 Hi, I'm Zor. Welcome to a new Zor application. All the previous lectures about trigonometry were introductory, explaining basically what is the trigonometry about, and what trigonometric functions exist, their properties, etc. Now, we are moving towards more important from my perspective and more useful for you guys. Part of these lectures which is related to problems basically. So, we are going to prove certain trigonometric identities, solve certain trigonometric problems, equations, etc. So, without further ado, let me just start with some very, very simple properties of certain trigonometric functions. These are the most basic trigonometric identities, which are the basis for basically all other more important and more difficult problems. Okay. Now, the first and very important and very known to everybody identity is this one. Now, what's important about this? X can be anything. It can be an acute angle, in which case this is basically a very simple example of the Pythagorean theorem. Because if this is your angle X, this is A, B and C, then A over C is sine of X, right? B over C is a cosine of X. And the sine square, if you will square these and add together, you will have sine square X plus cosine square X. And on the left, you will have A square plus B square over C square and A square plus B square is equal to C square because of the Pythagorean theorem. So that's a simple thing. Now, that's where in many cases, text books basically stop. Now, this is not where we are stopping. We probably should really examine this thing for any angle X, not only acute one, because the identity is true for any angle X. So let's go to a classical definition of the sine and cosine related to the unit circle. So on the unit circle, any angle is basically identified with the point on the unit circle, such as the angle from the positive direction of the X-axis, counterclockwise towards the position when the radius intersects this point, is actually our angle X. Now, and I'm talking about any angle X. So it can be acute angle, it can be obtuse angle, it can be actually something which is greater than the hundred and the eight degrees, et cetera, so any angle. Now, in this, considering this particular picture, it's also obvious that in most cases, considering that sine is, by definition, an ordinate and cosine is abscissa of this. So the coordinate of this is cosine X, sine X. That's the coordinates of this point. And don't forget this is the unit circle, which means that the radius is equal to 1. So no matter where this point is, if there is such a triangle, we can still apply the Pythagorean theorem, regardless of the fact that real abscissa, for instance, in this case is negative. It doesn't really matter because we're still squaring it. So even if it was negative, square of this will be the same as positive, right? So this particular thing is still true for any case when we can have this type of a triangle. For instance, this one. So abscissa is cosine, ordinate is sine. Sine is negative, true. However, being squared, it doesn't really matter that it's negative. The length is still satisfies the Pythagorean theorem. Sine squared plus cosine squared is equal to radius, which is 1. The only cases which we did not really consider yet are cases when we cannot really build the triangle. When either ordinate or abscissa is equal to 0, so there is no triangle. We cannot really refer to the Pythagorean theorem. What are these points? Well, obviously it's this one, this, this, and this. So in these four points, either ordinate or abscissa is equal to 0. So how can I prove this particular identity is true for these points? Well, there are only four of them. Let's just check it out. What's the coordinates of this point? 1, 0. This is 0, 1. This is minus 1, 0. And this is minus, and this is 0, minus 1. These are coordinates of these four points. And in any case, square of 1 plus square of another is always equal to 1. Square of 0 plus square of minus 1 is equal to 1. So we do have exactly the same identity for all cases. For cases of acute angle, we can use the equivalent definition of the sine and cosine as a ratio between the catatris and the hypotenuse. Or for any angle, we can use the unit circle and still refer to the Pythagorean theorem in most of the cases except these four. And these four are directly verified. That's it. That's the end of the group. Now we can say that this particular identity is true for any angle, not necessarily the acute one. Now, this is a very, very fundamental identity which I'm sure everybody who learned trigonometry and forgot everything else probably remember this one. All right, let's go to slightly less trivial, but very easy to prove. Sine square of x equals to tangent square x over 1 plus tangent square x. Now, what can we say