 Hi, I'm Zor. Welcome to a new Zor education. I would like to spend some time and analyze how a trigonometric function of sum of two angles or a difference between two angles can be expressed in terms of trigonometric functions of individual components. Now, it's actually quite involved and what I will do in this particular lecture is to explain this from the geometrical standpoint, which is not a general type of proof because in general angles can be anything. In geometry, angles are usually considered as part of geometrical figures and in this particular case part of the triangles, which means angles are basically restricted. So what I will do in today's lecture is to examine this problem only for acute angles so I can use right triangles as a geometrical proof of whatever the statements are. So what I will do is I will start with two formulas, cosine for difference between two angles, cosine for sum of two angles and again let's assume that my angles are acute, all of the angles alpha, beta and their sum, all are acute angles from 0 to pi over 2 greater than 0 and less than pi over 2 so I can construct right triangles using these angles. Let's start with the difference, let's say. Okay, so let's assume that I have the right triangle. I'll use the same letters as in my notes so it will be clear. So and this is an angle alpha. Okay, now angle beta would be here. This is beta. Okay, so this angle between these two rays is alpha minus beta. I assume that alpha is greater for definitiveness. Basically if it's other way around I'll just change the letters. But let's assume alpha is greater than beta. So this line is in between AB and AC. Again I heavily rely on geometry of this picture. Okay, so what I will do right now is I'll do two extra constructions. Number one, I am constructing this line which is perpendicular to AB and it intersects this particular ray which marks the angle beta at point D and from D I drop a perpendicular to AC which intersects at point E. Basically that's it. These two additional constructions allow me to express the angle alpha minus beta actually the cosine of this angle in terms of other angles. And here is how. I will use the definition of the cosine which is adjacent k-theta in the right triangle over the k-pottons. So for this angle which is alpha minus beta. Let me mark it as alpha minus beta. So for this angle considering triangle AGE, what can I say? What's the cosine of this angle? Well adjacent k-theta is AE and k-pottons is AD. So far so good. AE over AD. Adjacent k-theta over k-pottons and this is right triangle. By the way it's right triangle because I have dropped this perpendicular and this is the right triangle because I also construct it as a perpendicular. Now let's consider what AE actually is. AE can be expressed as AC plus CE. So AC plus CE forward AD. Now I will replace CE with FD because these are both perpendicular and these are perpendicular and this is a perpendicular again because that's how I draw it. Which means that this is equal to and I will divide separately. AC over AD plus and instead of CE I will use FD or DF rather. Over AD. That's the same thing, right? Now one more little thing. Now AC over AD. They are not related actually. So I will use something else which they are related to. I will divide by AD and multiply by AD. So instead of AC over AD, AC over AD, I divide it and multiply by AD. Now here DF and AD. DF and AD are also not very much related. However I can also do this trick. DF over BD times BD over AD. So DF AD and I multiply and divide it by BD. Now I have a much better situation because let's consider AC over AD. Well consider triangle ABC. This is an adjacent to alpha and this is a hypogenuse. So this is equal to cosine of alpha. Now what is AB over AD? AB over AD. Now this is an angle beta. So AB is adjacent to alpha and AD is hypogenuse. So this is cosine of beta. Now one little comment here. Consider this angle on the top. This is perpendicular to this and this is perpendicular to this which means this angle is also alpha. So angle FBD has sides BF and BD corresponding with perpendicular to sides of the angle alpha. So BD is perpendicular to AC and BD is perpendicular to AD. So both are acute angles and as two angles corresponding with perpendicular sides they are equal. So this is alpha and this helps me to decipher this thing. DF over BD. This is the right triangle. DF is an opposite to alpha and BD is a hypogenuse. So this is sine of alpha times. Now BD over AD. BD over AD. This is a triangle. AD is a hypogenuse. So we have a BD which is opposite to beta over hypogenuse. So again this is sinus but in this case for beta. That's it. This is the formula for a cosine of a difference between two angles alpha and beta where alpha is greater than beta. We have heavily relied by the way on the fact that these angles are all acute and that alpha is greater than beta. So this is this particular case and not pretending that I have proven this formula for all kinds of angles alpha and beta. I didn't. That would be in the future. But so far for acute angles this is the proof. Now let's switch to the sum. Again let's start with a triangle ABC. Right triangle. We will in this case, this is angle alpha. In this case we will add the angle beta. Not subtract. Okay. This is angle beta. And what we will do, very similar actually. We will put the perpendicular to this line and call it D. We will drop the perpendicular to this and call it E. And we also need this line. And again by the same logic this is also alpha. Because this is perpendicular to this and this... Oh sorry that's not alpha. Is it? That's alpha plus beta. Well I just drew it correctly. It should be perpendicular to this line. So this is the right end. Yes that would be better. So now this is alpha. Because this is perpendicular to this and this is perpendicular to this. Okay. So this is point D. It's just a little further. Okay. So this is my... Not very perfect but still a drawing. Which represents whatever is necessary. Now let's talk about cosine of alpha plus beta. Now what is the triangle where this is in the Q-tangle? A