 Hi, I'm Zor. Welcome to Unisor Education. I would like to present to you just one problem. The whole lecture will be about one particular problem. It's not an easy one. I would say it's challenging. It's very important for you to try it yourself first. So go to the notes with this lecture on Unisor.com. And this is about evaluating sine of 18 degrees. So we have an angle of 18 degrees. And the problem is find out the value of the sine in radicals actually. It's like sine of 30 degrees is square root of 3 over 2. So sine of 18 degrees is something which we can express in some good form. All right, so how can we approach this? Well, first of all, since I present this problem, it means it has a solution. And it's a solution which is not like going to a computer and find out what's the sine of 18 degrees in zero point, whatever it is. Or going to some internet site and find out. So I would like to derive the value of this particular sine as a logical conclusion of certain statements which I'm about to make. So how can we approach this problem? Well, we obviously need to do some creative thinking. So what's so special about 18 degrees? Well, what's special is that 5 times 18 is 90. So it's kind of a 90 degrees, which is a very kind of a nice basic angle, divided by 5. Well, we know about the half-ambles, right? We can find out where the half-ambles or double-ambles. So we can probably manipulate this in some way or another. So the first thing which is, I would say, well, a guess or good creative approach to this is to realize that 18 plus 4 times 18 is 90. So as I said, 90 is 5 times 18. So this is definitely the true equality. Now, if this is an angle and this is an angle, these are two angles which together add up to 90 degrees, which means sin of 1 is equal to cosine of another. This is one of the fundamental trigonometric identities. So we are going to use it, which means sin of 18 degrees equals to cosine of 4 times 18 degrees. Or, if you wish, cosine of 18 degrees equals to sin of 4 times 18 degrees. So both are true statements. And I'm going to use one of these statements by basically applying the formula for sin of two angles to convert this into basically sin or cosine of 18 degrees. I can do that, right? So let's apply twice the formula for sin of some. Now, you remember that, let me put it here, sin of alpha plus beta is sin alpha cosine beta plus cosine alpha sin beta. Now, if alpha and beta are exactly the same thing, so it's basically a sin of 2 alpha, then I can really put it this way. 2 sin and cosine of alpha. And this is 2 alpha. So that's what I'm going to use with alpha equals to 18. So on the left, I will still have cosine of 18 equals, and this one is double 36, right? So this 4 I'm basically dividing as 2 and 2. So it's sin of 2 36 degrees equals. Using this formula, it's cosine 18 degrees plus equals to 2 sin of 36, which is 2 times 18, and cosine of 2 times 18. So we reduced 4 times 18 into this. Now let's reduce it more. So the cosine of 18 degrees equals. Now, this again is a sin of 2 of double angle. So it's 2 sin of 18 degrees cosine of 18 degrees, right? Now cosine of double angle, okay, let me put it here again. Cosine of double angle is, this is the alpha plus alpha, so it's cosine times cosine minus sin times sin. So it's cosine square alpha minus sin square alpha, which is the same thing. I would like to have only sines, and since cosine square is 1 minus sin square, I can put it equals to 1 minus sin square of alpha. So instead of cosine of 218, I will have 1 minus 2 sin square of 18 degrees. Now, look at this. 18 cosine of 18 degrees I can reduce, right? Both sides of equation I can divide by the same number, obviously not equal to 0. And now I have an equation, basically. 1 equals, this is 4, sin times 1. So it's sin of 18 degrees minus 4 times 2 8 sin and sin square. It's sin in third degree of 18 degrees. So I've got an equation, actually. If I will be able to solve this equation, I'll find my sin of 18 degrees, right? Well, a small complication, this is an equation of the third degree, the cubic equation. But you can deal with it. So let's rewrite this equation in a little bit more typical format. We will add 8 sin cube and subtract 4. So I will have 8 sin cube of 18 degrees minus 4 sin of 18 degrees plus 1 equals to 0. And this is the cubic equation which I have to solve to get the sin of 18 degrees. Now, let me introduce a little shortcut and then I will talk about why I did it. If you go to any tax book or internet site which contains something like a derivation of sin of 18 degrees, you will basically see the following. Let's rewrite this in a more algebraic form. 8x cube minus 4x plus 1 equals to 0, where x is sin of 18 degrees, which we would like to find out, right? So what they are saying is, they're saying 8x cube minus 4x plus 1 equals 2x minus 1 times 4x square minus plus 2x minus 1. Now, don't ask how they came with this particular equality. Obviously, if you perform the multiplication, you will get this. 2x times 4x square is 8x cube. 2x times 2x is 4x square, but minus 1 and 4x square gives minus 4x square, so there is no x square. Now, 4x would be, so this is minus 2x and minus 2x, so it's minus 4x and minus 1 times minus 1 gives you plus 1. So this is true equality, very easy to check. I just multiply directly. Now, if you have an equation like this and the left side is represented as the product of these two, then obviously either this one is equal to 0 or this one is equal to 0, right? That's the only way how the product of two numbers is equal to 0, either one of them is 0 or another. Now, if this is equal to 0, we will get x is equal to 1 half. Now, sine of 18 degrees cannot be 1 half. 1 half is a sine of 30 degrees, if you remember, right? Sine of 30 is 1 half. So this is definitely smaller because sine is monotonically increasing function. So this does not give you the solution which you're looking for. However, this quadratic equation does. So by solving this quadratic equation, you can get the value of 8 minus 2 plus minus square root 4 plus 16, which is equal to... Now, minus is definitely not supposed to be considered because it gives the negative result. Now, this is definitely the positive thing. So the only right solution is x is equal to minus 2 plus square root of 20. 