about this identity? Well, even if I will prove it, the first thing we should really have to keep in mind that this is not necessarily true for any x. Why? Because tangent is not necessarily defined for any x. Tangent is, as you remember, sine over cosine. And in all those points where cosine is equal to 0, tangent is not defined. Where is the cosine equal to 0? Well, it's pi over 2 plus pi n. Because cosine is abscissa of the point. So where exact abscissa is equal to 0? That's these points, right? Everywhere else, abscissa is either positive or negative. But in this case, it's positive. In this case, it's negative. And only in these two points, north pole and south pole, abscissa is equal to 0, which constitute an angle of pi over 2 plus pi plus pi plus pi plus pi or minus pi. So that's the general formula. So this is a restriction on this identity. x should not be equal to this, where n is any integer number. Now, outside of these, we can actually prove it. And proof is very easy. Everything depends on the definition of tangent. Let's go to the right part. Now, from the definition of tangent, this is sine square over cosine square. That's the tangent square is, right? Over 1 plus sine square over cosine square, right? Well, we can use the common denominator, which is a cosine square. And instead of this, we can write this, right? That's the same thing. Cosine square divided by cosine square is 1. Now, as we have just proven before, cosine square plus sine square is equal to 1. So I can replace this with 1, unconditional for all angles. Now, we have sine square over cosine square and 1 over cosine square. Well, obviously, cosine square is reducible. And we will have only sine square. Because it's sine square x over cosine square x, 1, and cosine square x reduced, we have sine square. So that's the proof of this. And again, this is a true identity, but only to those cases where our tangent is defined. Very similar, even slightly simpler. Cosine square is equal to 1 over 1 plus tangent square x. Exactly the same restriction on tangent. x should not be equal to pi over 2 plus pi n. Because tangent otherwise doesn't exist. By the way, I didn't mention it the last time. We don't really have any problems with the denominator being equal to 0. Because tangent square is non-negative, 0 or positive, obviously, because it's a square, and then plus 1. So the denominator never equals to 0. So that's fine. So now we will do it this way. 1 over 1 plus sine square over cosine square. So as before, we replace it with cosine square x. As before, we replace this with 1. So it's 1 over 1 over cosine square, which is equal to cosine square x. OK, that's finished too. And the last one which I wanted to talk to is, again, it's like the first one, like sine square plus cosine square. It seems very simple. However, we do have to consider certain cases. So we kept actually two things. Sine phi equals cosine phi over 2 minus phi. And cosine of phi is equal to sine over phi over 2 minus phi. Now, these two are absolutely trivial in case of an acute angle phi. Why? Again, let's use the definition of trigonometric function for acute angle related to the right triangle. They are absolutely equivalent to the unit circle definition. So if this is phi, obviously this is phi over 2 minus phi, because the sum of two acute angles in the right triangle is equal to 90 degree, phi over 2, which means if this is a, this is b, this is c. Now, sine of phi is, by definition, in case of acute angles and right triangle, it's a over c. And cosine of phi is equal to b over c. This is opposite to the hypotenuse. And this is adjacent to hypotenuse. But relative to this angle, phi over 2, whatever was opposite to phi, now is adjacent to phi over 2 minus phi. Whatever was adjacent to phi is now opposite to, which means that the sine of phi over 2 minus phi equals b over c. And cosine of p over 2 minus phi is equal to a over c. So as you see, this is equal to this. And this is equal to this, which is exactly what we wanted to prove. So my point is that for the cases of acute angles, the theorem is actually trivial. And unfortunately, many textbooks actually stop here. They don't consider that this is actually the identities which are always true for any angle of phi, minus 3 pi over 2, for instance, or anything else, or 0. You can't construct the triangle with a 0 angle, right? But the theorem is true. So what we have to do right now is we have to expand this particular proof to all other cases. To do this, I will need something which I have already mentioned, the properties of the trigonometric functions, like this one. If you have an angle phi, and this is also phi, then this is pi minus phi, correct? Which is 180 degree, minus phi, minus the same. So these two are obviously on the same