 Hi, I'm Zor. Welcome to a new Zor education. Let's talk about sign of certain angles where we can really easily calculate the value of the sign. I call these angles basic angles. And primarily I'm talking about zero, P over six, which is 30 degrees, P over four, which is 45 degrees, P over three, which is 60 degrees, and P over two, which is 90 degrees. So let's call these angles major or basic or whatever you want to call it. And again, I'm considering sign of these angles because I can do it. I can calculate the value of the sign. It's not easy to calculate the sign of, let's say 19 degrees. But 30 or 60 or 45, that's easy. So that's what we are going to do. And I'm sure it's the basis for many different problems in trigonometry, which I hope I will eventually show you. All right, from the beginning. Let's return back to the basics. Our sign as a function is defined as an ordinate of point A on the unit circle where this angle counterclockwise from the positive direction of X towards the point A is equal whatever we want this calculation to be done with. Now, we are concentrating only on these values of angle which are in the first quadrant of the coordinate plane. We will talk about other angles a little later. So in this case, the angle X of A is one of these. So let's just calculate what is the sign for each case. Now, in case zero, well, if this angle is zero then A coincides with X, which means the ordinate, this position has coordinates one, zero, this is one, ordinate is zero. So in this particular case, obviously, the sign is equal to zero because ordinate is zero. Next, next is P over six, which is 30 degrees. So if this is an angle of 30 degrees, let's consider this particular right triangle. A, A, P over. Now, in this right triangle, so P is a projection from A to X axis, Q is a projection to Y axis. So this particular calculus is equal to lowercase P and this particular calculus is equal to Q because the coordinates are P and Q. Now, what do we know about this triangle? Well, if this is 30 degrees, then we know that an opposite calculus, which is Q, is half of a paper genus. Now, the paper genus is one because it's a unit circle. Q is our ordinate. So basically what we have come up with that the sign of P over six, which is 30 degree is one half of the ordinate. Next, P over four. Okay, let's consider this angle 45 degrees. Well, then obviously this is also 45 degrees, right? So we are talking about the right triangle with two-catchity congruent to each other. So P is equal to Q. Now, if P is equal to Q and P squared plus Q squared, which is a fifth variant theorem, is equal to square of a paper genus, which is one. Well, we are interested in Q, right? So we can substitute instead of P. We can substitute Q. We will have two Q squared is equal to one. Or let's multiply it by two. It's four Q squared is equal to two because I want to have a square root nice. So from four Q squared, if I do the principle square root on both sides, I will get two Q, right? Because two Q squared will be four Q squared. And on the right, I will have square root of two. So Q is equal to square root of two over two. So sine of P over four equals two this. Next, next, this is P over three over 60 degrees. Okay, so this is 60 degrees. Now, this is 30 degrees, right? Now, we have a very similar situation to before. We have an angle of 30 degrees in a right triangle, which means that the opposite catatus, which is P in this case, equals to half of the hypotenuse, right? So P is equal to one half. Now, hypotenuse is one. So again, let's use the Pythagorean theorem to find Q. We know that Q squared plus P squared is equal to square of hypotenuse, right? Or Q squared plus one quarter. P squared, right? It's one quarter equals to one. Q squared equals to three quarter. So Q is equal to square root of three over two because this squared would be three quarters. So we came up with sine of P over three equals square root of three over two. Last in this series is 90 degrees. When this angle is 90 degrees, it means we are turning all the way and A would coincide with Y. Now, it's ordinate, obviously, would be the ordinate of Y, which is one. The ordinate of point Y, which is an intersection of this unit circle with a Y axis, obviously it has a C sub zero and ordinate of one. So for sine of P over two, we have equals to one. Okay, so we have finished with these major angles. Now, how to deal with angles outside of the first quadrant? Well, let's remember the qualities of the sine. Number one, sine is an odd function, which means, so whenever angle is negative, from a positive direction of X, we go clockwise, for instance, this, which is exactly the same thing as this, by the way, because one angle is different from another by a period of two pi and sine is a periodic function. So for all these, we can use this. So for instance, sine of minus 30 degrees is equal to minus one half. Sine of minus P over three is minus square root of three over two, et cetera, so this is this. Now, what if it's in this quadrant? Let's say my angle is here. The third quadrant, and this is an angle. Well, let's continue this towards this point, so it's a diameter. And now, obviously, this angle is this angle plus pi. So we need a formula something like this. If this is X, this is X plus pi. So to calculate X plus pi, we can actually calculate sine of X plus pi. We have to calculate the sine of X and do something with it. Now, what should we do? Again, we know this property. We have discussed it when we were talking about properties of sine function. Now, if the angle is here, well, let's have a symmetrical relative to the Y axis. Obviously, this angle is equal to this angle minus this angle, right? So we need something like a sine of pi minus X. And again, we were discussing this property before. That's the same thing with sine X. Now, both cases are actually quite obvious from the geometry of this, because if these two points are symmetrical, it means their ordinates are the same. Now, if this point is centrally symmetrical relative to zero, then it means obviously that the ordinates are equal in absolute value and opposite in sine. So using these and these properties, we can find out any sine angle for one of the major angles which we were discussing. That's it for sine. I'm planning to do the same for all other functions. And I do suggest you actually to go to unizor.com and look through notes for this particular lecture, so you basically understand everything better. But it's a really simple material, and it's important only because we will probably have a certain number of problems related to these particular colleges of sine and other functions. That's it. Thank you very much.