 Hi, I'm Zor. Welcome to a new Zor education. Well, let's solve some problems. That's the purpose, actually, of the whole course, so you will train your mind to solve the problems. So, I do recommend you, if you didn't attempt to do these problems just by yourself, pause the video and try it right now. It's very useful to do things just on your own. I'm just helping you in case you cannot come up with a solution. And I do recommend, after this lecture is completed, try again just by yourself. Repeat the proofs of whatever the formulas, derivations are presented in this lecture. Okay, so we're talking about problems on sum of angles, and the first one is I would like to derive the formula for tangent of sum of two angles. Now, I assume that the previous lectures related to sine and cosine of the sum of two angles are clear, and I will use obviously the results of this. And the obvious beginning is the definition of the tangent as cosine, as a sine over cosine. That's one thing which we have to obviously use. Now, I do know the formula for sum of two angles and the cosine of two angles. So let's just use it. Sine alpha, cosine beta, sine beta, cosine alpha divided by cosine alpha, cosine beta minus sine alpha, sine beta. Okay, so what we have to do here to transform it into tangent of the angles, let's divide the numerator and denominator by cosine alpha, cosine beta. What happens is, because if I divide sine alpha, cosine beta, divide by cosine alpha, cosine beta, cosines will be reduced and the sine over cosine will be left, which is by definition tangent of alpha. If I divide this by this, my cosine alpha would be reduced and I will have sine beta over cosine beta, which is by definition tangent of beta. Now, in the denominator, I divide by the same product of cosines. So I will have one here minus sine over cosine and sine over cosine. So that's the product of tangents and that's the formula. This is the formula which expresses tangent of sum of two angles as tangents of the components of the sum. And the derivation is very simple. I just use the definition of the tangent, then expressions for sine and cosine of sum of two angles and the only little trick which I had to do is divide the numerator and denominator by the same cosine times cosine, cosine alpha times cosine beta. Now, obviously, this formula is true wherever alpha, beta and alpha plus beta are defined and you know that with a tangent, it's not always. Now, the tangent is not defined wherever cosine is equal to zero, which is pi over 2 plus pi n, where n is any integer number. So if angle alpha is like this, not equal to this and beta not equal to this alpha plus beta not equal to this, then the formula makes sense. Then the formula has every member defined. That's very important to remember that the formulas for sine and cosine, they are absolute in some sense. They are true for any two angles. The formulas for other functions like tangent, for instance, might not be true for all the angles but some angles tangent or other functions might not be defined at all. Okay, let's move on. Let's analyze this formula a little bit. Let me just repeat it here. So what's interesting is there is an additional restriction here in this formula. You see, it's not only dependent on alpha and beta and alpha plus beta so that their tangents corresponding with are defined. We also dependent on the denominator of this particular fraction not to be equal to zero, which means tangent alpha times tangent beta should not be equal to one, right? Because if it's equal to one, the formula has additional restriction on alpha or beta or alpha plus beta, whatever it is. Now, let's think about this particular thing. And I would like to derive some additional condition on alpha and beta so the denominator not equal to zero. Because again, that's another restriction on the validity of the formula. When is this equal to one? So let's go by definition. Sine alpha over cosine alpha times sine beta over cosine beta is equal to one, right? That's what it means. Now, we are obviously working in this area where tangents are defined. Alpha plus beta not equal to pi over 2 plus pn. So these are, as I mentioned before, necessary for all our tangents to be defined. So we are assuming that these inequalities are held. Now, we are adding this one. Now, obviously, cosine of alpha and cosine of beta are not equal to zero right now because of these restrictions, right? So I can multiply this equation by cosine alpha cosine beta and what I will have is sine alpha sine beta equals cosine alpha cosine beta, right? Or if I subtract sine alpha sine beta from both sides I will have cosine alpha cosine beta minus sine alpha sine beta equals to zero. Now, do you recognize this formula? Well, this is the cosine of alpha plus beta. So my restriction on the denominator here, which is translated into this, which is translated into this and this actually means that the cosine of alpha plus beta should not be equal to zero, which is already something which we have covered. So whenever our tangents, all tangents are defined in exactly the same area, we can say that this is not equal to one. Because if it's equal to one, it's exactly the same as this