 Hi, I'm Zor. Welcome to Unisort Education. I would like to present to you a couple of problems related to systems of trigonometric equations. As usually, my recommendation is try to solve these problems yourself first before you listen to this lecture. Now, as you know, I'm trying to position this course as the one which develops your mind, your creativity, your analytical abilities, etc. And it's the process of thinking about something. That's what actually develops your mind in as much as exercise in a gym develops your muscles. So try to think. I mean, that's the only recommendation which I have. If you listen to the lecture without previous thinking about the problem, that's just not even 10% of the work which you would like actually to get for your mind, for your abilities. So it's very important. Think about these problems first. It doesn't really matter whether you solve them or not yourself. It's important to spend time and think about how to solve them. And by the way, there are certain problems which I present in this course as examination problems. And I don't provide these lectures with answers. These are just multiple choice answers which you have to choose and just plain guessing will never give you the perfect score. And it's a perfect score on the exam, which means all problems must be solved correctly. That's what your goal is. Okay, without further ado, I have two problems. One is, well, actually, both seem to be easy. However, in both, there is certain trick or carefulness, if you wish, which you would like to exercise. Okay, one looks very simple. Sine x, sine y equals a, cosine x, cosine y equals b. So that's the system. Two variables, x and y, and two equations. This is the system. Now, I don't know about you, but looking at this, I immediately see the formula for cosine of a sum of triangles and cosine of a difference of triangles. So cosine of x plus y equals cosine x, cosine y minus sine x, sine y, and cosine of x minus y equals cosine x, cosine y, plus sine x, sine y. Now, what does it mean in our case? Well, I have the product of the cosines, and I have a product of the sines, which means immediately I can conclude that cosine of x plus y is equal to product of cosines minus product of sines, which is b minus a. And cosine of x minus y equals a plus b. Transformation from here to here is invariant, which means it doesn't add additional solutions and doesn't lose any solutions, because this is invariant transformation. Okay, so I have this system. Now, is it easier? Of course it is easier, because it actually falls into something like cosine phi is equal to whatever, p for instance. And we know the solution to this, and it's already actually addressed in the previous lectures, is that the angle phi is equal to plus minus arc cosine of p plus 2 pi n, where n is any integer. So from here, using this, we can derive the following. x plus y equals plus minus arc cosine of b minus a plus 2 pi n. And x minus y is equal to plus minus arc cosine a plus b plus 2 pi m. I use a different integer because it's completely different things. And we might actually use different integer, not necessarily the same. So in this case, I can add, for instance, 0 and in this case, 1. And it will be a solution. Or in this case, minus 3 and in this case, plus 5. That's the solution. All right, so that's what I have. Well, now this seems to be like a system of two linear equations with two variables, which is very simple to solve. However, let's be very careful about plus or minus in this case, and m and n in this case. I would like to address them, how should I say it, very carefully. Let me do it this way. Since x plus y can be anything from this, let me just represent it in two cases, separately. Arc cosine b minus a plus 2 pi n, or it can be minus arc cosine of b minus a plus 2 pi n. And x minus y can be plus arc cosine of a plus b plus 2 pi n. Or it can be minus arc cosine of a plus b plus 2 pi n. Now, in these two cases, they actually should be combined into system. So can I have plus here and minus here? Yes, minus here and plus here, yes. Plus and plus, yes. Minus and minus, yes. So it's four different systems I have. And four different solutions, and each solution is not actually a single solution, because there is this integer n or m, which should be added. So let me just, as one particular example, like one case, the case number one. I'll consider first and first, which is x plus y is equal to this, and x minus y is equal to this, from which I can conclude by adding these two. y and y will be reduced, it will be 2x here. So x is equal to 1f of arc cosine b minus a plus arc cosine of a plus b plus, well, pi times m plus m. And y is equal to 1 half. I have to subtract from this, I subtract this. So x and x will be reduced and will be 2y. So that's why 1 half here. So it would be arc cosine b minus a minus arc cosine of a plus b. And from this, I will subtract, and that will be n minus n. So b and a are constants, so there is no problem with this. Now, this is many pairs of solutions where m and m are any integer numbers. Now, it's actually natural to think that, hey, if m and n are any integer numbers, can I just replace m plus n as k and say n minus m as l and say that k and l are anything any integer numbers? Well, not exactly, because look at these, you see, x plus y should give me something which is multiple by 2 pi. And these two always give me something which is multiple by 2 pi. For instance, if I add them together, I will have 2n. If I subtract them, it would be 2n, right? So k and l also must be of that particular property. If I sum them together, they must be even. So I will have 2 pi here. Because if I will have, for instance, 0 and 1. So I will have 0 here and pi here. And my sum would be, sum