 Hi, I'm Zor. Welcome to a new Zor education. I would like to present a couple of problems related to conditional trigonometric identities. As a matter of fact, three problems and a couple of extensions. All right. So as you remember, the conditional trigonometric identity is basically a theorem where there is something, some condition, which is given to you. Usually it's about angles or some trigonometric functions of these angles. Then another trigonometric equality should be proven. All right. So number one, there are three acute angles, alpha, beta, and gamma related to this equality. So the angle alpha is equal to the sum of beta and gamma and there are all acute angles. Now, what we have to prove is the following, that tangent alpha plus cotangent beta plus cotangent gamma equals to their product. So sum of these three numbers equals to their product. Well, first of all, I said that these three angles are acute, which means we don't really have to worry about whether tangent actually is or cotangent to the same matter, defined or not defined. So we can manipulate with these functions tangent and cotangent quite freely. Obviously, I have to express tangent alpha in terms of beta and gamma. That's basically the whole thing is about. Now, I do not remember the expression of the tangent of sum of two angles. So tangent alpha, which is tangent of beta plus gamma, I do not remember the formula and I always brag about it. I don't remember the formulas, but I can derive them. Well, in this particular case, I will derive them from something which I do remember, actually sine and cosine I do remember. So let's just very briefly explain how I derived this type of thing. Now, sine of sum is sine cosine plus cosine sine. You see, I do remember something. Now, for the cosine, it's cosine cosine and then since this is a plus, the cosine has a minus. So cosine beta cosine gamma minus sine sine. Now, obviously I would like to express it in terms of tangent should be expressed in terms of tangent. But in this particular case, you see there are all others are cotangent. So I would like actually to express the tangent of sum of two angles in terms of cotangent of each individual one. Now, I do remember that if I would like to express in terms of tangent, I have to divide numerator and denominator by cosine cosine, in which case I will have here tangent of beta because this will be reduced. And in this case would be tangent alpha because this would be reduced. And here I will have one minus tangent times tangent. Now, since I would like to express it in terms of cotangent, I'd rather divide it by sine sine. Now, what happens is if I divide sine cosine by sine sine, my sine beta will be reduced and that will have only cosine divided by sine of gamma which is cotangent gamma. Plus, if I divide this by sine sine, now my sine gamma will be reduced and I will have cosine beta divided by sine beta which is cotangent beta. Here, cosine times cosine divided by sine by sine, this is the product of cotangents, minus one. Right, so that's what happens if I divide numerator and denominator by sine beta sine gamma. So I will use this expression instead of tangent alpha in this particular equality which I would like to prove. And again, considering all angles alpha, beta, and gamma are acute angles, I don't have to worry about when this particular cotangent exists whenever the denominator equals to zero, et cetera. This is all not really the case. All right, so, okay. Now let's just substitute this and see what happens. If we will get an integer, if I will substitute this as a tangent, then my work is done. All right, so if I put it here, I will have cotangent gamma plus cotangent beta divided by cotangent beta cotangent gamma minus one plus cotangent beta plus cotangent gamma. That's in the left side, right? Obviously I have to use the common denominator in this particular case, which is this. So if I will multiply this times this, I will have cotangent gamma plus cotangent beta plus. Multiply this times this, I will have cotangent square beta cotangent gamma. Now, by this I will have cotangent beta cotangent square gamma, and then minus one, which is minus cotangent beta minus cotangent gamma. That's what I have in the numerator. Now, denominator is exactly the same. cotangent beta cotangent gamma minus one. Now, as you see, cotangent gamma will be reduced and cotangent beta will be reduced. So I will have only these two divided by these. So that's my left side. So I will have, actually I can factor out cotangent beta cotangent gamma here. cotangent beta cotangent gamma times cotangent beta plus cotangent gamma divided by cotangent beta cotangent beta cotangent gamma minus one. That's my left part. Now let's talk about the right part. So instead of tangent alpha, I will put cotangent beta plus cotangent gamma divided by cotangent beta cotangent gamma minus one times cotangent beta and cotangent gamma. Well, lower than before, this is exactly the same as this one. So my identity has been proven. That's it, and the proof. Next, next is the following problem. I have a circle and inscribed in that circle quadrilateral with angles alpha, beta, gamma, and delta. I have to prove that cosine square alpha plus cosine square beta plus cosine square gamma plus cosine square delta minus two equals two. Two times product of their cosines, cosine alpha, cosine alpha, cosine beta, cosine gamma, cosine delta minus product of their sines. That's what needs to be proved. Okay, now this property of this particular quadrilateral. So it's a convex quadrilateral inscribed into a circle. Now, what's the, one of the fundamental properties of this type of angles of this particular quadrilateral? Well, obviously, some of the opposite angles is equal to 180 degrees pi, y, for very simple reason. This angle, let's call it A, D, C, D. Angle, let's say A, A, D, C is inscribed and this is part of the circle, the arc which is supported this angle. Now, the opposite angle is supported by the rest of the circle, right? So some, now every inscribed angle is measured as half of the central angle which is supported by the same arc. So basically you have obviously the sum of these two angles is equal to sum of the full angle divided by two. So alpha plus gamma equals pi, beta plus delta equals pi. That's the most important quality which I can, important characteristic of these angles which I can derive from the fact that this is an inscribed convex quadrilateral. Now, this actually allows us to express trigonometric functions of gamma and delta in terms of alpha and beta, right? All right, let's think about it. Cosine gamma is equal to cosine pi minus alpha equals two. Okay, what is a cosine? Let's just remember it's abscissa. This is the x coordinate. So if this is alpha, then this is pi minus alpha and abscissa is x coordinate and obviously they have an opposite sign and equal in magnitude. So it's minus cosine alpha. How about sine of gamma? Sine is