 Hi, I'm Zor. Welcome to Unizor Education. This lecture is part of the Advanced Mathematics course for teenagers. It's presented on unizor.com website and that's where I suggest you to listen to this lecture because the site contains very important notes, which basically can be started like a textbook. Today's lecture will be about it's kind of a synergy between trigonometry, geometry and the theory of limits. So basically it's about behavior of the function sine of x in the vicinity of the point where the angle is closer to zero. So basically it's the limit of this ratio sine of x over x, where x is measured in radians. That's very important, x is radians. So I'm going to prove basically that this limit is equal to 1, which means that if you take a look at the graph, this is the sine from 0 to 2 pi. So here close to the x is equal to 0, the behavior of the function sine of x is very much like the behavior of the straight line, which is y is equal to x. This is y is equal to sine of x. So they are very, very close and the closer to go to zero, the more the behavior of the sine of x looks like the behavior of the x itself. So that's what I'm going to prove. And again, I'm going to use the geometry and algebra and trigonometry, all of them together to derive this particular formula, this limit. Okay, so how am I going to do it? I'm going to derive it for positive x, which are going to zero. With negative, it goes exactly the same thing because the sine is the odd function. So the ratio actually is exactly the same for a negative x as for x because this is negative and this is negative if you change the sine of x, right? So let's assume that x is only positive and it goes to zero, which means I can actually consider only angles within the first quadrant. So remember for definition of trigonometric functions, I was using the unit circle, right? So this is one. This is also one and the angle, let's call the angle x and this is x regions again. Let's put this point A, so the angle x has a sine, which is basically an ordinate of point A on the unit circle. Cosine being an abscissa, right? So the length of this is actually a sine. Now, let's put this letter P. That's where the unit circle intersects the positive direction of the x-axis. Now this would be Q and let me connect A and P and let's consider a triangle O, A, P. O, A, P. Well, I would like to compare the area of this triangle and the area of a sector P, O, A, which includes this little piece between the chord A, P and the arc of a unit circle. Now, obviously the area of a sector is greater than the area of the triangle. Because of this little piece which I put some shades on it, right? So let's just calculate exactly what's one and what's another and we will get an inequality between these two areas. So the angle is x. The triangle I'm going to consider is A, O, P. Its area is base, which is O, P, which is equal to 1, by the way, because it's a unit circle, times height and height is as we know the sine of x, right? So the area is equal to one half of base times the height, the altitude. That's my area, right? Now, let's talk about the area of sector. Let's call P, O, A. So the angle is P, O, A because it's actually measured counterclockwise. That would be a better designation of an angle in the sector. Now, let's just think about it. This sector is part of the entire unit circle, right? Now, obviously it's proportionally smaller than the area of an entire unit circle and the factor between them is exactly the factor between this angle and the complete angle, right? The full angle, which is 2 pi in regions of 360 degrees. But we're talking about regions. So measured in regions, the whole full circle is 2 pi and this sector has x as a measurement of its angle. So the areas are exactly proportional to this ratio. This is the area, area of circle P, O, A. And this is area of full circle, right? So the proportionality between the angles is exactly as proportionality between the areas. Now, this we know and this we know and this we know. So how about this area? area of circle, sector P, O, A is equal to x times P r square, but r is equal to 1 because it's a unit circle. So it's just pi and divided by 2 pi, right? Equals to x over 2. So what can we say now? That the area of a sector is equal to x over 2 and the area of a triangle is sine of x over 2. Therefore, we derive the inequality between these two. The area of one-half sine of x is smaller than the area of a sector. We got that, right? Okay, let's go further and consider a slightly different triangle. Let's continue. Let's extend this up and this vertical up. So this is the perpendicular. So this is point A. So P r is perpendicular to OP. Now we know that the perpendicular at this point is completely outside of the circle. So the area of triangle P, O, R is greater than the area of this sector. Now the area of a sector we know it's x over 2 and the area of this triangle I'm going to calculate right now. Now let's think about. If you divide this by this, you will get a tangent of this angle, right? Now, this is a catechus of the right triangle. This is the right triangle because P r perpendicular to OP. So since Rp divided by OP is equal to tangent of x, Rp is equal to OP, which is 1, times tangent, which is tangent. Now this is a catechus and another catechus is OP. So the area of a triangle, since this is a right triangle, I multiply this catechus by this catechus and divide by 2. This is 1. This is tangent. So if you multiply it will be tangent divided by 2. And this is greater than the area of a circle. So I can do this. One half tangent of x. So basically, I'm