 Hi, I'm Zor. Welcome to the New Zor education. I would like to continue talking about trigonometric inequalities, but in this case, instead of solving the inequalities, which means, given an inequality, find arguments when this particular inequality holds, it's a slightly different twist in this particular topic. You have to prove that certain inequality is held if certain restriction on arguments is given. Well, there are certain unconditional inequalities. For instance, in trigonometry, you can always have that sin of x is less than or equal to 1. But something like this, just less without the equal sign, if x is not equal to pi over 2 plus 2 pi m. Now this is a conditional inequality, so if this condition for an argument is established, then you have this particular inequality held true. So, a few examples of this particular thing I'm going to present right now. And the first one is related to an interesting observation which you might or might not actually made before. If you consider the graph of the function sin, something like this. Well, obviously it goes here as well. It actually, it might actually come to deorize the resemblance of this particular direction of the function y is equal to sin x and a straight line which basically represents y is equal to x graph. And this is y is equal to sin of x. Well, so the question is on this microscopic level, whenever you are very, very close to the point 0, how actually this angle bisector which is y is equal to x and sin, how close they are. And here is a very interesting theorem. I'm not going to prove this particular theorem right now, that's for the later. But basically the point is that they are so close that the ratio between sin x and x, this is sin x and this is x, right? Sin x, x. So, this ratio actually goes to 1, the closer we get to the 0. Which means they are very, very much close to itself. However, this particular line y is equal to x is always above the sin. So sin goes underneath but very close. So close the tangential line to the graph of the sin actually is y is equal to x. It's like you have a circle and the horizontal line which is a tangential line to a circle, very close to the circle in this particular case. They are basically almost the same. So this is tangential to a circle and this is tangential to a sin in this particular point. So, from these qualitative observations I would like to go to a more analytical one. And basically as the first step in investigating the behavior of these two things, I would like to prove that sin is always less than x. If x is in this particular interval from 0 to pi over 2. So, I'm considering the sin only in this particular area from 0 to pi over 2. And in this area, sin is always below the line y is equal to x. That's what I'm going to prove to you. Okay, so that's what I want to do but I don't want to do it on this graph. I will use some kind of a geometry in this particular case. And, you know, sometimes it's considered to be like a bad case if you depend on the drawing very much when you're proving certain theorems. However, you can always make it a little bit more rigid and strict and robust, etc. That's not my purpose. My purpose is just to explain the reason why this particular inequality actually this helps. So, let me consider a unit circle. Does it look like a unit circle? Oh, this is a nice unit circle. And now I'm considering angles only from 0 to pi over 2. So, angles from this position to this position greater than 0 and less than pi over 2. So, this is an angle. And for any angle in this category, I would like to prove this. Okay, here's how I'm going to do it. Let me draw a chord AB. Let's say the angle is x. And I'm talking about regions, obviously. I mean, if you will change it to degrees, it will definitely be wrong. So, I mean, all right. So, we are talking about regions and that's the case in this particular case. Now, I consider the altitude of triangle AB AC. Now, this altitude is actually an ordinate of the point A, right? This is ordinate and this is obsessed. So, ordinate of a point on a unit circle is equal to sin of the corresponding angle. So, AC is sin of x, where x is an angle in regions. Now, OB is equal to 1 because it's the unit circle. So, the area of the triangle, the area of triangle AB is equal to the base, OB, which is 1, times the altitude, which is AC, which is sin of x, and one half of this, right? So, that's the area of triangle AB. Now, how about the area of a sector, or AB? Sector is piece of the, like a piece of pipe, basically. So, it's a slice of pizza, all right? Now, number one, obviously, it's bigger than the area of a triangle by this little piece, right? Corn is always inside the circle. So, the area of the sector is definitely greater than the area of triangle AB. So, this is the area of the circle. Well, we know that the area of the sector is proportional to the angle, right? So, the total angle in regions is equal to 2 pi, and the total area of the circle is equal to pi r square, right? Now, r is equal to 1, so it's pi. So, this is the total angle, this is the total area. Now, our angle is x, so what is, now, proportionality should be preserved, right? So, what is supposed to be here? Well, obviously, x over 2, right? So, this is angle, and this is angle. Our angle and the full angle. This is area, and this is area. Area of a sector and area of an entire circle, right? So, this is equal to x over 2. And, since area of the triangle is less than the area of a