and this is already taken care of. So the denominator equal to zero does not present any additional restrictions on the end goals. So the end goals are still restricted by these three inequalities. So none of them should be equal to pi over 2 plus pi m. Okay, that completes the analysis of this formula for the tangents of sum of two end goals. Let's move on. Okay, next couple of problems are related to one interesting task. Let's assume you have some kind of equation where you have sine and cosine and tangent and whatever other different trigonometric functions participate. What's the difference between this equation and let's say this one. Sine squared x plus 2 sine x minus 3 equals to zero. Now, this is a trigonometric equation, but you can always say let's assume that we assign y as a sine of x, a new variable in substitution. Well, we will have a quadratic equation for y, right, which we can solve and since we know the value of y using arc sine function, we can always find the x. Now, what if I have a mixture? Let's say this is a cosine. Well, now it's not that simple. You can't really have one particular variable to replace the sine or cosine or whatever to get to an algebraic equation of polynomial, right? So, you cannot easily transfer this into polynomial equation or something which resembles the polynomial equation at least. So, the next few problems are related to how to resolve this. Now, if I will be able to express sine and cosine and any other function of some variable x using one and the same expression, then using that expression and substituting this expression as a new variable, I can reduce it to algebraic equation or polynomial or something like this. So, and here is the rule. Actually sine x and cosine x and tangent x and cotangent x and second and cos second of x. All can be expressed in terms of tangent of a half angle. So, what I will do is I will derive a few formulas how to express, for instance, sine of some angle through tangent of half of this angle. And again, the problem is not to complicate the issue. The problem is to express all different trigonometric functions through the same. So, sine through the tangent of half angle, cos sine through the tangent of half angle, etc. If I will be able to do this, then the equation like I have just written can be rewritten first by replacing sine and cosine with the tangent of half an angle and then using tangent of half an angle as a new variable. Because that reduces the multiplicity of the trigonometric function to a single trigonometric function. Sine and cosine and tangent and all others are all replaced with the tangent of half an angle. Alright, so my first formula is how can I express sine of the angle as in terms of tangent of half an angle. Well, here it is. No doubts about that, right? I split the phi into 5 over 2 plus 5 over 2 and now I will use the formula for sine of some of two angles, right? Which is sine of 5 over 2 cos sine of 5 over 2 plus sine of 5 over 2 times cos sine of 5 over 2 equals to 2 sine equals to 2 sine 5 over 2 cos sine 5 over 2. Okay. We know this. Now, why is this better than this? Well, here is why. Let's multiply and divide by cosine of 5 square. What happens? Well, if I divide, one cosine will be reduced so I will have only one cosine in the denominator and sine in the numerator so I will have 2 sine over cos sine which is tangent 5 over 2 times cos sine square of 5 over 2 equals. Now, if you remember in some elementary trigonometric identities we used to have a formula. One plus tangent square of 5 equals to one plus sine square over cos sine square equals. Now, if you have it as one denominator in the numerator you will have cosine square plus sine square which is one. So, from here you deduce that the cosine square is equal to one over one plus tangent square. So, this is equal to 2 tangent 5 over 2 divided by one plus tangent square 5 over 2. So, that's the expression. Now, we have expressed sine as a function of tangent of the half angle. Now, again, just by itself it would be worse if we don't really need it. It's simpler than this. But now, the next problem is how to express cosine of 5 in terms of tangent again. So, this is cosine of 5 over 2 plus 5 over 2 which is cosine square 5 over 2 minus sine square 5 over 2, right? It's cosine by cosine minus sine by sine. And what I will do, I will do exactly the same. I divide by cosine square like here and multiply by cosine square. Now, if I divide by cosine square, I will have one minus tangent square. If I multiply by cosine square, that's the same thing as divided by one over one plus tangent square. Same thing as before. So, I was using the formula. The formula cosine square equals the jacket just derived. All right. So, now we have both sine expressed as a tangent of the half angle and cosine expressed as the function of the tangent of half an angle. Now, what's the advantage of this? I mean, this advantage is obvious. This is much more complex, right? The advantage is that both sine and cosine are expressed in terms of tangent of the same angle, right? So, if I would like to convert a trigonometric expression which involves both sine and cosine, for instance, sine x plus cosine x equals