 Hi, I'm Zor. Welcome to a new Zor application. I would like to present a couple of problems related to conditional trigonometric identities. These problems are slightly different than the ones which I considered in the previous lecture. You see, the previous lecture was about problems where the brute force method was actually working. These problems are different, and I don't want you to have an impression that all these conditional identities are just really hard work and no creativity at all. So considering the character of these problems, I do encourage you to try to solve them yourself first, think about them even if you didn't really solve the problems. It's still very, very useful if you think about them. So I will present my solutions. Now, I have not made out these problems myself. I borrowed it from one of the books which I have listed in the reference material on Unisor.com, and I had to solve them myself. So I will present my solutions and my train of thoughts, which might or might not be exactly the same as yours, but anyway, I think it serves some educational purposes, I guess, and I'll try not to make any mistakes on the way of solving these problems. There are some calculations which are really a little bit maybe involved, but still the creativity, I think is the most important characteristic of these problems. Okay, problem number one. I have equation. Actually, what's interesting is that this particular problem is the combination of the algebra and trigonometry, because there is an equation here. So equation is, that's the polynomial of the third degree, and it's not easy to solve it, right? Now, what's known about this equation is that the solutions are numbers which are tangents of three different angles. That's what's known. So these three numbers, where alpha 1, alpha 2, and alpha 3 are some angles, these three numbers are solutions to this equation. That's what's known. Then there is another equation. I'll use the same coefficients, but in a different sequence. And I know that some other three numbers, which happen to be tangents of three other angles, are solutions to this equation. So these three numbers, which are tangents of some three angles, beta 1, beta 2, and beta 3, are solutions to this equation, and these equations, they share the coefficients, but they are used in a different way. So that's what's given. That's what I meant, algebra and trigonometry combined together. Now, what is necessary to prove what kind of identity here? That some of these angles is pi times some integer number. Well, to tell you the truth, I was at loss when I saw this problem and it's all different things. Again, algebra and geometry and how they are related. Well, so I'll try to share my train of thoughts as I was solving this problem because that might actually help. I don't want just to present the solution, etc. So my first thought was, okay, these are all tangents. Now, the fact that some angle is equal to pi times some integer k reminds me that the tangent of this angle equals to zero. Because again, these are tangents, that's why tangent comes to my mind, and the pi k is exactly the arguments where tangent, this is zero, this is pi, this is minus pi, etc. So at pi at minus pi at zero, it's all the points where tangent is equal to zero. So proving this is equivalent to proving the tangent of this equals to zero. That's completely equivalent. So that's what I am going to prove. That's what I have decided basically. Because to prove that some of these is equal to pi times k is kind of a strange thing actually. It just doesn't resonate with me at least. But this actually means much more interesting. Okay. So how can I get that? Well, obviously, this is some of these three and some of these three, and I can use the formula for tangents of some of these angles. So this is equal to tangent of first sum plus tangent of second sum divided by something. Doesn't really matter what it is, because I have to prove that this is equal to zero. So I have to prove that this is equal to zero, the numerator. Again, I hope the denominator is not equal to zero, etc. So there is a proper definition. So these are minor details, which obviously you don't really think about, at least in the first stage. In the first stage. So this is what I have to prove, right? Okay. Now these are tangents of sum of three angles. So I still have to really have a formula for tangent of sum of three angles, which can be obviously very easily derived. But now, how can this be used? We don't need this anymore, right? This is given, this is what I have to prove. Now how this can be used? Well, this is my algebra part. Now, the fact that I know that these three are solutions of this particular equation, what does it actually mean? It means the following. That this is equal to, that's what it means. X cubed has the coefficient of one, so