 Hi, I'm Zor. Welcome to Unizor Education. This is the third lecture in a series of how trigonometric functions behave in case when the angle is one of the basic angles. We were talking about basic angles in a general lecture. When I introduced them, it's right in the beginning of the trigonometric course. Now, I would like to basically calculate what exactly are the values of different trigonometric functions when the angle is one of these basic values. Now, today's lecture is about tangent, and we do have lots of different basic angles. Obviously, I don't remember the value of tangent for all of them, but what I do suggest to you is to be able to derive these values from the coordinates of the points on the unit circle, which represent these basic values. So that's exactly how I approach this problem, and that's how I suggest you do. So, considering I don't remember much about exact values of trigonometric functions, what do I remember? Well, I remember actually very little, something like if you have an angle of 30 degrees, then the opposite catch-a-tives in the right triangle is equal to half of the hypotenuse. That I do remember. Now, the other catch-a-tives, I can really derive using the Pythagorean theorem. So, if this hypotenuse is 1, we are talking about the unit circle, this catch-a-tive is 1 half, and if this is x, then x square plus 1 half square is equal to 1 square. From here, x is equal to 1 quarter, 3 quarter, square root of 3 quarter is this. So, I know that my coordinates of this point are, abscissa is equal to square root of 3 over 2, and originate is equal to 1 half, and this is 30 degrees or pi over 6, that I know. Now, I also know something about 45 degrees. But in this case, again, I don't remember anything. I do remember that if one angle is 45, then another is 45. So, these catch-a-tives are equal to each other. So, what's their value of hypotenuse is equal to 1? Well, again, Pythagorean theorem, if catch-a-tives is x, then x square plus x square, the catch-a-tives are the same, right? Is equal to 1. From here, x is equal to square root of 2 over 2, which means square root of 2 over 2 comma square root of 2 over 2. Both abscissa and originate are coordinate at this point, which represents pi over 4, which is 45 degrees. The third basic angle in the first quadrant is 60 degrees. Now, 60 degrees is basically, if you drop the perpendicular, this triangle is exactly the same as this triangle, because if this is 30, this is 60. If this is 60, this is 30, right? So, this particular catch-a-tives is equal to half of the hypotenuse, which is 1 half. And this one is equal to square root of 3 over 2. So, the coordinate at this point would be this. And this is pi over 3 angle. So, these are basic angles in the first quadrant. And using the coordinates of points which represent the basic angles, I can find out the value of the tangent, because the definition of a tangent, if you remember, is sine divided by cosine, or originate divided by abscissa. So, I have to divide the second member of the sequence by the first. Okay, so, it's very easy to calculate. Now, let's go to the angles which I am asking you to calculate. These are angles in other quadrants. Again, every one of them has three, in the middle and one-third, in the middle and one-third. So, let's start. 2 pi over 3 is this one. 2 pi over 3. Now, look at this. This is 90 degree plus 30 degree, right? Because this is 120. That's easier this way. Now, this is also 30 degrees, because it's 90 degree minus 60 degree. So, these angles, this one and this one, are symmetrical relatively to the y-axis. Now, if the angles are symmetrical, I have proven this theory in the beginning of this course of trigonometry when I'm talking about when I introduce basic angles. So, if angles are symmetrical, then these points are symmetrical relatively to the same y-axis. Now, if they are symmetrical relative to the y-axis, it means that their origins are the same because they are projecting into the same point on the y-axis, but their abscissas are opposite in signs, but the same in absolute value. So, the coordinates of this point now are abscissa is the same, and sorry, ordinances are the same, abscissa is opposite sign. Yes. So, if I know this, what's the tangent of this angle is coordinate divided by abscissa. So, it's minus square root of three. That's it. Next. Next is three pi over four, three pi over four, which is 135 degrees, and the coordinates of this, look at this thing. This is 90 plus 45, and this is 90 minus 45, which is 45. So, these are symmetrical, so their ordinances are the same, and the abscissas are opposite in sign. So, it's minus square root of two over two, square root of two over two. So, I replace the abscissa with an opposite sign, but retain ordinate. So, what's the tangent, which is a ratio of ordinate over abscissa? It's obviously minus one. Now, next is five pi over six, which is 150 degrees, and obvious symmetry is with 30, because 30 is 90 minus 60, 150 is 90 plus 60. So, the angles are symmetrical, relatively to the y-axis, so these points are symmetrical. So, the coordinates of this point are minus square root of three over two and one half. So, the tangent would be minus, obviously, one over square root of three, or which is the same square root of three over three. Okay? Next, I have pi, which is this one, and obviously since my ordinate is equal to zero, and abscissa is equal to minus one, my tangent is equal to zero. Okay? Next one, minus pi over six is this one. So, it's minus pi over six, which is minus 30 degrees. Well, now you can use both approaches. Number one, you know from the properties of the tangent that this is an odd function, which means that the value of the tangent for minus pi over six is minus the value at pi over six. From another perspective, these angles, plus 30 and minus 30, are symmetrical, which means these two points are symmetrical, in this case, relative to the x-axis, right? They are projecting to the same point, which means they are abscissa is the same, but ordinates are opposite in sign, but equal in absolute value. So, what's the coordinate of this point? Well, again, we retain the abscissa and change the sign of ordinate, which is square root of three divided by two over two comma minus one half. So, what's the tangent? If you divide this by this, ordinate by abscissa, it's minus, obviously, one over square root of three. Which is the same as minus square root of three over three. Next is minus pi over four. Minus pi over four, which is minus 45 degrees. Now, for minus 45 degrees, again, from the considerations of symmetry, you can say that abscissa is the same as in this case, but ordinate is different by sign. So, it's square root of two over two and minus square root of two over two. And if you do the ratio ordinate over the abscissa, it's minus one. That's the tangent. Next is minus pi over three, minus pi over three, which is minus 60 degrees. Obviously, symmetry was plus 60 degrees. So, the coordinates are one half. I retain the x and I change y coordinate to the opposite. And their ratio, y coordinate over x coordinate is minus square root of three. That's the tangent for minus pi over three. Next is minus pi over two, which is this one. The coordinates of this point are abscissa is zero and ordinate is minus one. Now, you cannot divide by zero. Tangent is not defined at pi over two. Next, minus two pi over three, which is here, which is minus two pi over three, which is minus 120. Abvious symmetry is plus two pi over three. So, I know this. So, now I have to retain the abscissa and change the ordinate to opposite. So, it would be minus one half, minus square root of three over two, and their ratio would be minus and minus, it would be square root of three. Now, this is minus three pi over four, which is the same as minus 135 degrees. Now, the symmetry with this guy gives me, I have to retain abscissa and change the sign of ordinate. It's minus square root of two over two, minus square root of two over two, and the sex, the obvious ratio would be one. So, the tangent of this angle is one. Next is five pi over six, which is this one, five pi, five pi over six, which is minus, which is minus one 50 degree. Abvious symmetry with this guy is plus five pi over six, and they have to change this. So, it's minus square root of three over two, minus one half. And the ratio is one over square root of three, which is square root of three over three. And minus pi is this point exactly the same as before, so the ratio is zero. I have zero here. So, these are the values of coordinates of the points which represent basic angles, and based on the coordinates and the definition of tangent, which is ordinate divided by abscissa, or if you wish, sine divided by cosine, which is the same thing. Based on this definition, I calculated the values of the tangent in each of these basic cases. Well, I will continue doing exactly the same for the other functions, so the next one will be cotangent. And, well, that's it for today. That's it for tangent. Thank you very much.