 Hi, I'm Zor. Welcome to User Education. Somebody brought to my attention that I have missed one particular lecture about problems, trigonometric problems. So, I'm returning basically to the mass proteins course right now to solve these trigonometric problems. Now, before you, which this lecture, I suggest you to look at these problems just yourself and try to solve them. It's definitely the most intended actually purpose of this course is to basically develop your mind and tune it towards solving certain unknown to your previous problems. So, it doesn't really matter what kind of problems I'm solving right now. What is important is for you to solve them because that would prepare you for any kind of unknown in your real practical life, not related to mathematics or physics or anything like that. Because the problems which I will present right now have absolutely nothing to do with your future. You will definitely not be in a position of solving these problems, exactly these problems. But you will have a lot of new problems which you will have to basically solve by yourself. And solving these mathematical problems basically prepares you for this particular task. Okay, so I have six trigonometric problems. I'll just go one by one. And in some cases, I might actually miss certain calculations because they will be trivial, but I will try to go through all of them. Okay, number one is I have to prove that two times sine to the sixth of angle plus cosine to the sixth of this angle plus one equals to three times sine to the fourth plus cosine to the fourth. Okay, now this is a trivial exercise in basically algebraic manipulation. The only thing which you have to really know is you see there are sines and cosines here. It would be nice. It would be all dependent on one particular function. For example, sine. For this purpose, we will do very simple thing. We know that sine squared plus cosine squared is equal to one for any angle phi. So I'll just express cosine in terms of sine and substituted there. And check if I will have the same thing on the left and on the right. Because angle phi can be any angle, right? So cosine squared equals to one minus sine squared. Cosine to the fourth of angle phi is equal to square of this, which is one minus two sine squared plus sine to the fourth of phi. And six cosine six is multiplication this by this square by fourth degree, right? Which is, okay, let's multiply this by this. It would be one minus two sine squared phi plus sine four phi minus sine squared phi minus sine squared multiplied by one. Multiply by this would be plus minus and minus would be plus two sine to the fourth phi and minus sine to the sixth phi. So this is cosine to the fourth, which is this cosine to the sixth, which is this. If you will substitute this into this and this, well, obviously you can make it a little bit easier. It's one minus sine squared would be minus three sine squared. Sine four would be also plus three sine four and minus sine to the sixth. And if you will substitute this into this and I'm not going to do it. So if you will do it, now I did it on a paper, actually. You will have exactly the same on the left and on the right, basically. So all you need to know is basically the approach in this particular case. The approach is to basically unify. So everything is expressed in terms of sine, let's say, or cosine. It doesn't really matter. You will have exactly the same result. So this is how this particular problem and maybe some others which is equivalent or looks like analogous, whatever. This is an approach. Just unify these particular equations, bring them into some common denominator, if you wish. Not in particular sense of arithmetic common denominator, but in this case sine seems to be like a natural base for everything. Or cosine, whatever, but not both of them. That's what's important. What's kind of difficult in this particular case is because you have two different functions, sine and cosine. If you have one function, everything would be fine. Okay, that's problem number one. Problem number two is very, very analogous. I have to prove that sine to the eighth minus cosine to the eighth equals four sine to the sixth minus six sine to the fourth plus four sine square minus one. Okay, it's exactly the same case because right now you have more sines than one particular cosine. So you have to do exactly the same thing. You have to do the cosine square is equal to one minus sine square. You have to have cosine to the fourth, which is equal to minus two sine square plus sine to the fourth. And you have cosine to the eighth of five, which is square of cosine four. So square of this, which means this multiplied by itself. Okay, multiply by itself. Let's multiply by this would be minus two sine square five minus sine to the fourth five. Multiply by this would be minus two sine square five plus four sine to the fourth minus two sine to the sixth. And then multiply by this sine to the fourth by one times this minus two sine to the sixth plus sine to the eighth. So that's my cosine to the eighth. So if I multiply, if I subtract, if I do this with a minus sign, so all signs will be with a minus and sine square, a sine to