 Hi, I'm Zor. Welcome to a news or education. Let's talk about certain trigonometric equalities. Right now, I think it's very important to start with something relatively simple, because we are going to use whatever I'm presenting today in subsequent lectures to derive some more sophisticated trigonometric identities, equalities, et cetera. So today, we will talk about cosine of a double angle. I would like to express this cosine of a double angle in terms of cosine or sine of a single angle. So what happens with the angle, actually, with trigonometric function cosine of this angle if I double the angle? Well, let's start in a simple picture. I'll derive some formula, and then I will criticize it. So for instance, this is our angle phi, and this is another phi. So angle A, O, C is 2, 5, exactly what we need. So my question is, how to express the cosine? OK, now let's connect C and A and consider a triangle A, O, C. This triangle. Now, O, D is a bisector of this angle, because that's how we constructed it. Now, the triangle is obviously isosceles, because this is the radius, and this is the radius, and both radiuses are equal to 1, because this is a unit circle, the usual tool we're using with trigonometric identities. So as is well known, O, D is perpendicular to A, C, and also if this is a point M, it's the midpoint of A, C, which means A, M equals to M, C. Now, all these properties are relatively simple geometric properties of a circle and a chord and a perpendicular to a chord. I will use this. So now, I will use another thing. I will use the distance between two points on the coordinate plane. Now, what's this point coordinate? Obviously, it's 1, 0, right? Because the abscissa is equal to 1. That's the radius. And ordinate, this is the y-axis, and this is the x-axis. So ordinate is obviously 1, 0, sorry. Now, how about C? Well, we know that by definition of a cosine and by definition of a sine, if this angle is 2 phi, then C has coordinates cosine 2 phi and sine 2 phi. This is, by definition, the cosine and the sine, right? So whenever this is an angle, this is the point which corresponds to angle, then its coordinate is sine and its abscissa is cosine. So I know the coordinates of this and I know the coordinates of this. And that's why I can find this distance. On another hand, from a right triangle O, M, A, I can find A, M, because I know the coordinates, and then M, I know the angle. So A, M is half of this. So by doubling this and equating this to this length, which I calculated based on the coordinates, I will have some equation which will help me to connect cosine of 2 phi and some trigonometric function of the single angle phi. So let's do this. This is the plan, basically. So from this triangle O, M, A, I know that A, M equals 2. Well, let's think about it. This is the right triangle. In the right triangle, we know that sine is a ratio of opposite calculators to paper alternatives and cosine is the ratio of adjacent calculators to paper alternatives. Now the opposite to angle phi is A, M. So A, M relative to the hypotenuse, which is equal to 1, equal to sine, right? So A, M ratio to 1, which is a hypotenuse, is a sine of phi. That's why A, M is basically equal to sine phi, which means that A, C is equal to 2 sine of phi. And the square of a distance, A, C square, from this point to this point, according to this, is just 4 sine square phi. So again, sine square phi means sine phi parentheses power of 2. Now on another hand, square of a distance from one point to another, according to the formula for, well, square of a distance between one point and another on the coordinate plane is, so A, C square equals the distance between x-coordinates square plus the distance between y-coordinates square. I hope you remember it. If you don't, you can refer to the lecture about coordinates in the mass concepts. And that's where you will see the derivation of this formula. So again, distance on the x-axis, which is cosine 2 phi minus 1, right? This is cosine 2 phi. This is the x-coordinate. And x-coordinate of this is 1. So the distance square plus the distance between y-coordinates square between sine phi 2 phi and 0. So it's just sine square 2 phi. And these are equal. And that's basically the equation which will help me to define cosine 2 phi. And here is a very simple thing. If I will open the parentheses, I will see that this is cosine square of 2 phi minus 2 cosine 2 phi times 1 plus 1 and plus sine square 2 phi, this one. Now, there is a fundamental trigonometric identity that sine square plus cosine square of the