sector, I can conclude from this that 1 half sine of x is less than x over 2, which means that the sine of x is less than x, which is to be proved, right? Now, that's it. That's the first trigonometric inequality I wanted to present to you. And, again, don't forget that I'm talking right now only about angles x from 0 to pi over 2. I'm talking about acute angles only. Well, that's the number one. The second inequality which I would like to talk about is about tangent. And I would like to prove that tangent x is greater than x. Again, if x belongs to interval from 0 to pi over 2. Okay, how to prove that? Well, I will use basically the same approach. What I will do in this case, just like this out, I will continue this line and this is the perpendicular. So, this is A, B, C, D. Okay? Now, the triangle O, D, B is greater than the sector O, A, B, right? The area of a sector, so area of a sector O, A, B, we have just concluded this is x over 2. Now, what is the area of O, D, B? Area of triangle O, D, B is equal to base, which is O, B, which is 1, times altitude, right? Now, what is altitude? Think about tangent. Tangent is, in any right triangle, is a ratio between the opposite characters to the adjacent characters. So, tangent of x is equal to D, B divided by O, B. O, B is 1. So, D, B is a tangent and we have to divide it by 2, right? Base times tangent of x and divided by 2. Now, from the same geometric consideration, since the area of a triangle O, D, B is greater than the area of a sector, we have come up with tangent of x over 2 is greater than x over 2, which means tangent of x is greater than x. So, x is in between sine and tangent of x. Let me go back to the graphics. It's very interesting to see it graphically. What I'm talking about is the following. If this is y is equal to x, then this is y is equal to sine of x and this is y is equal to tangent of x. That's how they behave locally, very close to zero. They are coming into the same thing, but sine from the bottom and tangent from the top and they are all kind of straightening their graphs so at zero they are basically parallel to each other. That's what's happening. That's the behavior of these options. Now, from this particular point which we have just established, this one, we can very easily derive a couple of other things. For instance, since x is less than tangent of x, now they are all positive. We are talking about x always from zero to five over two. So, we know that if we invert numbers, then the relationship between them is also inverted. If x becomes a denominator and tangent becomes denominator, since this denominator is greater than this, then the fraction correspondingly would be smaller. Now, what is one over tangent? Well, it's cotangent of x. Remember, this is cosine of x divided by sine of x. That's the definition. Tangent is sine of x over cosine of x. Now, what else can be said around here? Well, from this, you can always multiply it by sine and you will have that sine of x divided by x is greater than the cosine. So, cosine is smaller, so I put it on the left. Now, at the same time, since sine of x less than x, let's divide it by x and we will have x is positive, so nothing actually is changing as far as the inequality is concerned. So, sine of x over x is less than 1, right? If I divide it by x. So, I have this. And here is what's interesting thing. As x is getting smaller and smaller, the angle, so in the graph, we are getting closer and closer to zero. So, sine of x divided by x. Now, this thing, if x goes to zero, cosine of zero is one, right? So, it goes to one basically. Now, this one also is not changing, that's one. So, sine of x divided by x on both sides, smaller and greater, have boundaries which are gradually going to one. Now, what does it mean? What's the behavior of sine of x over x? It's also supposed to go to one. So, and that's exactly what I meant, that around zero, sine of x is straightening and it's almost the same as y is equal to x. But I'm not proving this right now. This is just an explanation, so you have a feeling of how sine over x behaves around the x. And same thing with tangent as well. But it's just straightening from the top. Sine is straightening from the bottom and tangent is straightening from the top. But around zero, they are all very, very, very close to y is equal to x. Okay, that's it for sine and cosine. Now, let's move on. What else is interesting? Sine of x times cosine of x is less than or equal to 1 half. How can we prove this? Basically for any x, there are no restrictions here. It's unconditional. Well, I'm sure we remember the formula for sine of 2x, which is 2 sine x cosine x. Right? So, obviously, sine of 2x is less than or equal to 1 because sine of anything is less than or equal to 1. And since sine of 2x is this, you basically divide it by 2 and you get this inequality. So that's simple. Next, another inequality, which is also unconditional. Sine of sine and cosine for any angle is less than or equal to square root of 2. How to prove that? Actually, a very similar problem was solved before for a different purpose for solving basically trigonometric inequality. And what I did in that case, and I definitely suggest you to do, let's just multiply both sides by square root of 2 over 2, square root of 2 over 2, which is actually the same as sine of p over 4 and the same as cosine of p over 4. So, if I will multiply this inequality to this particular constant, it would be invariant transformation. It