origin, originates are the same. So sine of pi minus alpha is equal to sine of alpha. So I'm going to use obviously these things to convert cosine gamma and cosine delta in terms of alpha and beta. So what I will have is I will have cosine alpha. And now this is minus cosine alpha, but this is square. So it's exactly the same. Same thing is beta. So I will have two cosines of alpha and two cosines of beta. Square alpha plus two cosines square beta minus two. That's on the left side. Now, what do I have on the right side? This is left. What do I have on the right side? Okay, this cosine is cosine gamma. It's minus cosine of alpha. And this is the minus cosine of beta. So I have two negatives which gives me the positive. So basically it's cosine alpha square and cosine beta square. So it's two times cosine square alpha times cosine square beta. Now this sine is the same. So basically I can replace sine of gamma with sine of alpha. Same as sine square alpha. And same thing with beta and delta. Sine square beta. That's what I have on the left side. All right. Well, obviously in this case, I have to replace sine was one minus cosine square. And one minus cosine square. Since sine square plus cosine square is equal to one, right? So what do I have in this case? I have two times cosine square alpha cosine square beta minus. Now I have these to multiply. One times one is one. One times minus cosine square is minus cosine square what I have minus here. So it's plus cosine square beta. Minus cosine alpha square alpha by one. It's minus and this is minus. So again, it's plus cosine square alpha. And cosine square alpha and cosine square beta plus minus is the minus side. Okay. Doesn't help. Well, this is reduced, right? And what do I have? I have two times cosine square, two times cosine square and minus two. That's exactly the same as this one. So if I multiply two by one, two, three, I will have exactly this one. So that's the end of the proof. You have exactly the same identity. Now, as an exercise for those who are not afraid of long calculations, I can tell you, but I'm not going to prove it here as part of this lecture. I can tell you that exactly the same theorem is true not only when alpha plus gamma is equal to pi and beta plus delta is equal to pi. Not only this, there is a weaker condition which also results in exactly the same identity, namely alpha plus beta plus gamma plus delta is equal to two pi. Now, this is weaker than the combination of these two because from these two, this one follows obviously, but not the other way around. So the theorem, however, is true even for this particular condition. So try to prove it yourself. Next, pi is a q-tental. Then two cosine pi over two equals square root one of sine plus square root minus sine. So for an acute angle, this is a true formula which express the cosine of a half an angle through sines of the full angle. All right, now, pi is acute angle which means sine and cosine are positive and they are from zero to one. They're less than one actually. So these are all positive numbers. And under the square root, I also have the positive number always. So if that's true, I can safely say that this particular equality is completely equivalent to the square of this equality. I can square both sides of the positive numbers and then we'll get exactly the same thing. So it's two cosine square pi over two is equal to square of this, which is one plus sine pi plus two product, this times this would be one minus sine square pi plus square of this, right? Now, obviously this is cosine square and again, considering that the pi is an acute angle, square root of a cosine square is a cosine. So now, sines are reduced. So the right part would be two plus two cosine pi. Now, is this true? Is now we can reduce basically by two. So I will have cosines where pi is equal to one plus cosine pi. Is that true? Well, I don't remember again all the formulas, but I do remember the formula for cosine of the sum of two angles. Now, cosine of pi over two plus pi over two, which is pi, this one, equals to cosine square pi over two minus sine times sine, which is sine square pi over two equals. Now, sine should be replaced with one minus cosine. So we'll have two cosine square pi minus one. Is that right? Or I didn't make a mistake or something. So two cosines, oh, I'm sorry, this is my mistake. It's four, it's not true. I'm squaring this thing, I see my mistake now. Okay, which actually means, this is cosine pi, which actually means, no, what the, what the, sorry. Okay, now this is the right. Cosine minus, yes, okay. So from here, I get two cosines square pi over two is equal to one plus cosine pi, which is exactly what we have. Okay, so by the way, is this the end of the proof? Well, quite frankly, not exactly, because what I have actually done, I derived from this, this. Now, this is a true statement, I agree with this. But if I can derive from God knows what, a true statement using strict logical transformations, it doesn't mean that this is actually true. Because from false, I can always derive a true through a correct logical conclusions. Like for instance, if minus one is equal to one, this is a false statement. I square both sides, and I will have one equals to one, which is correct statement, right? And by the way, I did square both sides. So that's not, so that's not really a true proof. Now, what is a true proof? Here is what I should say. All transformations which I have done are reversible, which means from here I can get here, here, here, here. And what's most important from here, I can always get here, because both sides, I consider as a positive angle, right? I mean the Q angle, so all members are positive. So I can actually say that if a square is equal to b square and a greater than zero and b greater than zero, then from this false, that a is equal to b. Then I can extract the square root from both sides. That's what's very important. For positive numbers, that's true. So now I can say that since all these transformations are reversible from obvious identity, we derive this one, and that's the end of the proof. And by the way, what I wanted actually to suggest here again is a kind of self-study thing. The same, well, similar theorem actually is true for a sign. So try to prove this one yourself. Well, this is the end of this lecture. I have presented you a few problems on conditional identities and how to prove them. I might actually suggest some other problems I'll see, but that's the end of it, thank you very much. And don't forget, you have to really sign to unizor.com, try to organize your educational process as a process, which means you will have somebody like a supervisor or a parent also signed in as a supervisor, enroll you in the program, and then you will be able to take exams, and your supervisor can actually examine the results of your exam, and well, consider this course as completed or any component of this course. Thanks very much and good luck.