that's why I put x's here without anything. And since I know the tangent of alpha one is the root of this equation, the solution of this equation, it means, if you remember our algebra lectures, that this thing is divisible by x minus the solution of this equation. And same with the second solution and same to the third solution. Now, same thing is here. Okay, does it make my life easier? Not yet, because I still don't know how to use it in this. Because for this, I need the values of tangents, right? And I'm working to get the values of tangents. I can't. Because a and b and c are just some numbers. All right, but what's interesting is that these coefficients, you see, they are equal to each other. Why was it given to me? Well, I've decided the following. If I will open these parentheses and multiply whatever is necessary, then I will get the value of a and b and c. Now, here, I also can multiply open all the parentheses and I will get values of c, b and a and it will be exactly the same c as c here and a as a here as b as b here. So that's how I will get relationship between tangents of alpha and tangents of beta, right? So, let's do it. Now, my coefficient a is here. It's coefficient at x square. Now, x square, I can get multiplying these by x times x times tangent a3 or x times x and tangent a2 or x times x and tangent a1. So, this is how I can get the coefficient a from the first equation. So, x square times tangent one but it's minus sign, right? So, it's minus tangent a1 minus tangent a2 minus tangent a3, right? So, that's how I can get x square. I might multiply this by this by three number, this, this and this three number, this and this and this three number. All of them have a minus sign. That's why I put this. This is a from this equation. What's a from this equation? Well, this is the free one, which means there are no, no x's. No x's can only be obtained from multiplying three members, right? So, this is equal to, from the second equation, the multiplication tangent b1 times tangent b2 times tangent b3 but it should be a minus sign here because we have one minus, two minus, three minus. So, it's minus sign. Now, instead of minus sign, I will do plus here, if you don't mind. That's the same thing, right? So, I have this important relationship between tangents of alpha and tangents of beta. That's number one. Okay, next is b. All right, b is equal to, from the first equation, where can I get the coefficient at x? If I will take this and this times x or tangent alpha one times tangent alpha three and this x or tangent alpha one times this x. So, it's three members. Now, the sign will always be positive because I have one x and two minuses, right? So, b from the first one is equal to tangent alpha one times tangent alpha two plus tangent alpha two times tangent alpha three plus tangent alpha one times tangent alpha three. One, two, two, three, one, three, equals. Now, from the second equation from this one, I have the same b here. So, it's coefficient at x to the first degree, but everything is the same except angles are beta. So, I'll just put exactly the same, but with beta. So, this is my equation for b. And finally, c. c actually here and c and here, they're very much like a here and here. The only thing is my a and b should be, alpha and beta should be changed because the coefficient at c is all the tangents with beta like here for a, right? Which means, so again, I will use minus c equals, minus c equals tangent beta one plus tangent beta two plus tangent beta three. And that's equal to, that's for this one, but to this one, c is multiplication with a minus sign, but reverse, I put minus sign here. So, it's tangent b one, tangent, sorry, alpha one, tangent alpha two, tangent alpha three. That's where it is. So, I don't really need alphabet and gamma, I mean a, b, and c, sorry, anymore. All I need is these, this one, this one, and this one. I have three equalages between these different tangents and their expressions with them. And what I have to prove, based on this, that this is equal to zero. Now, obviously for this, I need to transform the tangent of the sum of three angles into something more palatable. All right, so let's do it this way. Now, I don't really need this anymore. It serves its purpose. And what I will do here, I will derive the formula for tangent of three angles. Let's use u, v, and w equals. Well, let's combine u plus v as one particular angle. So, the formula would be tangent of u plus v plus tangent of w divided by y minus tangent u plus v and tangent w. So, that's what the formula of the tangent of sum of two angles, one angle being u plus v and another angle being w. But now I have to represent tangent of u plus v in the form of tangent u and tangent v, which would be, well, very, very similar. So, let me retain this. This would be tangent u plus tangent v divided by one