 Hi, I'm Zor. Welcome to a new Zor education. I would like to start a series of lectures dedicated to calculation of trigonometric functions for different basic angles. I would like to refer you back to the lecture about what are these basic angles. In the first quadrant, obviously, it's 30, 45, 60. And then every other quadrant actually symmetrically repeats these values. So, let's just start from sine. That's my first lecture out of six for each function, each trigonometric function. Sine, cosine, tangent, cotangent, secant and cosecant. I will basically use the same methodology and the same values of angles, where I will calculate the value of trigonometric functions. And always I will use the unit circle and coordinates of the points, which are representing these particular angles. I do suggest you to do it yourself first and then listen to whatever the lecture I have. So, today is sine. To calculate the value of the sine for all different angles, basic angles, I usually use this unit circle. And for each point, each angle, I have to find exactly where is the point on this unit circle, which represents this particular angle. And then all I have to do is to determine the coordinates of this point. Now, obviously I don't remember coordinates for all these basic points. It's three points on each quadrant plus the diametrically opposing 0 and 1. So, I don't remember. So, everything what I'm doing right now is basically a derivation from something which I do remember. And the only thing which I do remember is 30 degree, which is pi over 6. That's the only angle I remember. And the only thing which I remember that the calculus opposite to 30 degrees angle is equal to 1 half of hypotenuse, and hypotenuse is equal to 1 because it's the radius of the unit circle. Everything else I'm going to derive. So, 1 half is ordinate of the point A. Now, what's the abscissa? Obviously I'm using the Pythagorean theory. This square plus this square is equal to 1. So, as it's easy to x square plus 1 half square is 1 quarter equals to 1. So, x is equal to 3 quarters and x square, I'm sorry. And x is equal to square root of 3 over 2. So, I derived that. Fine. My first angle is 45 degrees. I don't have to remember the value of ordinate and abscissa because they are equal. And again, from the Pythagorean theorem, I know that x square plus x square, both of them are x, is equal to 1. So, x is equal to square root of 2 over 2. So, I know these as well. Square root of 2 over 2. And finally, the 60 degrees. Now, this triangle is very much like this one because if this angle is 60 then this one is 30 degrees, right? So, the abscissa in this case of this point would be half of the hypotenuse. So, this would be half and then this would be square root of 3 over 2. Now, basically that's it. That's all I need. And again, I remember only the one half. Everything else is derived. Now, let's talk about angles which I have. 2 pi over 3. Now, 2 pi over 3, pi is 180 degrees, right? So, 2 pi over 3 is 120 which is here. This is 120 degrees. Now, this is 90 degrees. Now, these are symmetrical, right? If this is 60, then this is 90. This is 90, this is 60, so this is 30 degrees. And this is 30 degrees which means that these points are symmetrical. I proved that theorem in the lecture about basic angles which means that their abscissa and ordinate are the same as far as their absolute value is concerned. As far as the signs, well, ordinate projection onto y is the same but abscissa projection on the x is different by sign from this particular point. So, this particular point has square root of 3 over 2 and abscissa 1 over 2, right? Now, this one is square root of 2, square root of 2 over 2. That's actually this one. Let me put it a little bit higher. Okay. Square root of 3 over 2 comma 1 half, square root of 2 over 2 comma square root of 2 over 2, and actually it's twice worse, I'm sorry. This is ordinate. It's 1 half square root of 3 over 2. And here we have square root of 3 over 2 comma 1 half for this point. So, this point has exactly the same ordinate as this one which is square root of 3 over 2. And that's what we are interested in. We are interested in the sign. The whole lecture today is about sign. So, the sign is an ordinate and ordinate is the same. So, basically, the sign is equal for this square root of 3 over 2. Next, 3 pi over 4 is 1, 35, it's this point. Now, what's the symmetry here? 45, this is 90 minus 45, and this is 90 plus 45. So, these points are symmetrical, which means they are on the same ordinate. So, ordinate in this case is equal to square root of 2 over 2. 45 pi over 6 is this, it's 150 degree. And the symmetry is obviously between this point and this point. This is 90 minus 60, this is 90 plus 60. Symmetry is so I have the ordinate 1 half. Now, that's exactly the methodology which I'm using, and I will continue to use for these angles and for all other functions. So, today is a sign, so I'm interested only in ordinate. But basically, it's exactly the same for anything else. I start from the first quadrant and then I use the symmetry to spread information around. Now, pi. Pi is this, 180. Well, you don't have to really do anything for this now symmetry because you obviously understand that position of this point has coordinates minus 1, 0. Ordinate is 0, so it's 0. Sine is equal to 0. Minus pi over 6. Now, minus pi over 6 minus 30 degree means we go clockwise. Clockwise, by 30 degrees. Now, here you can use, this is minus 30 degrees. Now, here you can use two different methodologies, if you wish. Number one, you can always remember that the sign is an odd function, which means it changes the value of the function if argument changes its value. I mean, it changes the sign of the function if the argument changes the sign. The absolute value is retained. And here it's actually visible quite well. Because if you are from the plus 30 to go to minus 30, then these points are symmetrical relative to this diameter. And since they are symmetrical relative to the x-axis, they are projecting from the same x-point, which is exactly the same, but the coordinates are equal in absolute value, but opposite in sine. Which means if I'm talking about sine and sine is an ordinate, all I have to do is change the sine of the ordinate. So it used to be 1 half for 30 degrees for pi over 6. Now, for minus pi over 6, it will be minus 1 half. Pi over 4 with a minus, again, minus pi over 4 is this. Minus 45 degrees. And obviously this point is symmetrical to this one. So I have to change the sine of its ordinate, minus square root of 2 over 2. Next one is minus pi over 3, which is here. It's minus 60 degrees. And this symmetry was this point. So the ordinate is changing the sine. So it's minus square root of 3 over 2, minus 2 pi over 3. It's minus 120 degrees. Well, a little bit more, something like this. Now, this is 120 degrees clockwise. Now, this is 120 degrees counterclockwise. So, again, these points are symmetrical. So I have to take this one, take its ordinate, and change its sine. And the sine was, and the ordinate was, oh, I missed 1, okay, I missed minus pi over 2. I missed this point. Minus pi over 2, which is minus 90 degrees. Obviously the obsessed is equal to 0. Ordinate is equal to minus 1. So it's minus 1. Okay, now minus 2 pi over 3 is this one. It's minus 120 degrees. This is minus 90. So that would be square root of 3 over 2 with a minus sign. Minus square root of 3 over 2. 3 pi over 4 is this one. It's minus 135. Obvious symmetry was plus 135, which is square root of 2 over 2. So here we have minus square root of 2 over 2. Minus 5 pi over 6, that's this guy. It's minus 150. Symmetry is with this one. Now, 150 pi over 6 was 1 half, the ordinate. And so I will have minus 1 half. And finally, minus pi. Minus pi is exactly the same point as plus pi. It doesn't really matter. And this ordinate is equal to 0. This is minus 1. Ordinate is 0. So that's how I deal with all the calculations for all the basic angles for the sign. So first, I'm implementing this little picture. I know a couple of values here. Whatever I don't know, I calculate based on the Pythagorean theorem. Everything else is done through the symmetry. And that's how I encourage you to basically approach these problems. I don't think you can remember all these values for all these angles for all the different functions. Six functions, whatever number of these values, and 10, 12, whatever. It's a huge number of numbers that you have to remember. There is no need for this. You need a system and a system using one or two basic principles and the symmetry gives you everything. Well, that's it for a sign. Next will be cosine. Thank you.