same angle equals to 1, right? So this is 1. And this is 1. So it's 2. So it's 2 times 1 minus cosine 2 phi. So that's very convenient. Now we have almost established whatever we wanted to. So we have 1 and we have another. And they are equal to each other, almost them. Obviously, we have to reduce by 2. So I will have 2 sine square phi equals to 1 minus cosine 2 phi. So the cosine 2 phi equals to 1 minus 2 sine square phi. So this is the formula we wanted to derive. The cosine of the double angle equals to 1 minus 2 sine square of a single angle. Now, I have derived this formula using this particular drawing. Well, that's not really a very rigorous way to derive this formula for any angle. Obviously, this is true if I can't draw this triangle. I can draw the perpendicular, et cetera. So there are some cases when it's not really easy to do. So this case really encompasses all the cases when my angle phi is in the first quadrant from 0 to pi over 2. Then my 2 phi would be from 0 to 2 pi. And these are strict less than signs, not equal. Because if, for instance, phi is equal to 0, there is angle 0 double. It will be also 0. So I can't really draw any triangle. I cannot rely on this particular way of proving the things in case phi is equal to 0. Now, same thing for phi is equal to pi over 2. Because if phi is pi over 2, 2 phi is pi. So again, these three points do not form a triangle. And they're not even talking about something greater than pi over 2. Because then my double angle will be somewhere here. And triangle will be completely in different place. I mean, there are some problems with this proof. That's what I wanted to talk about and pay you to pay attention to. So what I would like to do is to make it a little bit more rigorous. Now, how can I make it? First of all, I think all these extreme cases when my three points, o, k, and c, are aligned on the same line. So I cannot really draw any triangles. Which means for phi is equal to 0, pi over 2, pi, and 3 pi over 2, I have to really prove this separately. Now, how can I prove it? Well, the easiest way, considering I know the angle, just substitute it. Now, let me just, as an example, actually, let's take pi over 2. Sine of pi over 2 is equal to 1. Remember, this is the graph of the sine. So this is 0. This is pi over 2. This is pi. So sine is equal to 1, which means the right side is 1 minus 2, which is minus 1. Now, the cosine of 2 pi. Now, 2 pi now is 2 times pi over 2, which is pi. Now, the cosine is this. So this is 0. This is pi over 2. This is pi. So it's minus 1. So the cosine of pi is equal to minus 1. And here we also got minus 1. So that is a sufficient checking for the formula in case pi is equal to pi over 2. Now, I'm not going to check for all these four points. I did put in notes for this lecture on Unisor.com, real verification. And you will see that this is actually true. So all these cases are covered by explicitly checking. OK, that's done. Next, well, now we can say this, right? No problems with this. Now, let's move with the phi beyond pi over 2. Well, if pi is somewhere here, let me check. Let me get another column. OK. If this is phi, so this is point B. And my point C would be somewhere here. This is a phi. And this is a phi. So this is 2 phi. Now, in this case, which I mark as phi from pi over 2 to pi. Now, in this case, I can always say that this angle, which is pi minus pi, does belong to this particular interval. So if phi is in the second quadrant, then pi minus phi is in the Q-tangle. And it belongs to the first quadrant. So if I will use this as a new angle, I use the letter chi for this case. Then for letter chi, which is an angle in the first quadrant, I know that my theorem is true. So I can actually write that cosine of 2 pi is equal to 1 minus 2 sine square of pi. Because I have actually proven this particular theorem for angle in the first quadrant. OK. Now, let's check what chi actually is. So it's cosine 2 pi minus 2 phi. 2 times chi is 2 pi minus 2 pi equals 1 minus 2 sine square of pi minus phi, because that's the value of pi. Now, let's change the left part. Now, 2 pi is a period for a cosine. So I can actually drop this. Now, cosine is an even function, which means if you change the sign of the argument, it does not change the value of the function. So this minus actually is also not needed. So I have actually 2 pi here. Now here, sine of pi minus phi, remember, sine of pi minus phi is this angle. Sine is ordinate, right? Ordinate of this and ordinate of symmetrical this ordinate. They are the same. So sine of pi minus phi