of these, my sum would be like r cosine of this, because r cosine of a plus b will be reduced. Will be plus 2, plus pi, just plus pi, right? Because k is equal to 0, l is equal to 1, it will be plus pi. And that is not right, because the period of the cosine is 2 pi, not pi. So if my cosine of x plus y is equal to b minus a, I cannot have pi multiplied by something, or like 1, for instance. I have to have 2 pi by something. So k and l can be replaced m plus n and n minus l with integers k and l. However, there is one particular requirement. There are some and their difference must be even, which means either they're both even or they're both odd. So any way you would like to express this, either with m and n saying m and n are any integer numbers, or you can say pi k and pi l, but then the condition is that both k and l should be either both odd or both even. So their sum and their difference is always even, okay? So that's the little twist which you have to really think about it. Now, I can, exactly in the same fashion, I can have the second choice. If I will have, let's say, plus here and minus here. And you will see exactly the same type of manipulations. Well, let me just do it, just one. So it's this and this, right? So their sum would have a minus. So x is equal to 1 half arc cosine b minus a minus arc cosine a plus b plus pi n minus m, actually m plus m, because I'm summarized. Yes, I'm summarized. And so let me wipe this out, so it doesn't confuse. And y is equal to, if I subtract, it would be sum. So 1 half arc cosine b minus a plus arc cosine of a plus b plus pi n minus m. So basically, I exchange the x and y. And looking at the original system, original system, you remember, right? Sine x, sine y is equal to a, cosine x, cosine y is equal to b. Obviously, if I change x and y values, it will be exactly the same thing, because it's symmetrical relative to x and y. So I should expect, actually, that if pair x, y is a solution, then the opposite, y takes the value of x and x takes the value of y is also a solution. And then other two solutions, when I have minus here and plus here, similarly it would be minus here and either plus here or minus here. It doesn't really matter. So these four different formulas, 1, 2, and then 3 and 4 would be with minus here. They represent four different sets of solutions and sets because m and n can be any integer numbers, which would actually go through all that. These four formulas, each of them represents a set of solutions, represent a complete set of solutions to this particular system of equations. All right? Well, that's it. Now, in the beginning it was simple. I just recognized the cosine and the cosine of sum and cosine of difference. But then at the very end, you really have to be very careful with these cases. That's what probably very important. So sometimes there is an easy solution, but you really have to watch every step so you don't lose anything or you don't add anything, etc. And the next problem would be even more educational in this particular way. Okay, this is a system of three equations with three unknown variables. Actually, I was thinking about examining this particular system is, hey, it would be nice if I can express tangent of sum of three angles in terms of tangents of each angle. Then I will have only tangents, which means I can replace tangent of x with some variable tangent of y with some variable and tangent of z with some variable. And have another system of three equations with three variables, but it is not trigonometric equations anymore. So these will be just a regular algebraic system of three equations with three variables. But obviously, I don't remember. I don't know the formula tangent of three, the sum of three angles. I do remember for two, and I do remember that tangent of sum of two angles is expressed in the terms of tangent of each one, right? So you remember this, so tangent of phi plus psi is equal to tangent of phi plus tangent of psi divided by 1 minus tangent phi tangent psi, right? This is the formula, and again, I do remember. The formula of sum of two angles, I do remember actually, sine, cosine, tangent. But tangent, I don't, but it's very easy to derive. Same thing with secant and cosecant, but I do remember sine, cosine, and tangent of sum of two angles. These formulas I do remember. So this is one of them. So if the sum of two angles can be, tangent of sum of two angles can be expressed in terms of tangent of each one of them. Then, obviously, I hope that the tangent of three angles summed together also is expressed, but let me derive this formula. And obviously, that's just step by step, first, two, and then the third one. So, tangent of x plus y plus z equals. So x plus y is one variable, right? One angle, tangent of x plus y, and tangent z is another. One minus tangent x plus y, tangent z equals. Okay, now let's express tangent of x plus y in terms of tangent, x and tangent of y. So I will have tangent x plus tangent y over one minus their product. Plus tangent z divided by one minus, same thing here. Tangent of x plus tangent of y over one minus tangent of x, tangent of y times tangent of z, right? So what I will do is, in the numerator, I will multiply this by this. And it will be one longer numerator. In the numerator, I will have a fraction where numerator will be this plus this multiply. Now, and the denominator in the big numerator will be one minus product of tangents. Same thing here, also I will have, in the big denominator, I have, if I will multiply this by this, and this by this. I will have numerator whatever I will have, and denominator one minus product of tangents. And I will just drop both denominators, right? I will use only numerator from here and numerator from there. Because denominators