level, and I did prove it before when I was talking about sine. So their ordnance, these two, are exactly the same, which means sine of phi equals to sine of pi minus phi, right? How about cosine? Well, cosine is subsistent. These are equal in absolute value, but opposite in sines, right? So cosine of phi equals to minus cosine of pi minus phi. So these are things which we are going to use, as some kind of auxiliary things. Now, let's consider that our phi is not acute. Well, first of all, let's just immediately get out of the way cases when phi is equal to this, this, this, and this. So we don't have these cases the same as with the sine square plus cosine square. I have to consider them separately. We can consider them separately as well, so it doesn't really disturb us. And it's obvious. If phi is equal to 0, sine is equal to 0, cosine of pi over 2 is equal to 0. So they are the same. Now, cosine of 0 is 1, sine of pi over 2 is also 1. So it's the same. And similarly, you can check all these points that they are all exactly the same as far as these two identities are satisfied. So now, let's consider that the phi is somewhere here in the second quadrant. Well, if phi is in the second quadrant, then this angle, which is pi minus phi, is obviously acute. And for acute angles, we have already proven this, right? So we know that for pi minus phi, it's true. Well, let's just define this angle as chi. And the chi is in the acute angle. So if chi is in the acute angle, I know that sine of phi is equal to cosine of pi over 2 minus pi. Now, phi is pi minus phi. So let's just substitute it here. Sine of pi minus phi is equal to cosine of pi over 2 minus phi, which is minus pi plus phi. So it would be pi minus pi over 2, right? Pi minus 2 minus pi plus phi. That's what it will be, right? Now, sine of pi minus phi is the same as sine of phi, regardless of the angle, right? So we can substitute this instead of this. Cosine is an even function, which means if I change the sign of the argument, the function doesn't really change. We did talk about this before. So I'll change the sign. And I have my identity. Well, how about the cosine? OK, let's change it. So now we're proving the second identity. I know that cosine of phi is equal to sine of pi over 2 minus phi, because phi is in the acute angle. Now, let's substitute cosine of pi minus phi is equal to sine of pi minus phi over 2, right? Now, cosine of pi minus phi is minus cosine of phi. So it's minus cosine of phi. Sine is an odd function. It changes the sign if I change the sign of the argument. So it's minus sine of pi over 2 minus phi. So I change the sign of the argument. So I change the sign of the function. So minus and minus, obviously, can multiply by minus 1 both sides. And I have this equality. So we have actually proven for the second quadrant. Now, knowing that it's true for these, how can I derive it from these guys? Well, we can do it in a few different ways. The first way, which is exactly like this one, if phi is somewhere here, if this is phi, I can consider phi minus pi as an angle. So let's consider phi minus pi as an acute angle. And everything actually will be exactly the same. That's basically the way how I can do it. On another hand, and by the way, that's a good exercise for you. I mean, you will have exactly similar kind of things. So another way to approach it is, you see if you can get to this angle not from counter clockwise, but also clockwise with a minus sign, right? You remember that angle measured counterclockwise is positive. Counterclockwise is negative. So we can say that this particular angle is negative, but still it's absolute where it is from 0 to pi. So my approach in this case, and I would also leave it for you as homework, is let's just consider changing the sign of phi. What happens? It will be minus phi here. It will be plus phi here. Now we can use the odd or even properties of the cosine and sine. And that's how you can actually continue and prove this thing. I would like actually to leave it for you to just try it yourself. That's basically very easy and repetition and things like this. So that's all the elementary properties of the trigonometric functions which I wanted to talk about. I do recommend you to do the proof of these for angles greater than pi, greater than 100 degrees yourself. That will be very helpful. And next lecture will be just one much more difficult, well, not even difficult. It's probably just a little bit more within the elementary range of identities. And then we will gradually start increasing the difficulty of these things. So that's it for today. Thank you very much. Don't forget this proof which I left for you to do as a sub study is very important. I do recommend you to complete it. Other than that, thank you very much and good luck.