write that, but I think in degrees. If you struggle with fractions, take out your calculator and punch it in. You can ignore the pi. You can do zero plus one fourth. I showed you how to do fraction arithmetic. One fourth plus one fourth. It'll reduce it for you. Is that all right? But that's how you get those points. Okay, I think now we can delete out these graphs. Now we are going to do a phase shift, and I'll do an easy one. I think not too easy. Y is equal to 3 sine 2x minus pi. So this time, I know my amplitude is 3. I know my period is pi over—sorry, equal to pi, and I know I'll be doing pi over 4, pi over 4. We just did that, right? But now I have a phase shift, which I have remade as PS, and that is going to be minus a negative pi. I got to take the negative sign divided by 2, minus the constant divided by the coefficient of x. Is that all right? So this time, and now if you would like to watch the board, I'll give you a few minutes to graph it before we—my—oh, this I should reduce, that simplifies to pi halves. My starting point is no longer at the origin. It has moved over to pi half on the x-axis.

Is that all right? So I move it over, and then I know I'm going to have four quarters, and because there's no reflection, it's going to go up, it's going to go down to zero, and that's going to be my graph. Yeah. If it was tangent, I would take the tangent graph and draw it there with the asymptotes at exactly the same point, right in the middle. Yeah. If it was cosecant, I would take the secant graph, put it there with my asymptotes at the first and third quarter points. Yeah. All I need to do is I just figured out that the starting point has moved there. I've got to figure out what these are. How much do I have to add this time? One fourth pi, right? So I think in my head, and again, yeah, trouble with this, use your calculator. I think in my head as this being two pi over four. Yeah, same denominator. Each of them has the same denominator, right? Two pi over four plus one is three pi over four, right? Three pi over four plus another one is four pi over four, which is pi. Is that okay? Four pi over four plus one is five pi over four. Five pi over four plus one is six pi over four, which is two seventy— I'm sorry, three pi over two. Yeah, whole radians die hard. And I don't fire it through in my mind as 90 degrees, and I keep adding because I find it easier to add degrees, yeah? But I train my mind to think that way, but my pen to write radians, because I cannot write degrees on there because those are real numbers. Is that all right? And then if you wanted to identify those points, clicking the points off the graph. But there's your graph. Is that all right? That is pretty much what you do with every single equation out there. In the first one, I just asked you to do one cycle. In the second one, I will do a few of them. You basically reproduce the cycle and just select the portion that you're asked for, yeah? Making sense? And we'll do a few of them. I'll give you a minute to copy it down, and then we look at these three in Desmos. Same with this in Desmos, just to get a feel for this, yeah? And they haven't yet got the scaling right so that you get the right marks on the axes, and I haven't figured out how to force it to do it a particular way. I used to use the old—this would be one of my old ones, to learn how I can force Desmos to give me specific markings on the x-axis. In that old program, I could do it, which is what I used to create that handout. But it no longer works on Windows 11. Can I put up the screen? Have we all got on that graph? And then after this, we take a break. As I change, so those have become minus two and two. And because of the reflection, I'm going to go first quarter, then second quarter, and then come down here. Is that okay? I know the first quarter ends on the x-axis. The second quarter ends, in this case, because of the reflection, at the highest point. Third one ends, and then I finally end up there. Because I'm starting at the origin, my period is four. I know that this point would end at four. And four is a nice number to divide by four. Four divided by four is one. So this would be easy count. One, two, three. Is that okay? And since I made rules in class, the these four points are important, so I need to label this. This is two, two, and this is off the x-axis. This is four minus two. If it's on the x- or y-axis, you don't need to label them. Four minus two. Next step? Is that—did I have all that worked out? Okay, let's do one. Now let's do five. And then we come back to four. Yeah, because I want to do four, because you can see there's a problem in four.