would be inequality which will have exactly the same solutions. But now, if I will do it this way, then sine x times sine p over 4 plus cosine x cosine p over 4 is basically cosine of x minus p over 4. Now, on the right, I have square root of 2 times square root of 2 and divided by 2, which is 1. So, my invariant transformation actually converted this inequality into this inequality. And this is always true because the cosine of anything is less than or equal to 1. So, which means that since it's an invariant transformation, this is true as well. So, all I did here, I just kind of decided to multiply it by this particular constant and it immediately transformed into obvious inequality. Okay, then the last little problem which I have here, it involves tangent and cotangent. Well, it cannot be unconditional. Well, obviously, condition is that the tangent and cotangent exist, right? Now, the tangent is defined for everything, but that's when the cosine is equal to 0, right? So, it's pi over 2 plus pi n. And the cotangent is cosine divided by sine, but it's not defined if sine is equal to 0, which is this, right? So, we have to exclude actually these and these. And, well, if you're interested, I can probably combine it, x not equal to pi k over 2. So, if this is true, then this is defined. If this is defined, then it's basically not even a trigonometric equation. It's an algebraic equation because since cotangent of x is equal to 1 over tangent of x, considering, again, this restriction, so everything makes sense. Everything is defined, right? This is cotangent is cosine over sine, tangent is sine over cosine. So, if you reverse, you will get this. And now, if I will just consider tangent to be any value y, whatever the value is, then my inequality actually means y plus 1 over y should be greater or equal than 2. And this is always true because if, well, again, I'm talking about, yeah, I forgot to mention it. We actually are considering whenever a thing is positive. So, because obviously if y is negative, it's not true. But if y is positive, so for positive tangent, that is obvious true because if you multiply it by 2, you will get y squared plus 1. Greater than 2y, right? Which is y squared minus 2y plus 1 greater than 0. And this is obvious because this is y minus 1 squared, right? So, for every y, not necessarily tangent, I mean, whatever it is, but for every positive y, this is true. So, for all positive tangents, it's also true. So, I did not mention it before, but I also have to mention it right now then. So, it's not only for those x which are such that tangent and cotangent are defined, obviously for those when they are positive. Now, you can always restrict it to, let's say, from 0 to pi over 2 and for all acute angles, and that would be true. But obviously, it's not only for these angles. It's also everywhere, wherever tangent is positive. And you have to add the period, actually, right? So, it's from, let me put the graph on so we will be more precise. So, the graph of the tangent is minus pi over 2, pi over 2. And the tangent becomes like this, right? And then it's periodic. So, from 0 to pi over 2 plus period. Now, the cotangent is this. So, again, we can add the period pi k here and pi k here. So, it would be on this from 0 to pi over 2 and this, this is pi. This is 3 pi over 2 and this when you have it here. So, the sum of these on this and on this and on this, et cetera, et cetera, will always be not only positive but greater or equal than 2. So, this is the positive for both of them and this is the positive for both of them. And I think I have covered all these cases. So, what was it about? It was about certain inequalities which are held for certain values of the arguments. Well, pretty wide area. So, it's not like something which we were trying to solve in equation for. It's like certain very, very narrow intervals where the whole inequality was held. In this case, it's a more general kind of inequality. And in these more general cases, you can always talk about relatively general intervals where the argument is changing. For instance, all acute angles like from 0 to pi over 2. That's pretty wide area of the different angles. All right. This is actually the third lecture dedicated to problems for inequalities. I think I might actually add something more. But so far, I think I'm satisfied with the number of problems presented to you. What's very important is to try to solve these problems yourself. Try to go through all these inequalities by yourself. Just consider only acute angles from 0 to pi over 2. Don't go into the details. But for those angles, all these inequalities actually are supposed to be held. Okay. That's it for today. Don't forget that unison.com contains the whole course of mathematics. Not actually completed yet, but sufficient enough. Almost like 300 lectures are there right now. And you are encouraged to sign in as a student as well as somebody else or maybe you yourself signing in as a supervisor or parent who can enroll you into this or that particular topic, which will enable you to take exam. Now, exams are very important. It will allow you to basically quantify your knowledge, your accomplishments. And until you get the perfect score on the exam, I would strongly recommend to go through exam again and again. You can take it as many times as you want, basically, until you will get basically the perfect score. I think it's very, very important. Okay. Thanks very much and good luck.