20 is 4 times 5, so I will do this. That's the square root of 20, right? And divided by 8, which is equal to minus 1 plus square root of 5 over 4. And that's the solution, that's the sine of 18 degrees. Now, well, we've got this solution, right? But I had this shortcut magically appeared in front of me. So the rest of this lecture, I will actually devote to how I came to this particular... Now, how I can come to this particular derivation, because the way how I came, I just basically look at it somewhere in the internet or something like this. However, if you don't have any idea about how to really derive it yourself, it would be a very interesting lesson which I'm trying to present right now. So we have defined actually, we have determined what's the sine. So basically the problem is solved. So what I'm talking right now is a purely algebraic part of this trigonometric problem. And algebraic part means how can I represent this in this format? Okay, so let's scratch all sinuses. Sines, not sinuses, sines. Everything related to trigonometry is finished. Now we are talking about purely algebraic problem. How to represent the polynomial of the third degree as the product of these ones? How can I find it? Now, here is my first logical consideration. Well, obviously any polynomial like this, if it's equal to zero, you can find solutions using some again known formula like cardano, which was like very, very long time ago invented by this particular mathematician by the name cardano. So you can definitely solve it. Now, another theorem which was from the course on algebraic equations was that if you have an equation which is the polynomial of n's degree equal to zero and you know the one particular solution, then you can represent this polynomial of n's degree, this particular equation. And you know that x is equal to c is a solution. Then you can represent it as x minus c times polynomial. I'll use different letters, b, zero, x, n minus first degree. So if you don't remember this, you can always refer to one of the lectures which is about these particular equations where I explain why this is true. So you can always divide the polynomial by x minus the solution of this equation. So I know that the solution exists. This is the polynomial of the third degree. Therefore, it must be represented in this format. Now, since I have presented this problem to you, I would say good guess would be that this representation which has integer coefficients on the left must have integer coefficients on the right. Now, if I'm lucky and I hope I am lucky and this particular representation does exist where all the coefficients are integer numbers, then I will try to find it using exactly this consideration that these are all integers. Because most likely, if it's something much more complicated than this, this problem would not be presented to you. I present the problem which has a nice solution, so if it's not really as nice as this, I would probably not even talk about this. So you can consider it some kind of very human quality of these problems. Problems must be human, right? So the human side of this problem is that I can always look for this representation as representation with integer coefficients. All right, so let's assume that I don't know this and I'm looking for this where A, B, C, D and E are integer coefficients. This is my approach to find this particular representation. Now, what can I say from this representation? Well, let's just think about it. What's the coefficient at X cubed AC, right? So AC is equal to 8, right? What's the coefficient at X squared? It's BC plus AD. BC plus AD and X squared here is equal to zero, right? Doesn't exist. Now, what's the coefficient at X to the first degree? It's A times E plus B times D and this is equal to minus 4. And what's the coefficient at X to the zero's degree which is a constant? It's BE. Now, this is a system of one, two, three, four equations with one, two, three, four, five variables. Well, you can say that, hey, the number of equations must be actually equal to the number of variables, otherwise most likely we will not be able to find it. Well, true. However, we do have one extra consideration. I said that I'm looking for integer solutions. That greatly helps to resolve even the system of four equations with five unknowns. And here's how. Most important is, the last one, you see, BE is equal to one. Now, B and E are integers, right? So how can the product of two integers be equal to one? Well, we have two cases. Case number one. B equals one and E equals to one. And case two, B equals to minus one and E equals to minus one. Which already gives us two from one equation, we already got two unknowns. Well, granted, we have two different cases which we are going to consider as two different cases. But in each case, we already reduced the system of four equations with five variables to the system of three equations with three variables, right? Because from one equation, we got two variables, alright? It helps, right? Okay, so let's consider the first case. B and E both are equal to one. So what do we have? Here, they do not participate. So this remains as is. B is one. So it's C plus AG is zero. Now, B and E are one, so it's A plus G equals to minus four. Okay. This is my system of three equations with three unknowns. Well, first of all, it's much easier, right? Now, what can we do about this? Well, the easiest way is probably to resolve it for D and substitute into this. Thus, reducing the system of three equations with three unknowns to two equations with two unknowns. So D from this, D is equal to minus four, minus eight, right? So we substitute to this and we will have AC is equal to eight. C plus A times D, which is minus four, minus eight, is equal to