is equal to sine of phi. I did pay attention to this identity in the previous lectures. So I can actually, instead of pi minus phi, put phi. So using this intermediary angle chi and the properties of evenness of the function cosine and the fact that the pi minus phi sine is exactly the same, using this and these properties, I have proven this formula for phi even in the second quadrant. OK, is this everything? No, not yet, because we still have third and fourth quadrant. But here is, again, a little logic which helps to prove it in this case as well. Now whenever the angle is in the third and fourth quadrant, I can always represent exactly the same angle with a negative sign, right? So instead of going this way, I can go this way. Now, this is counterclockwise, which is a positive direction of the angle. This is counterclockwise, this is clockwise, which is a negative direction. So if this is a positive angle, I can use this one, which is negative. Now, for any angle in the third of the fourth quadrant, the negative part of this angle would be in the first or the second quadrant, right? In this case, for instance, this is the angle in the fourth quadrant, but this is the angle which is in the first quadrant, just with a sign minus. And same thing for any other. So whenever my angle is in the third or the fourth quadrant, so my angle is from pi to pi to 2 pi. Actually, I don't need this, because we have already proven for all the equals. OK, when my pi is in this, I can use pi equals to minus pi. Now, in this particular case, I know that pi is negative, but it's negative, and it's in the first or the second quadrant. So for the first or the second quadrant, we have actually proven this, right? So we know that cosine of 2 pi is equal to 1 minus 2 sine square of pi. We know that for pi in the first and second quadrants, it's true. OK, now, what if I would change pi to minus pi, which is actually high? So I definitely know that if I change the sign, now cosine doesn't really change. So the cosine is an even function. So it doesn't really matter what sign of my angle here, positive or negative. Sine does change the sign of the function if argument changes the sign. But this is a square, which means we don't really care whether it's plus or minus. Square will always result in the plus. So again, in this case, using the negative angles, we have also proven. So basically now, we can say that we have covered all the different cases. We covered all the fourths, yes, all the four quadrants. And we covered the basic angles, 0 pi over 2 pi and 3 pi over 2. Because in these cases, we cannot really build a triangle. That completes the proof of this particular formula. And there is one more little detail. So if you just wipe this, so you will have a clean view. Let me rewrite this formula again. So the cosine of 2 pi equals 1 minus 2 sine square pi. Now, I also know that sine square pi plus cosine square pi is equal to 1. Now, using this identity, I can substitute instead of sine square, I can put 1 minus cosine square. And that gives me cosine 2 pi equals 1 minus 2 1 minus cosine square of pi, which is equal to 2 cosine square pi minus 1. So not only we can express the cosine of 2 pi in terms of sine pi, but also in terms of cosine pi. On another hand, I can replace this 1 with a sine square plus cosine square. And I will have the next representation equals. This is sine plus cosine and minus 2 sines. So it will be cosine of square pi minus sine square pi. So this is 1, this is 2, and this is 3. Three different incarnations basically have the same formula. How to express cosine in terms of double angle in terms of sine of single angle, in terms of cosine of a single angle, or in case of a mixed sine and cosine of a single angle. These are trivial consequences from the formula which I have already proved. Well, that's it. I do recommend you to read the notes to this lecture, which basically explain in writing whatever I was just talking about. By the way, in general, I think it's very, very useful for you personally to write down with your own hand or your own computer type in logical statements which really clearly demonstrate that you are in command of this logic. Now, in this case, proof, for instance, if you can really put in writing everything which you are actually leveraging as your logic which proves this particular theorem or another theorem, it's an extremely useful exercise, and I do suggest that you use it a lot. That develops your concentration, your logic, your creativity. That's it. Thank you very much, and good luck.