minus tangent u, tangent v. That would be tangent u plus v. Plus tangent w divided by one minus. And again, this thing, I will put this. Tangent u plus tangent v divided by one minus tangent u tangent v times tangent w. This I don't need anymore. I hope I didn't make any mistakes because if I did, that would be unfortunate. All right, now, this should be simplified somehow, right? Okay, I think one minus tangent times tangent might be dropped from the numerator and denominator. And again, I hope that I don't have these undefined angles, u, v, w, or their sums, et cetera. So, we are excluding all these bad cases. Leaving common is a good one. So, what will be here is tangent plus tangent plus this times this, which is also another tangent, and minus the product of u, v, and w. That's what would be tangent u plus tangent v plus tangent w minus tangent u times tangent v times tangent w. That's what would be in the product, in a numerator. Now, denominator would be, I know what I would just put it here. I would have more space in this case. So, it's sum of tangent minus their product. Okay, that's the numerator. Now, denominator is one minus this minus this times this. So, it would be one minus tangent u times tangent v and minus tangent u times tangent v and minus tangent u times tangent w minus v times tangent w. That's what it would be. One minus, and then pairs of tangents multiplies to each other. One minus, and this minus. Right, it would be this. One minus tangent u tangent v minus tangent u tangent w minus tangent v tangent w. That's what the formula actually is. That's the formula for a sum, tangent of sum of three n plus. Now, if u, v, and w are alpha one, alpha two, and alpha three, I will get this member. And then for beta, I get this member. And then I can actually use these formulas and try to prove that sum of these is equal to zero. Okay, but what's interesting is that if my u, v, and w are equal to correspondingly alpha one, alpha two, and alpha three, I will use, I will put alpha one, alpha two, alpha three, alpha one, alpha two, alpha three. But I know that my alpha one, alpha two, and alpha three product of their tangents is equal to some of these. So when I substitute instead of u, v, and w, alpha one, alpha two, and alpha three, instead of these, alpha one, alpha two, and alpha three, I can put some of the tangents. So in my numerator, I will have tangent alpha one plus tangent plus tangent alpha two plus tangent alpha three, minus, so I'm talking about this, only about this number. But I can put minus, and instead of tangent, product of tangents of alpha, I will use sum of tangents of beta divided by, now here I have tangent alpha one, alpha two, alpha one, alpha three, and alpha two, alpha three. I have something similar to this, and this is this expression. What's interesting is that this is exactly the same as these pair products of beta. So whenever I put this one, I can put it here in the same way, alpha one, tangent, alpha two, minus, tangent, alpha one, tangent, alpha three, minus, tangent, alpha two, alpha three. So that would be in this denominator. Now, we're talking about this piece, right? This is what I have done so far. Now, this thing, now what's interesting is, this thing is pretty much similar, but instead of alpha, I have beta. So it's plus, tangent, beta one plus tangent, beta two plus tangent, beta three, and again, instead of product of three tangents for beta, I will use sum of three tangents for alpha and divide it by the same thing for beta. Now, what's interesting is that denominators, I was saying that you see these sum of pairs of products are the same for alpha and for beta. Here, look at this. So denominators are the same because these are alpha, these are beta, but we know that some of these pairs are the same, which means since denominators are the same, I can just add numerators, and here I have sum of tangents for alpha with plus and here with a minus. Beta with a minus and here beta with a plus. So obviously, this is equal to zero, which was to be proved. This is the end. Well, I think it's interesting and I was quite surprised actually to see this problem and I had fun solving this, but again, what you have to really think about, well, you have to think out of the box basically, that's what it is, because if you are given an algebraic equation and it's known that the tangents of some angles are solutions, quite frankly, I don't really know how to approach this problem. Well, again, you have to think about this and you have to try to find the proper way to use the conditions which you've got. So this is one problem which I wanted to discuss with you today and there is another one. Also, I would say a little bit out of the box. Now, what's known is that alpha, beta, and gamma are angles of a triangle. What's also known is that sine alpha, relative to sine