the eighth, sine to the eighth will disappear. And then basically, well, let's maybe simplify it. So we have sine to the eighth. Now sine to the sixth minus two and minus two. So it's minus four sine to the sixth. Sine four, one another, another six. So it's plus six sine to the fourth. Second will be minus four minus four sine to the second degree. And plus one, right? Plus one. Okay. Now with a minus sign, you will have minus plus minus plus plus minus. Then you will add sine to the eighth. This will disappear. That will be four sine with a plus. Okay. Minus sine to the fourth plus sine and minus one. Yes, exactly. So that's the second problem. And approaches exactly the same. You unify both sides to have only one particular function, sine, instead of two sine and cosine. So these are trivial problems and they require only accurate manipulation. As soon as you have decided what is the approach and the approach is to express cosine as a sine, after that it's just trivial manipulation with numbers. Number three. Okay. So you have a unit circle. Now, let's take the angles which are multiple of 30 degrees, which is this one, this one, this one, this, this, this, this, this. And they have to find sum of sine square of these angles, phi. And angle phi is actually k times 30 degrees. Where k is from zero to everything. So how many of them? Twelve, right? 360 divided by 30. So it's 12 points. So from zero to 11. That's what basically needs to be calculated. Now, let's just consider only the first quadrant. And we will take this one, this one, and this one. These three. So it's zero, 30 degrees and 60 degrees. Now, why? Well, because then the next one would be very similar. Actually, I think I should better include this one as well. Because this is zero anyway. Sine of zero and sine of 180 is zero. So these are not contributing to the sum at all. So let's just take these three. And then corresponding with this, this, this, this, this, this, and this. Okay. So let's talk about these first three. Sine of 30 is one-half. Square we have one-quarter, right? Plus 60 is square root of three over two. Plus this is one. So what do we have now? We have one-quarter, three-quarters, and two. So that's two. So these are two. Okay. Now, these are also two. Why? Because the difference between sine and cosine relative to these angles, like plus 30, minus 30, plus 60, minus 60. The difference will be in the sine, but since it's a square, sine will be unimportant. So we will have these also equals to two. So what remains, this and this. Now, this is one-half, square, and square root of three, square. So it's one and one. So what do we have? One, one, two, and two. So it's six. So that's the sum. So what you have to remember is basically the sine of 30 degrees is one-half, and that's something which you do have to remember. It's easy. And 60 degrees is square root of three over two. I just remember it, but obviously you can find it out yourself without any problems. Because 30 degrees is, this is 60 and this is 30, right? So this is an equilateral triangle. So this is also 60. And this thing, this particular piece is half of this, right? This is half of the side. This is the whole side. So that's why the sine, which is ratio between this to this, is one-half. And if this is one and this is one-half, then by theorem of Pythagorean it will be square root of three over two, that side. So that would be sine of 60. So that's it. That's a simple thing, direct consequence from Pythagorean's theorem. So that's it. That's my problem number three. Okay. Problem number four. Okay. So again, you have unit circle and we are considering only the first quadrant. We have two points, point A and point B. Now A has coordinates xA, yA. Point B has opposite coordinates yA, xA. Okay. What needs to be proven is that their sum, sum of this angle and this angle is equal to 90 degree. Okay. So what we will do is we will drop these perpendicular. And what I'm basically staging right now that these triangles are congruent. Question is why? Well, very simply. What is xA is this and yA is this? Now, I'm saying that coordinate of B is yA, xA. Now this is yA and this is xA. So as you see, these triangles are absolutely, have absolutely the same size. xA, xA, yA, yA and hypotenuses are equal to one. This is the unit circle. So these are congruent triangles. Then the angles which are opposite to equal sides are equal. So let's take, let's say, this side and this angle in the triangle OA. Let's put it M, M. So in the triangle OAM, this angle AOM is opposite to y. In this triangle opposite to this equal side is this angle. So whatever this angle is, this angle is. But now OBN is the right triangle and the sum of these two angles is supposed to be equal to 90 degrees. So that's why angle AOM which is equal to OBN and OBN plus BOM equals 90 degrees. That's why that angle plus this angle, angle AOM plus BOM equals 90 degrees. That's it. Solve equation. Sinex equals to cosine x. Okay. Let's do it geometrically. It's kind of easier this for me. So this is a unit circle. Now what is a sine? Sine is the ordinate of this angle. What is cosine? Cosine is abscissa of this particular angle. Now if they are