zero. Well, let's simplify it a little bit. So AC is equal to eight and C is equal to, well, it's four plus A times A. So it's four A plus A squared. I think I'm right. I think so. Now, we can substitute this to this times A. So what do we have? We have A cubed plus four A squared minus eight equals zero. We have another cubic equation, right? Well, but here is a very important problem. The original cubic equation had, God knows what kind of solutions. I mean, we do know what kind of solutions. It's not integers. Here we are looking for integer solutions. And that makes our job much easier. Because if you can remember, integer solutions must be among factors of the three coefficients, right? So factors are basically one plus minus one, plus minus two, plus minus four. And basically, well, we can try plus minus eight, but that probably won't work. Now, plus minus eight doesn't really make any sense even to check because it's too big a number. Eight cubed is, what, 512? Plus sign or minus sign doesn't matter. We will never be able with four times 64 to bring it to zero. So this is definitely not a solution. Now, one and minus one, we can very easy to check. One means it's one plus four minus four doesn't equal to zero. Minus one, this is minus one, plus four minus eight also. So that's out. Now, plus two, eight times 16, no good. Minus two, minus eight, plus 16, minus eight. Okay, minus two is root. Four. Four cubed is four times four times 64. 64 plus, well, another 64, no good. Minus four. So it's minus 64, plus 64, also no good. So, and plus, no. So the only solution is minus two. So we've found an integer solution just by checking what's the factors of the three coefficients. So we know that minus two is a solution. Fine, we got it from this. So we know this is minus two. Unfortunately, I wiped out these little equations, so I will repeat it again. AC is equal to eight. X square is BC plus AG equals to zero. X to the first degree is AE plus BG equals to minus four. And BE is equal to one. All right. So we are considering this case. And we found out that A is equal to minus two. Now, this is one. This is one. This is one. So this is minus two. So G is also minus two. So this is minus two. G is also minus two. So A times G is four. So C is minus four, right? Now, this is one, and this is one. That's it. Now, let's check. Minus two X plus one times minus four X square minus two X plus one. So if you will multiply, let's think about it again, minus two times minus four gives me a coefficient with XQ, which is minus two times minus four is eight. Correct. Now, X square. X square would be three member by this. So it's minus four X square. And AG is plus four. Zero out. So there is no X square. Good. Now, X is AX times E, which is minus two X. And B times GX is also minus two. So it's minus four X, OK? And one by one is one. So this is a right representation. Now, if you remember the representation, which I gave in the first was two X minus one times four X square plus two X minus one. Now, this is actually exactly the same as this one, except these coefficients are multiplied by minus one and these coefficients are multiplied by minus one. But since it's product, minus and minus gives you plus. So it's exactly the same as I have here. Now, if I will use minus one and minus one as B and E, I will have a slightly different system, which will result in basically very, very look-alike cubicle equation for A, in which case we will have the coefficients with an opposite sign. So this would be two. This would be minus one. This would be four. This will be plus two. And this will be minus one, which is exactly, as I said in the beginning. So we don't really have different representations. It's exactly the same representation of the one which I told you, which has, so either it's this or both multipliers are multiplied by minus one, which is exactly the same thing. And therefore, I have actually came with this particular representation, which represents our cubicle equation. And now, solving this particular cubicle equation, using these representations very easy. I said it before. This gives me x is equal to 1 half, which is not good. And the second one, the quadratic equation, gives one of the solutions is positive, and that's exactly what sign of 18 degrees is. Well, I shouldn't say it's an easy way. But I think it's interesting, and I think it's challenging. And let's not forget that some people who really came up with the problems like that, there were some people who really were smart enough. They realized that you can use certain known formulas and approaches to solve this problem. And the purpose why I'm dedicating this particular lecture to this one problem is precisely because I would like you to get into the mind of people who were trying to solve the problem. Now, they didn't know how to get sign of 18 degrees, but they realized that four times 18 plus 18 gives you 90 degrees. So they were using this particular peculiarity of 18 degrees. Then, after that, they used the known formulas of double angle, and they came up with the cubic equation, the one which involves this particular polynomial of the third degree, which obviously they didn't know how to solve. But they assumed that maybe, just maybe, it has the representation like this with integer coefficients, and they wanted to basically find out, well, let's try. So they tried, they did exactly the same way as I did, or some other way. I don't really know how they did it. This is something which I came up with. But in any case, you have to have this inquisitive mind. You don't know how to solve this problem. You're just trying to create this solution in your head. Try this way, try that way. I'm not presenting you anything which you can, you know, use in the future. But you will go through problems like this and the solutions many, many times, your mind will be tuned to seeking the solution using one of the existing approaches. Well, that's it. Try to do it yourself again. That's my general recommendation. And good luck.