will be the same. So what will be the numerator in the top one? So it's tangent x plus tangent y plus this one minus tangent by tangent, multiplied by this. So it would be one by tangent, so it's plus tangent z. And it will be minus their product. Tangent x, tangent y, tangent z. That's my numerator. Now, what will be in the denominator? So the numerator part of this thing, it's one minus this. One minus tangent by tangent is times one, right? So it's one minus tangent x, tangent y minus this times this. Which is tangent x, tangent y, z, and minus again, tangent y, tangent z. So that's my answer. Now, as you see, this is symmetrical relative to x, y, and z, right? We can transform them in any way we want. And this is also symmetrical, x, y, z, x, y, z. And this is x, y, x, z, and y, z. So it looks like we have the right formula. Now, using this, let me just take the tangent of the last equation, and I will have tangent of x plus y plus z, and I can express it in this way. I will have minus 3 on the right, because tangent of r tangent is, obviously, whatever the argument is from definition of the r tangent. And then I will have the three equations where only tangents of x, y, and z are present, right? So I can actually replace tangent x with a variable u, tangent y with a variable v, tangent z with a variable w, and I will have an algebraic system of equations relative to u, v, and w, because only tangents are present. However, and here is a very important twist which you really should not lose your focus from. Yes, x plus y plus z equals something. That's fine. But then I take the tangent of both sides. Is this an invariant transformation? So let me ask again, if p is equal to q, is tangent of p equivalent to tangent of q? Are these two invariant? Are these solutions exactly the same as these and these exactly the same as those? No. Why? Because the tangent is a periodic function. And as we know, we can add to this angle q, for instance, pi k, and it will be exactly the same equality. Not only p is equal to q is the solution of this, but also p is equal to q plus pi k, where k is an integer. So this thing has an infinite number of solutions. So if p is a variable and q is a constant, let's say this is 3. Then from this, we derive p is equal to 3 plus pi k, not just p equals to 3. So it's not invariant, which means that we have to be very careful at the very end when we will derive some kind of solutions to our system after we have taken tangent of both sides. We have to be very careful examining our solutions, because we might introduce certain solutions which actually do not belong, right? So keeping this in mind, I would still like to proceed with taking the tangent of both sides and replacing my variables with these new variable substitutions. So I will have u times w equals to 1, v times w equals to 1, right? The tangent of y is my new variable v, tangent of z is still w. And on the left, I have this thing, and it's equal to minus 3, right? Because I took tangent of both sides. So let me just put it, which is u plus v plus w minus u v w divided by 1 minus u v minus u w minus v w equals to minus 2. This is the system which we can attempt to solve, but this is algebraic. It's not trigonomic, which I hope makes it simpler. Now let's see if we can solve it. Because if we can, then from u, v, and w, you will just know how to derive x, y, and z from each of these guys, right? OK, so what's my proposed plan to solve this particular system of equation? Well, let's consider u and v are kind of symmetrical in this particular case in the entire system, right? I would attempt to use w as a base variable, and I will express u in terms of w, v in terms of w, and substitute everything in the third. And then I will get only one equation with 1w. Now, obviously, I have to think about what if my denominator equals to 0, which means no solution. So I have to really be careful about this as well. So there are a lot of things to be careful about this, right? So the last equation I actually can rewrite as this numerator is equal to minus 3 times denominator, right? So I can just multiply both sides by denominator to have a linear result, any kind of fraction. So from this, I will have u is equal to 1 over w, v is equal to 1 over w. Again, I'm doing it kind of freely, but I have to really think about what if w is equal to 0, right? Is this a solution? Well, each one of these cases must be examined separately. What if w equals to 0? What if 1 minus uv minus uw minus vw equals to 0? So these are separate cases, which I have to examine separately. Don't forget it. Even if I forget it, when I'm presenting this lecture, you should not. All right, so having these, I will substitute it into this equation. And what will I have? Well, let me just write u is 1 over w plus v is 1 over w plus w minus u times v. It's 1 over w square and w, so it will be 1 over w. Now equals to minus 3 times denominator, which is 1 minus. Now, uv is 1 over w square, right? Minus uw, which is 1, right? And vw, which is 1, is 1. So that's my equation. Now, let me just do some reduction here. This and this, right? Now, let me rewrite it. So on the left, I have 1 over w plus w. On the right, I have this minus and this minus and this minus. So it's basically 3 over w square plus 3, right? Minus minus, so it's 3 over w square and minus minus plus 3. And now, I will multiply it by w square. And I will get w plus w cube equals 3 plus w square. I multiply everything by w. And I already kind of consider that this case should be separately. And let me rewrite it like real equation. w cube minus, now I think I made a mistake. It's 3w, yes, it was 3w, so it's 3w. So minus the 3w square minus plus w minus 3 equals to 0. So w minus 3, q minus 3w, right? OK, now I have