equal, it means that two quantities of this triangle are equal to each other. So it's right triangle with equal quantities, which means this angle is supposed to be 45 degrees. So I will change my picture and I put it at 45 degrees P over 4. That's my angle. And in this case my sine and cosine abscissa and ordinate of this angle are equal to each other. Now where else? I mean obviously I have to really do something like this 45 degrees, but actually the opposite sine would be the same. Because in this case both are negative. This would be cosine and this would be sine of this angle. This angle is what? It's 180 plus 45. It's 225 degrees. So it's pi over 4 and pi over 4 plus pi plus 180. So this is 45 plus 180, which is 225. So these are two angles which basically have this particular property. At the same time let's not forget that in trigonometry angles can be greater than 360 degrees. So you can rotate your radius vector more than 360. So basically these two points correspond to a little bit broader number, well infinite number of solutions. So it's pi over 4 plus pi times k, where k can be any integer number. If it's 0, it's this. If it's 1, it's this. If it's 2, it would be p over 4 plus 2 pi, 2 pi 360 degrees. So we'll have again the same point here after one full circle. And then again 180 degrees pi and then again and again. So this is a general solution to this equation. Don't forget this is very important. So it's not really sufficient to find all the solutions to trigonometric equation within one full cycle from 0 to 25, from 0 to 360 degrees. What is important is to figure out what exactly other angles which are greater or less than. So k is basically any integer. It can be negative too. Because we can go backwards. Counterclockwise is positive. Clockwise is negative direction of changing the angle. So k any integer, 0 plus minus 1 plus minus 2, etc. So this is a general solution. Don't forget this particular piece. Very important. And one more similar but slightly more different. Solve the equation. Tangent x equals cotangent x. You remember the definition of tangent is sin over cosine. Which means we are talking only about angles where cosine not equal to 0. Now cotangent is inverse. Cosine divided by sin. Or if you wish 1 over tangent. The same thing. So sin should not be equal to 0. Alright fine. So we have to really just remember that if for whatever reason we will have a solution something like x is equal to let's say pi over 2 90 degrees. That's not a good solution because cosine will be equal to 0 and tangent would not basically exist in this case. Okay so what I will do is the following. First of all I will simplify it. I will use cotangent equals to 1 over tangent. And let's say tangent is z. What do I have? z equals 1 over z. Right? That's my equation now. I will solve it for z. And then knowing what is z I will find my x tangent of which is equal to z. So here what's the solution to this? So I can say z not equal to 0. So solution z is equal to 0 is not good but it's not a solution anyway. So I will multiply by z both cases. Both sides I will have z squared is equal to 1. Which means z is equal to plus or minus 1. So my tangent. So that's my basically simplified equation. Okay let's solve it. Again I will use geometry. So tangent is sine over cosine. So it's this is sine, this is cosine. So if they are equal to 1 that means they are equal to each other. Right? Which means it's again pi over 4. 25 degrees. Where else on this unit circle my sine and cosine would be equal to 1. Again opposite. Both of them are negative but their ratio is exactly the same. So this one also. So this would be my sine and this will be my cosine. They are equal to each other. And sorry. So this angle is also good enough. So it's pi over 4 and pi over 4 plus pi 180. Now minus 1. What would be minus 1? Minus 1 would be here. When lengths would be exactly the same. But in this case the cosine which is obsessive would be negative. And sine which is ordinate would be positive. Then it would be ratio would be minus 1. They are equal in lengths but since one of them is negative the ratio of minus 1. And similar to this. So these four points represent solutions in a unit circle. Now we have to express it trigonometrically with considering that angle can be basically greater than 360 degrees. So this is pi over 4 plus 90 degrees, 90 degrees, 90 degrees, 90 degrees. So it's plus pi over 2 times k where k can be any integer number. So that's a solution to, complete solution to this particular equation. Again don't forget that this represents all the solutions in this format. And you can put like comma k is any integer number something like this. Okay, well that's it basically that's all six problems which I wanted to solve. And again don't forget that it's your attempts to solve these problems which make some benefits actually. I mean it's good that you're listening to me, you're watching this lecture etc. But what's more important is to solve the problems yourself. Even if you don't really come up with a real solution, try it. Even the attempt to solve the problem is beneficial for you. Well that's it for today. Thank you very much and good luck.