 Hi, I'm Zor. Welcome to a user education. Today we will talk about trigonometry, primarily about trigonometric identities. A couple of them, and simple actually, but still very necessary. I believe if you are basically trying to study mathematics in general and trigonometry in particular. Now, this particular lecture is part of the course called Mass Plus and Problems. This is the continuation of the course called Mass for Teens, and it's all presented on Unisor.com. All is completely free, and I even came up with some kind of a slogan recently. We don't sell knowledge, we share it. Okay, so, going back to trigonometry. What's interesting about trigonometry is, one of the interesting things, is its relationship to complex numbers. Remember how we defined trigonometric functions? Well, I would say main trigonometric functions, sine and cosine. We had some kind of a unit circle on the plane, and so the radius is equal to 1. Now, any point has basically an obsessive and ordinal. So, if this is angle phi, then this particular piece, which is obsessive, which is x, was cosine of phi, and y was sine of phi. Now, at the same time, complex numbers, which have real and imaginary part, where i is i squared is equal to minus 1, imaginary unit, is also represented as a point on the plane where a is obsessive and b is ordinate. So, there is some kind of a relationship. And this relationship is perfectly represented by Euler's formula, which is e to the power i phi is equal to cosine phi plus i sine phi. Now, this formula is very interesting in one particular aspect. We had no idea what is complex or imaginary in this particular case, exponent of number E. And basically this is a definition of the operation of raising into imaginary exponent. At the same time, it's a very convenient way to represent complex numbers to represent sine and cosine. And I'm going to use this formula to prove a very simple elementary trigonometric identity, which is a sine and cosine of a sum of two angles. I think the similar problem was presented in the main course, Math for Teens. But I would like to repeat it again, because it's very important to basically feel how complex numbers intimately related to trigonometric functions. So, using this Euler formula, I'm going to show, not to prove, but to show that sine of x plus y is equal to sine of x times cosine of y plus cosine of x times sine of y. And cosine of x plus y is equal to cosine x cosine y minus sine x sine y. So, why did I say that I'm going to show and not to prove? Well, because actually this is a definition and the way how we continued to define the exponent, in case exponent is a complex number, was actually trying to find out, what happens if I will do e times i phi plus psi, for example. And then I was using actually these formulas to prove that this makes sense and it's equal to times e i psi, which is basically the characteristic property of raising into some power. So, using this type of law or identities of cosine and sine, we have proven that this makes sense, which means it makes sense to define the raising into some complex power exactly in a similar way to this one. So, that's why when I'm trying to say that I would like to illustrate actually, that this is actually fine if I'm using Euler's formula as given. So, that's why I don't want to say it's proof because I would have some kind of a logical loop, but it's an illustration. Okay, if it's an illustration, so let's just do the following. Now, e to the power i x plus y, considering this is actually some kind of function which satisfies all the regular rules of raising into some power, it's equal to e i x plus plus i y is equal to e i x times e i y equals. And now I'm using the Euler's formula for this and for this. So, e to the power i x is cosine x plus i sine x and e to the power of i i i y is cosine y plus i sine y equals. Well, let's just multiply. Now, cosine times cosine. This is cosine x, cosine y. Now, then I will use this times this. Now, this is the sine times sine and i square. i square, as we know, i is an imaginary unit, so i square is minus one. That's why I put minus. Sines x, sine y. I put the rest, plus. Now, multiply this times this. It will be i times sine x cosine y plus. This times this would be i. I will put i around out of the parentheses. And that would be sine cosine x, cosine x and y, cosine x, sine y. So, what do we have here? On one time, on one hand, this is cosine x plus y plus i sine x plus y. Again, Euler's formula. If this is the exponent, then it goes here. This is Euler's formula. And then I use the properties of exponential manipulation and convert it into this. So, I have two complex numbers. Real part should be equal to real part. And imaginary part should be equal to imaginary part. And here is both formulas. Cosine of x plus y is this, which is this. And sine of x plus y is this, which is exactly this. So, again, it's not a proof, but it's a good illustration. And if you have forgotten something like this, if you forgot this formula, then perfectly valid to use this type of derivation, which is, again, it's illustration, not the proof. But it's a good way to, if you forgot something, to come up with this formula very easily. One, two, three, four, four lines of manipulation and you got it. Because the original definition of x plus y and sine of x plus y and cosine and how it's represented using sine and cosine of x and y separately, the geometric proof of these formulas is really a little bit cumbersome and you might not really remember it, which is fine. But this, you don't really have to remember much, except one and only one Euler's formula, which connects this, this to this, or this to this. So, Euler's formula is very important for trigonometry and it allows actually to come up with many other formulas relatively easily. So, that was my first problem with trigonometry data integers. Sine and cosine of sum of two angles. Okay, next. Next on my agenda is tangent and cotangent of sum. Okay, now this is actually easy and this is a plain manipulation with no facts. So, this is equal to, by definition of what is tangent. Tangent is sine divided by cosine. Obviously, in all those points where cosine is not equal to zero. Now, now I'm just using the formulas which I have derived just a second ago. So, it's sine x cosine y plus cosine x sine y divided by cosine x cosine y minus sine x sine y. Okay, sorry. Oh yeah, that's correct, right. Okay, now how can I get from this back to tangent? I would like to use tangent for tangent. So, tangent of sum of two angles, I would like to express as tangent of some function of tangent of these two angles. Okay, how can I do it? Well, one of the things is let's divide everything by cosine x and cosine y. What happens? Both, numerator and denominator. Well, this denominator is easy because if you divide this by this, you will have one. If you will have this, divide