to solve this equation of the third degree, right? So it's not such an easy thing. However, in this particular case, I immediately see a very important thing. If I will factor out w square from the first two, I will get w minus 3. And this is w minus 3. So it's equal to w square plus 1 times w minus 3 equals to 0, which means that either this is equal to 0 or this is equal to 0. Now, this is not equal to 0 because it's square plus 1, right? Square is always non-negative, plus 1 is always positive. So w is equal to 3 is my solution. And obviously, from here, my u is equal to 1 third and my v is equal to 1 third. These are all solutions. Am I almost done? Well, look, I mean, if tangent of x is equal to 1 third, then x is equal to arc tangent of 1 third plus p pi k. y is equal to arc tangent of v, which is 1 third as well, plus pi m. And z is equal to arc tangent of 3 plus pi n. Is that it? Well, unfortunately, no. Well, first of all, I did not check all these cases. Remember, if w is equal to 0, well, w is not equal to 0. It's 3. How about 1 minus u v minus u w minus v w? Is that it equal to 0? I hope not. So this is 1 third and 1 third, 1 third, 3. And 1 third, 3 equals to this is 1 minus 1. So it's minus 1 and 1. No, something like this. OK, so that's good. Well, at least I did not, by multiplying by this, I did not introduce new solutions, which is good. So far, so good. However, there is one DJL which I did mention before, and it should actually play the very important role right now. Remember this? When I took the tangent of both sides, I was talking about this as a non-invariant transformation. So if k, m, and m are any integer numbers, obviously not all the numbers fit this particular equation, because arc tangent of minus 3 is always between minus pi over 2 and pi over 2 without including the boundaries, right? So whenever we sum these together, k plus m plus m must be in such particular value that altogether it falls into this particular interval. Is it possible? Well, let's just think about it this way. Let me just make some kind of a variation of the values. What is arc tangent of 1 third? Well, remember the function tangent looks like this, right? This is minus p over 2. This is pi over 2. This is 0, y, x. Now, tangent of pi over 4, which is 45 degrees, tangent is sine over cosine. It's equal to 1, right? So this is 1, and this is pi over 4, right? Well, it's not exactly half of this distance. Let's make it a little bit better. This is pi over 4, and this is 1, all right? Now, as the angle increases, the tangent increases. The angle decreases. The tangent decreases to 0. So I need 1 third. So from 1, I should go to somewhere here. So it's less than p over 4, less than 45 degrees. By how much less, I don't know. But well, considering the tangent is diminishing in some way, well, maybe it's 20 degrees. Maybe it's 10 degrees, something like this. Let's just assume for argument's sake that this is around 20 degrees, right? So this is around 20 degrees plus pi k. Y is also about this. And 3 is somewhere here, right? This is 1, this is 3. So it's greater than pi over 4. And again, this is 45, this is 90. I don't know how it is. Maybe 80, approximately, right? I mean, it doesn't really matter whether I'm exactly evaluating the value or really approximately. Can I choose k m and n in such a way that the sum of these would be arc tangent of minus 3? Now, what's arc tangent of minus 3? That's somewhere here, right? So let's say it's minus 80 degrees, something like this. Well, probably yes. I can put something like, I don't know, minus 3 here, or minus 2, for instance, here and 1 and 1 here. And I will be somewhere close. So I think I can play with these numbers in such a way that their sum would be within this particular interval. Again, being a little bit more precise with my variation, maybe I will have a calculator or something like this, I will definitely be able to find integer k m and n, which would satisfy this particular equation. And it's not just one triplet of k m and n, because I can increase one of them and decrease another, and that would be exactly the same thing. So basically what I'm saying is that there are a lot of solutions that are expressed in this general formula with a condition that k m and n such should be such that the sum of these three angles is within this particular interval somewhere. Well, basically that concludes the full solution of this particular equation. So what's important? I think it's very important to understand that taking the tangent of this particular equation is not an invariant transformation. Now, everything else, whatever we did, was, well, really some kind of technicality, I would say. Taking, for instance, deriving the formula for a tangent of three different angles sum together, that was trivial, basically. I didn't know the formula, I just derived it on the fly. And then when I have this converted into algebraic system of equations, it was a little bit maybe unusual to have the system which resulted in the equation of the third degree. But it was such an equation of the third degree where you very easily see the solution. So basically that was not something unusual. But I think the first one, the tangent of both sides, is not invariant transformation. It's very important for you to think about. What I would suggest right now is take these two problems and try to solve them yourselves. And again, think about what kind of solution you get, what kind of integer, KMM, whatever, in both cases for both problems which I have presented, what kind of values are allowed, what's important. Well, basically that's it. Thank you very much, and I will try to put maybe some more problems with some interesting twists if I can. Thanks again, and good luck. Good luck to you.