beta, relative to sine gamma is equal to four, five, six. Now, what's necessary to prove is that the cosine alpha ratio to cosine beta ratio to cosine gamma is equal to 12 to 9 to 2. Again, a little bit unusual problem and it's not something which you can solve just brute force. Now, I see sines, I see cosines, I see the triangle. What does it remind me? So I was immediately thinking about the law of sines and law of cosines. Now, the law of sines, let's remember, a to sine alpha is equal to b to sine beta, c to sine gamma, right? Where this is a triangle, a, b, c, and opposite angles are alpha, beta, and gamma. That's what it is. At the same time, this is the law of sines. I do remember that. Law of cosines. a squared equals to b squared plus c squared minus 2bc cosine alpha, b squared equals to a squared plus c squared minus 2kc cosine beta, and c squared equals to a squared plus b squared minus 2ab cosine gamma, right? This is an extension of the Pythagorean theorem. Pythagorean is for the right triangle and this is for any triangle and the square of any side is equal to sum of squares of two other sides and there is also a correction minus 2 product of these and cosine of that angle in between. So I think this is something which I probably can use. Okay. Now, from this, I can say that, well, let's consider, for instance, this. What does it mean? Well, if you reverse it, you will get sine beta divided by sine alpha s equal b to a. So what I want to say is that the sines of the angles are proportional to lengths of the opposite sides, opposite to these angle sides. Same thing with b and c and a and c obviously, right? So I can say that from this, I can say that a to b to c is also equal to 4, 5, 6, where abc are sides of the triangle opposite to alpha, beta, and gamma, right? So this is immediate consequence of the law of sines. Okay, that's good. We don't really need law of sines anymore. We've used it. We've got this proportionality among sides. All right. Now, if a to b to c is 4, 5, 6, and if I will assign the value of, let's say, p equals a divided by 4, then a is equal to 4p, b is equal to 5p, 3p, and c is equal to 6p. Now, why do they do it? Because it's easier to deal with integer numbers instead of fractions, right? I can always assign a as the unit of measurement, in which case b would be 5, 5, 4s of a. I just don't want to deal with fractions. I would like to deal with integers, and that's why I decided to have as a unit of measurement a divided by 4, in which case a is 4 of these units, b is 5, and abc the ratio a to b is 4 to 5, and b to c is 5 to 6, et cetera. So this is what I have right now, and this is what I will use in these particular equations. Now, let's substitute these numbers into the first one. What will I get? Instead of a, I will get 16p squared. Instead of b squared, I will get 25p squared. Instead of c squared, I will have 36p squared, and bc is 30 times 260. So here I will get 60p squared. And obviously, I can reduce everything by p squared, and that's what I've got. I've got the value of the cosine of alpha, right? So what is the cosine of alpha? Cosine alpha, cosine alpha equals 36, 50, 61, minus 16 is 45, 45, 60. All right? Now, here, b squared is 25. You know what? I will not use p anymore, because the p will be obviously reduced. I'll just use the numbers. Now, a squared is 16. 16p squared, and p squared will be reduced. c squared is 36, and 2ac is 24, 4g8, 4g8. p squared, and the gate p squared is reduced. So what's left? 16 plus 36 is 52, minus 25 is 27. So cosine beta is equal to 27, 4g8. You know what? Let me reduce this thing. This is by 15, right? So it's three quarters. This is by three is 9 sixteenths, and cosine gamma is equal to c squared is 36, a squared is 16, and b squared is 25, a b is 22 times 2, 4g. This is 4g. So my cosine of gamma is equal to 5, 4g1 minus 5. 5, 4g8. Well, obviously, I don't need this anymore. So you know what? Probably it would be better if I have sixteenths everywhere. So it's 12 sixteenths, 9 sixteenths, 2 sixteenths. And as you see, their ratio is 12 to 9 to 2. That's exactly what's necessary to prove, which I unfortunately wiped out, but that was what was necessary to prove. Well, that's it. That's the solution to the second problem and at the end of this lecture. These two problems are very interesting again in that respect that it really kind of wakes up your creativity and ingenuity. Think out of the box. Well, I think that's the purpose of all these math lectures which I'm trying to represent to you. Mathematics teaches the creativity. It's not necessary and it's not really interesting, actually quite boring to do the same thing using the known methodology. We did not know how to solve these problems. We just invented it on the way. So that's what the math is for. That's the training for your brain. Thanks very much and good luck.