 Hi, I'm Zor. Welcome to Unizor Education. I would like to continue solving problems related to usage of trigonometry in geometry. So it's geometrical problems, but a big part of it is supposed to be using the trigonometry. There are many, many problems, so I'm sure I will not cover everything. However, I'm trying to put together some material, some educational material, which should be relatively universal as far as different aspects of trigonometry is concerned. So here are these problems. If you did not spend any time solving them yourself just using the Unizor.com, this is the problem number, problem, set of problems number one in the trigonometry geometry chapter. So if you didn't do anything just by yourself, I do recommend you just to pause this particular digit and spend time trying to solve these problems yourself. They're very helpful, educational, they develop your creativity, your intelligence, so it's definitely a must. Because if you don't do anything yourself, if you just consume information which I'm presenting, then the usefulness of these lessons is really limited. Anyway, so let's just solve all these problems. Okay, here it is. Okay, problem number one. Prove that in any triangle A, B, sine square of gamma over 2 equals to minus A, P minus B. Now, I'm using the symbolics which are introduced in the previous lecture, and basically A, B and C are sides of the triangle, which lie opposite to vertices which are similarly called by using the capital letters. Now the angles are alpha, beta and gamma, again, alphabetically, and lower case P is half the perimeter. So that's the abbreviation, that's the symbolics which I'm using here, and I will try to use it more or less universally. So we have to prove that in any triangle this is true. Well, what am I thinking about the approach to solve this particular problem? First of all, nobody likes something in square if there is some way to bring down the power to one. And there is a way, actually. Remember cosine of 2 phi equals to cosine square phi minus sine square phi. Or if we express cosine square as 1 minus sine square, it will be 1 minus 2 sine square phi from which I can solve sine square phi is equal to 1 minus cosine 2 phi over 2, right? So I'm going to use this formula to convert sine square of gamma over 2. Now phi is gamma over 2, so it's 1 minus cosine of double angle which is gamma over 2. That's something which looks much simpler, right? So I will substitute it over there and see what happens. So 1 minus cosine gamma over 2 instead of this. So what happens is A B times 1 minus cosine gamma over 2 equals. Now I will express the capital of the perimeter with whatever the definition actually is. So P minus A would be minus A which means minus 2A would be minus A plus 2. So it would be minus A plus B plus C over 2. Now if it's P minus B, it would be A minus B plus C over 2, all right? If it's minus B and then using the counter-denominator 2, it would be minus 2B, so it's right, exactly. All right, now, so what we can do is we can multiply everything by 4, right? So we will get rid of the denominator. So on the top I will have, so if I multiply by 4 and this is 2, so I still have 2 here, so it's 2AB times 1 minus cosine. So I will multiply it separately, gamma equals. Now this, I don't have any denominators anymore, right? So what is this? It's C, okay, it's C plus A minus B, which is this, and C minus A minus B, which is this, right? This is C plus A plus B and this is C minus A plus B. Why did I do it? Because it's easier for me. Right now it's, remember, X minus Y, X plus Y, it's X square minus Y square, right? So X is C and Y is A minus B. So my right part 2AB minus 2AB cosine gamma equals, my right part is C square minus A minus B square. Or I can open up this square and I will have what? A square minus 2AB plus B square, right? And everything with a minus sign. So I will have minus A square, now plus 2AB and minus B square, right? So what's left? Well 2AB is immediately out and I have, if I will bring A square and B square to the left by adding this to both sides, I will have A square plus B square minus 2AB cosine gamma equals C square. Now what is this? Well you have to recognize this, this is a law of cosine, right? This is the law of cosines. And I did present this law in one of the previous lectures about trigonometric identities. So this is a true statement. So from this, using these transformations, we came to this. This is true. Does it mean that this is true? Well not exactly. What we have to do really is to say this very important phrase. Since all transformations are reversible, they are all invariant transformations. We didn't lose anything, we didn't add anything. This is all invariant and reversible transformations. So from here I can go to here, to here, to here and here. And that's the proof because from true statement we came to this one. And that's the end of the proof. Okay, next. Actually it's very similar. AB times cosine square of gamma over 2 equals P, P minus C. Where again A and B and C are sides. P is half a perimeter and gamma is an angle which is opposite to C side. Now again, I will do exactly the same in the beginning. I don't like the cosine square of gamma over 2. I would like to convert it into the first power of something. So how can it be done? Well again, let's remember cosine of 2 phi is equal to cosine square of phi minus sine square of phi, right? That's the only thing we should remember. Everything else is derived. Now I will express sine as 1 minus cosine. So what I will have is minus, so it will be cosine square phi minus 1 plus cosine square phi. Which is 2 cosine square phi minus 1 from which we conclude that cosine square phi is equal to 1 plus cosine 2 phi over 2, right? And that's what I'm going to use here. So phi would be gamma over 2, 2 phi would be gamma. So my next formula is this. AB times 1 plus cosine gamma over 2 equals, and again from the formula for P, P is A plus B plus C over 2. And if I will subtract C from this, it would be A plus B minus C over 2, right? So I have to prove that. Again multiply by 4. So this would be 2. So it's 2AB plus 2AB cosine gamma. And here it's A plus B plus C. So it's A plus B plus C and A plus B minus C, right? So it's A plus B square minus C square, right? It's A plus B square minus C square. So 2AB plus 2AB cosine gamma equals A square plus 2AB plus B square minus C square. 