almost finished and let me explain you why. Because now from now on it's just pure trigonometry. Well, first of all, I get rid of 2 in the denominator. So it would be sine of x less than x less than tangent of x. Now secondly, I will replace tangent with its definition as sine of x over cosine of x. Next, since I'm talking about angle, which is a very small positive angle in the first quadrant, which is sine is equal to, sine is positive. I just divide everything by sine. So I have 1 less than x over sine of x. Less than 1 over cosine of x. Now, if these two things are in this particular relation, if this is less than this, if I will invert, I will have an opposite inequality, right? It's just easier for me to deal with opposite because this one. And here I will invert as well. So I invert every number and I change the inequality to an opposite, right? Because if 2 third is less than 3 quarters, then 3 second is greater. Sorry, that's around. 2 third is less than 3 quarters. Then 3 third, 3 seconds, excuse me, is then 4 third. Right? It's obvious because this is 1 and 1 half. This is 1 and 1 third. This is greater than this. So this is a very easy kind of a property between the inequalities. And that's actually enough for me because now let's just think about it. I'm going to do to prove this particular thing when x is going to 0. Now, let's think about what happens with all these 3 when x goes to 0. When cosine obviously goes to cosine of 0, which is equal to 1, right? So this goes to 1 as x goes to 0. Remember the graph of the cosine is this. So at 0 it's equal to 1. Now, this is a constant. So it's always equal to 1. So what happens right now? This particular variable, we don't know how it behaves, but we know that its lower boundary goes to a certain limit and its upper boundary is exactly the same limit. So basically there is no other way for this but to go to exactly the same limit. We have already considered this in one of the problems when we were dealing with limits. This is just one of the theorems, which I have proven in in one of those lectures dedicated to limits. If you have two different sequences, actually three different sequences in this particular case. If you have a and you have xn and you have bn and an is always less than xn and xn and this is less than bn. Now this has certain limit and this has exactly the same limit. Let's call L. Then this sequence has no other choice but to follow, to exactly the same limit. So I have proven it there and then I and I'm using it here. So that means that sine of x divided by x has a limit of 1 as x goes to 0. So basically that's it. That's the proof. What I would like to bring your attention into is, you see this particular theorem is not only a trigonometric theorem. It's not only algebraic. It's not only theory of limits. It's not geometrical theorem. It's basically a combination of everything, well including the graphs of the functions as I was explaining basically that the sine behaves in the beginning close to origin of the origin of the coordinates very much like a straight line at four to five degrees. So all of these things are brought together in this particular statement, in this particular theorem. And sometimes you know, personally, I'm always fascinated when things considered from many different viewpoints are synergized, so to speak. You see that they're related in some way. Well, everything is related in this world. But this is just a very important signification of this principle. It's philosophical principle and I will have a lot actually of these maybe presented to you in different lectures. But sometimes it's really unexpected. I mean you, I did not expect the first time when I see, when I saw something like this that that that's actually a true statement. It seemed to me unrelated. One is a sign which is completely, you know, trigonometric function and another is basically some real number. Apparently this real number is an angle in regions, then there is such a beautiful synergy between sign of X and X itself. So I just wanted to, you know, to pay attention to this little I don't know, aesthetics of mathematics. It's one of those pieces where mathematics is presenting some harmony, some beauty in its own construction. And according to Albert Einstein, by the way, the beauty of the theory is one of the criteria of its truthfulness. So this actually proves that to measure angles, for instance, in the regions is it's really the best way to do it because then the function sign of X is presenting itself in a very harmonious way because this particular limit is equal to one. Now consider if the angle would be measured in degrees. Well, then it would be, you know, something like what would it be? I mean, if X is degrees, then you have to convert degree into regions. You have to multiply by 57, whatever it is, every region is equal to. It's kind of messy, but this looks much more beautiful. So if you had any doubts that regions are a perfect measure for the angles, this is the proof that it is. So thanks very much for listening to this lecture. I do encourage you to use this website Unisore as a source of educational process, not just some set of lectures which you can go to for finding certain things, because the site has this educational process part in it. If you sign in as a student and somebody else is like your parent, for instance, is your supervisor, then your parent can enroll you in certain classes, then you can take exams because the exams are built into the system and that would allow you basically to measure your mathematical strengths. That's it. Thanks very much. Good luck.