to 1. How to resolve this particular equation? Well, it's not obvious, at least not immediately obvious. But if you use these formulas which transfer both sine and cosine into the tangent of the half an angle, then you will have two tangent x over 2 over one plus tangent square x over 2 plus one minus tangent square of x over 2 one plus tangent square x over 2 equals to 1. Now, you multiply both sides by one plus tangent of x over 2 and you have minus tangent square x over 2 plus two tangent x over 2 plus one. I have changed the order, so I will have second degree, first degree, and zero's degree of the tangent equals to 1 plus tangent square x over 2. Now, this can be reduced. And then we will add this and subtract this to this side and we will have two tangent square x over 2 minus two tangent x over 2 equals to one. Sorry, equals to zero. We can reduce it by two. And basically now it's very easy to resolve it because, well, in this particular case, it's tangent of x over 2 times one minus tangent of x over 2 equals to zero. So either this is equal to zero and that's when sine of x over 2 is equal to zero. Tangent is equal to zero when the sine is equal to zero. So the angle is pi n. This is for x over 2. And for this case, x over 2 equals to, when the tangent is equal to one, that's 45 degrees, right? This is 45 degrees and the period is pi n. So it's pi over 4 plus pi n. Now, considering this is x over 2, the solution for x would be 2 pi n and pi over 2 plus 2 pi n. So on the unit circle, it's this angle and this angle, right? This is zero plus pi n, 2 pi n, and this is pi over 2 plus pi n. In this case, sine is equal to zero and cosine is equal to one and the sine is equal to one. In this case, sine is equal to one and cosine is equal to zero and the sine is equal to one. So, you see, that's how we have resolved these particular things. That's how you can solve an equation which contains both sine and cosine or some other trigonometric functions because all these trigonometric functions can be expressed in terms of one, tangent of the half an angle. Now, as a continuation of this problem, let's express, so we've done sine and cosine in terms of tangent of the half an angle. Let's do tangent. What if this equation contains not only sine and cosine but also tangent? We have to be able to express any trigonometric function in terms of tangent of the half an angle, right? So, this is by definition this. In terms of tangent of the half angle, this would be two tangent phi over two divided by one over tangent square. Phi over two, that's the sine. A cosine would be one minus tangent square phi over two divided by, so it will go to the top. Now, this is reduced, so we will have two tangent phi over two divided by one minus tangent phi square. So, that's the answer for tangent. Now, for cotangent, this is reverse, so this is reverse, right? Because cotangent is one over tangent. Now, for secant, that's one over cosine, right? So, you use the formula for a cosine, which is one minus tangent square divided by one plus tangent square and you just invert it. So, it will be one plus tangent divided by one minus tangent. And cotangent is one over sine correspondingly, you know what I mean. Okay, fine. So, that's all for these formulas. What I would like actually to point out that you don't have to remember all these formulas. These formulas are not for, I mean, the whole purpose of this lecture was not actually to convey you the formula which you have to remember, but to talk about how to derive these formulas and to encourage you to try to do it yourself. So, again, let me repeat my recommendation after this lecture is completed, try to do exactly the same just by yourself, derive these formulas yourself. Now, the last problem which I have is what is a sine of 15 degrees? Now, there are a few problems and the next one will be much more interesting. It will be 18, 18 degrees. Now, 15 degrees is really very easy. Why? Because it's half of the 13. And 30 is a nice basic angle. We know sine and cosine and all this. So, if we know this sine and cosine for 30 degrees, how can we derive the value for 15 degrees? Well, think about it this way. Remember cosine of 15 plus 15 degrees, which is cosine of 30, which we know, actually. Well, this is what? Cosine squared 15 minus sine squared of 15 degrees, right? This is the cosine of the sum of two angles. In this case, angles are the same. Or, if you wish, you can replace cosine with cosine squared as 1 minus sine squared. So, it will be 1 minus 2 sine squared 15 degrees. So, here is your equation. 2 sine squared 15 degrees equals to 1 minus, square root of 3 over 2, or 2 minus square root of 3 over 2. So, sine 15 is equal to 2 minus square root of 3 by 4. So, that's how easy we can derive something like sine of 15 degrees. Well, that completes this particular lecture. I think trigonometry presents a lot of different challenges and interesting problems, actually, which I will try to convey as much as possible in terms of equations, identities, conditional equations, conditional identities, et cetera. So, there will be many problems, and I think that's exactly why we have this course. So, thanks very much and good luck.