D T. Right. So the cosine of alpha plus beta is equal to A E. Which is adjacent. Over A D. Which is hypotenuse. And in this case A E can be represented as a difference between A C and E C. Right. So this is equal to A C minus E C over A D. Okay. And as before we braid it in half. So it's A C over A D minus E C over A D. And E C is obviously equal to B F. Because this is the rectangle. And now A C to A D minus B F over A D. And again as before we will multiply and divide by something very convenient. In this particular case it's A C over A D. Times A D over A D. A C A D, A C A D, multiply and divide by A D. In this case minus B F over A D. I will do B F over B F B D times B D over A D. And what is it equal to? A C over A D. A C over A D. That's adjacent coordinates over hypotenuse. So this is the cosine of half. A B A D. A B A D. Okay. So in the triangle A B D, A D is a hypotenuse. And A B is a characteristic to beta. So it's cosine beta. Minus B F over B D. B F over B D. This is an opposite character to alpha over hypotenuse. So that's sine of alpha. And B D A D. B D and A D. In this triangle A B D. Again B D is opposite character to beta. And A D is a hypotenuse. So it's a sine of beta. And this is the formula which I wanted to derive from this. So this is for cosine. The difference between cosine of alpha plus beta and cosine of alpha minus beta is this sine. For this plus it's minus. For a minus it would be plus. So let me just summarize it in one expression. And that would be the end of this particular proof. I will use this logic. Cosine of alpha plus minus beta equals cosine alpha cosine beta minus plus sine alpha sine beta. Plus minus corresponds to minus plus. For a sum it's a difference between these two. For a minus it's sum of these two. Now what's the interesting lesson which I would like you to actually come up with from this particular lecture? It's not really the formula itself which you should or should not remember but frankly I don't really care. What is important here is remember all these constructions which I made. Important to be able to do this type of thing. People somehow came up with these constructions to prove this formula. I mean somebody invented it. So somebody was the first guy who came up with idea. Okay let's draw this and draw this and that's how it looks like it works. This creativity is the most important part of the study mathematics because the more you are familiarizing yourself with different ways how different people approach different problems the more you develop yourself and your own abilities to do this type of thing. That's extremely important. So whenever you see a problem which you don't know how to solve that's great because it means that you really have to apply your creativity. Try to draw this line, try to draw that line, try to approach it from this angle or from that angle. This is actually the whole course which I'm trying to convey is all about. That's number one lesson. Number two lesson, you really have to realize how restricted this particular proofs were. They were restricted only to acute angles, alpha and beta and their sum also supposed to be acute because I explicitly used the right triangles for alpha plus beta and from alpha and beta. So I still have to prove it for any angles and I will do it in the next lecture completely different using a completely different approach. This however is something which is presented in most textbooks on trigonometry. Actually, I think if I'm not mistaken that only the plus actually is addressed and I mean this plus, alpha plus beta with a minus here. Only this formula is actually explained as a geometric proof of the formula. Quite frankly after that they're using this formula for all kinds of angles which means it's not really exactly the right way to approach it. You really have to prove the formula for all cases and there are really different cases for instance if one of these angles is equal to zero for instance then there is no right triangles where it can be included into or if angles are negative. I mean all these cases must be considered separately. Anyway, that's what it is and I will actually pay attention to this more general approach in the next lecture. In this lecture, I just wanted to present the geometric approach to formula for cosine of alpha plus beta and alpha minus beta. And as a conclusion, I will not go into the details but I'll just explain it very briefly how to find out this value. Well, if you will go to my notes you will have exactly the way how I did this and any other manipulations like alpha minus beta, it's all in the notes and you really recommend you to read it more carefully but I'll just explain the approach. You know the fundamental cosine of phi is equal to sine of phi over 2 minus phi. I have already proven that for all kinds of angles. So instead of phi I will use alpha plus beta which means that if I want to do this I will use this minus alpha plus beta which is equal to cosine of phi over 2 minus alpha minus beta. Now this is a difference between two angles and the cosine of the difference we know the formula is expressing this cosine of a difference in terms of sines and cosines of individuals. Now individual also will have sine or cosine of phi over 2 minus alpha which can be converted again into sine or cosine of alpha. So this is the approach and all the formulas for sine are derivable using this particular approach. Go to the notes in this lecture on Unisor.com in trigonometry sum of two angles and you'll just see how it's all done in the details. Okay, that's it for this particular lecture. Again, don't forget creativity when you are trying to prove the theorem by constructing conditional lines, circles, maybe whatever is necessary. This is extremely important and this is a great in geometry. Geometry is even better in this case than algebra because you really have to draw something from your imagination. And also another lesson is don't consider this to be a complete proof. This is not, it's only for acute angles which we can use to construct right triangles. That's it for today and good luck.