2AB goes out. I add C square to this and I subtract 2AB cosine to there. And I will get C square equals 2AB square plus B square minus 2AB cosine gamma, which is exactly the same law of cosine. And this is a true statement for any triangle. And from this we can reverse all these transformations to get to this and that's the end of the proof. So in both cases we were using some simple transformation, actually two different transformations. We expressed sine square or cosine square of half of gamma with the cosine of gamma. That's number one. So we were using basically the formula for cosine of double angle. And the second transformation was rule of cosines, right? Okay, next. We are going to prove the following formula. S, which is the area of a triangle, equals to one fourth of A square cotangent alpha plus B square cotangent beta plus C square cotangent gamma. We have to prove this formula. Well, we have many different ways actually to express the area, right? It's, let's say, a side times an altitude which falls onto this side divided by two. Or we had some formula which I proved in the previous lecture about using the radius of the circumscribed circle. Another formula for area of triangle was the Heron's formula, which I have derived long, long time ago in the geometry chapter, which was dedicated to the area of triangle. So this is yet another formula for the area of triangle. So we express the area in terms of sides and angles. Okay. Let's prove it. Let's circumscribe the circle around our triangle, okay? Now, this is a perpendicular from the center onto this side. And this is the radius. Now, this is R. Now, this is B. So this is B over 2 and this is B over 2. Right? You remember that the perpendicular in the socialist triangle to the base is actually not only the altitude but also the median and the bisector of the angle. Now, the AOC angle is a central angle supported by this arc. ABC, which is beta, is an inscribed angle supported by the same arc. And we know that the central angle twice as big as the corresponding inscribed angle if they are supported by the same arc. And this is half of the central angle. So this is also beta and this is beta. Now, let's consider this particular triangle. Let's say this is triangle AOG. What do we know about this triangle? This is a right triangle. This is AG is one catheter and OG is another catheter. Now, AG is equal to B over 2. OG is equal to, well, think about this way. If you divide OG by AG, this is a cotangent of beta of this angle. From which we can conclude that OG is equal to AD, which is B over 2, times cotangent beta. So the area of this triangle, the area of triangle AOG is equal to one catheter, which is AD, B2. Times another catheter, which is B2 over 4 and cotangent beta. Now, exactly the same way, if we draw this, we can have the area of AOG, triangle AOG is equal to, this is C and this is A, right? So this would be C2 over 4 cotangent gamma and area of BOC is equal to A2 over 4 cotangent alpha. And obviously, if you add all these together, you will get this formula. So what did we use here? Well, we used a couple of geometric properties, like for instance, that inscribed angle is half of the corresponding central angle if they are supported by the same arc. Then we were using the property of a socialist triangle AOC, and by the way, it's A socialist because A is always a radius and OC is a radius, right? Always a center. So in this A socialist triangle, the altitude is the same as median and angle bisector. And that's what actually made triangle AOG simultaneously, the right triangle and the one which has this angle equals to beta. Everything else is basically the definition of a cotangent and area. Everything else is simple. And I have the last problem. Prove that R is a radius of circumscribed circle times hA times hB times hC. These are three altitudes corresponding with two words, side A, B and C, equals to A squared B squared C squared. So we have to prove that this is true for any triangle. Now, let me draw exactly the same picture as before. Now, this is angle beta and this, as we have concluded, is angle beta as well. Now, this is its opposite to B, so this is B over 2 and this is B over 2. And this is R. So R is equal to, well, let's think about B over 2 divided by R is a sign of beta, right? So R is equal to B over 2 divided by sign of beta, right? Which is exactly similar to, if I will use this or this, I will have R from this particular triangle equal to C over 2 sign of gamma and in this particular triangle, very similar. So this is alpha, now this is alpha and this is gamma, equal to A over 2 sign of alpha, right? So this is simple. This is again from this drawing, it's kind of obvious and we were using this many times before. Now, using this, instead of R cube, the first R from the R cube I will replace with this, the second with this and the third with this. So what will I have? I will have A, B, C divided by sign A, alpha, sign beta and sign gamma. 2 by 2 by 2, it will be 8 and this 8, it will be 1, it will be reduced times HA times HB times HC. So that's what I will have on the left. So instead of R cube, I put this is R, this is another R and this is the third R. This will give me ABC on the top and the product of signs on the bottom and the 2 by 2 by 2 and this 8 will be reduced to 1 and I have to prove this. War, this. War, this, right? For everything is equivalent, reversible, so instead of proving this, I will prove this. Now, let me just draw another picture where it would become kind of obvious. If you have a triangle, C, this is HA, now this is gamma, alpha, beta. Now HA, HA, this is angle beta. So if I divide HA by C, I will have a sign of beta, right? This is an opposite calculus. So HA is equal to C times sine, wait, okay. How about HB? HB is A times sine gamma, right? HB divided by A is a sine of gamma. So HB is equal to A times sine of gamma. Great. And the last one, this is HCB times sine of alpha, right? Because HC divided by B would be the sine of gamma. So it's B sine of gamma, sine of alpha. And if you multiply these three, you will get this. I'm restoring. Well, these are three relatively simple problems. And what I recommend you to do is just by yourself go through all these four problems and try to do it yourself again after you have listened this lecture, just to make sure that you remember everything and it's fresh, etc. And by all means, don't hesitate to sign in as a student to unizord.com, have somebody else or maybe yourself under a different name, sign in as a supervisor and make sure that you are enrolled in some course and then you can take exams, you can take as many times as you want. Everything is completely free and no checking is done. So just verify whether you are mastering this material. So thanks very much. That's it for today. Good luck.