by this, you will have sine x divided by cosine x, sine y divided by cosine. So, that would be basically a tangent x times tangent y, right? Sine divided by cosine and sine divided by cosine y. Now, in the numerator, let's just think about it. If I will divide this by this, cosine will cancel out. I will have sine divided by cosine of x. So, that's tangent x. Plus, if I divide this by this, my cosine will cancel out. And I will have sine y and cosine y. So, that will be tangent y. So, here is your formula. Tangent of sum of two angles is this. Simple, right? Alright, let's go further. Now, next is kind of a useful thing in case you have some kind of equation. If you have an equation, simple one, something like sine x plus 5 cosine y, no, x. Let's talk about equation. Equals, I don't know, 1, whatever. Doesn't really matter. It's not easy to come up with solution to this particular equation because this is sine and this is cosine. You cannot operate with two different things, basically. You would prefer to have something, one unit, one function, if you wish, of x. So, here is a very interesting property. There is something which is called tangent of half an angle. Apparently, any other trigonometric function can be represented as function of tangent of half an angle, represented in algebraic sense. And I would like to demonstrate it right now. Okay, so I will just represent sine and cosine, but basically in my notes, I have a little bit more detailed. I have tangent and cotangent of some angle also represented with tangent of half an angle. Because again, you can continue this particular complexity and put something else here. Tangent of x equals 1 or something like this. So, it's different things and with different things, it's difficult to operate. But if each one of these can be represented as an algebraic function of tangent of half an angle, which is true, which can be done, then you will basically have an equation of tangent of half an angle. So, you can resolve that particular equation. And then, since you know tangent of x divided by 2, you can find out x divided by 2 itself as arc tangent, for example, or something like this. So, it's very useful in solving some kind of a mixed equation where different trigonometric functions participate. And again, I will use only, I think for sine and cosine, the rest of the expression is in my notes for this lecture. And by the way, I didn't really say that every lecture on Unisor.com has video part and notes. Notes are like a textbook basically. So, I put something more and trivial things I will just use only for the notes and I will not present so they don't really waste much time on this. It's all analogous. So, what I will do is I will express sine and cosine in terms of tangent of half an angle. Here's how. Okay, let's start with the sine. Sine x. That's the beginning. I mean, obviously I can represent x as sum of two halves, which is equal to... Now, let's use the function sine of sum of two angles. So, it would be sine of x divided by two times cosine of x divided by two plus cosine times sine, so I can easily put it as two. Okay? Fine. Now, I'm going to convert it into tangents, right? So, I will divide it and multiply by cosine square. So, it would be two. If I will divide by cosine square, I will have sine divided by cosine, which is tangent of x divided by two times cosine square of x divided by two. Now, here's very important things. What is one plus tangent square of phi? It's one plus sine square of phi divided by cosine square of phi equals... Let's get to the common denominator. I will have cosine square of phi plus sine square of phi divided by cosine square by phi equals... Sine square plus cosine square is one. So, I have this. Which means that cosine square of phi is equal to one over one plus tangent square of phi. So, instead of this, I put this because of this. All right. This is it. We have expressed sine of x as a function of tangent of x divided by two. Now, why do we have to go through these whole operations to come up with such a relatively complex formula for a sine? For very simple reason, because the cosine I can also represent as a function of tangent of x over two. So, let me just put this function here. Two tangent x over two divided by one plus tangent square x divided by two. Now, let's talk about the cosine. Very similar. Cosine x is cosine of x over two plus x over two equals cosine x over two times cosine x over two, which is cosine square of x divided by two minus sine square of x divided by two. So, cosine times cosine minus sine times sine equals... Do exactly the same thing. Multiply and divide by cosine square. So, I will have cosine square of x over two times. If I will divide it, that would be one. And this would be sine square divided by cosine square, which is tangent square of x over two. And again, I will use instead of this, I will use one over one plus tangent. So, that's equal to one minus tangent square of x over two divided by one plus tangent square of x over two. As you see, we have expressed both sine and cosine. Okay? Now, similarly, you can express tangent of x in terms of tangent of x over two. And we have already, well, basically done that, because that's easy. We have already expressed tangent of sum of two angles in terms of tangent. Remember, we just did it in the previous. Tangent of x plus y was equal to tangent x plus tangent y divided by one minus tangent x times tangent y. Now, instead of x plus y, we will put x over two plus x over two. So, here we will have two tangent of x over two. And here we will have tangent square of x over two. So, here is another function. So, you see everything in terms of all trigonometric functions. The cotangent will be similar. So, all trigonometric functions can be expressed as functions of tangent over half of the angle. That's what's very important, because now some kind of a mixed equation, like I was just saying, something like sine x plus five cosine x minus tangent x equals to one. What do you do in this particular case? You express each one of them as functions of tangent over half an angle. And you will have equation where you can... Equation of one argument only, which is tangent of x over two. You solve it somehow. Maybe it's solvable, maybe it's not solvable, I don't know. That's besides the point. But if it is solvable, then you will have tangent of x over two, and then you will have the x. Okay, so basically these are some kind of trigonometric identities which I believe are very useful. And that's why I have decided to spend some time presenting them. All right, so I suggest you to read notes for this lecture. I might have missed something during my lecture. Notes are a little bit more detailed, and I don't usually skip any kind of calculations. I just put it as they're supposed to be. Other than that, well, that's basically it. I wish you luck and have a good time.