WEBVTT 00:00:06.352 --> 00:00:12.115 In preparation for evaluating trigonometric functions of general angles, 00:00:12.117 --> 00:00:16.364 it is important to group the special angles into families. 00:00:16.366 --> 00:00:34.115 The first family we will look at is the quadrantal family. 00:00:34.117 --> 00:00:42.643 The first angle in the quadrantal family has a terminal side along the positive x-axis. 00:00:42.645 --> 00:00:49.640 This angle has a measure of 0 radians, also 0 degrees. 00:00:49.642 --> 00:00:58.408 The next quadrantal angle has a terminal side here, and it has a measure of 00:00:58.410 --> 00:01:04.656 pi over 2 radians or 90 degrees. 00:01:04.656 --> 00:01:12.908 The next angle has a terminal side here, and it has a measure of 00:01:12.910 --> 00:01:17.924 pi radians or 180 degrees. 00:01:17.926 --> 00:01:24.425 And then the fourth quadrantal angle we will look at has a terminal side here, 00:01:24.427 --> 00:01:33.195 and it has a measure of 3 pi over 2 radians or 270 degrees. 00:01:33.197 --> 00:01:39.946 Now there are infinitely many quadrantal angles, but they are all coterminal 00:01:39.948 --> 00:01:44.450 with one of these four angles. 00:01:44.452 --> 00:01:55.467 The next family we will look at is the pi over 3 family. 00:01:55.469 --> 00:02:05.224 Alright, the first angle is pi over 3. So pi over 3 sits about here, has a 00:02:05.226 --> 00:02:15.223 terminal side in quadrant 1, and has a measure of pi over 3 radians or 60 degrees. 00:02:15.225 --> 00:02:26.493 The next angle in the pi over 3 family, let's draw it, has a terminal side about here, 00:02:26.495 --> 00:02:37.259 and it has a measure of 2 pi over 3 radians or 120 degrees. 00:02:37.261 --> 00:02:44.516 Now watch what happens next. If we add another pi over 3, we would 00:02:44.518 --> 00:02:50.765 get an angle that has a terminal side on this axis, but we can see that that angle 00:02:50.767 --> 00:02:57.517 already belongs to the quadrantal family. So we skip 3 pi over 3, 00:02:57.517 --> 00:03:10.267 and we move to this angle which has a measure of 4 pi over 3, 00:03:10.269 --> 00:03:23.016 which is equal to 240 degrees. Now we add another pi over 3. This angle 00:03:23.018 --> 00:03:34.314 here has the measure of 5 pi over 3 or 300 degrees. 00:03:34.316 --> 00:03:42.070 And we can check this because if we add another pi over 3 we get right back to 00:03:42.072 --> 00:03:48.582 one rotation, which is 2 pi. So now we have the four major angles in 00:03:48.584 --> 00:03:54.584 the pi over 3 family, and then there of course are infinitely many others, 00:03:54.586 --> 00:04:00.842 but they are all coterminal with one of these four. OK. 00:04:00.844 --> 00:04:30.360 Next, we look at the pi over 4 family, first angle pi over 4 or 45 degrees. 00:04:30.362 --> 00:04:36.122.873 Now if you add another pi over 4 you would get to 2 pi over 4, but when you 00:04:36.124 --> 00:04:42.134 reduce that, you get pi over 2, and once again that angle is already a quadrantal angle. 00:04:42.134 --> 00:04:46.383 So now we have 1 pi over 4, 2 pi over 4, 00:04:46.385 --> 00:05:03.626 so the next angle in this family would be 3 pi over 4 which is 135 degrees. 00:05:03.628 --> 00:05:08.127 OK, now we add another pi over 4 and you see what's going to happen again. 00:05:08.129 --> 00:05:14.659 We get an angle that's coterminal with any angle on this axis, and of course 00:05:14.661 --> 00:05:20.923 once again that's here. So we add another pi over 4 and you see the routine. 00:05:20.925 --> 00:05:27.170 So we have pi over 4, 2 pi over 4, 3 pi over 4, 4 pi over 4, and now this is 00:05:27.172 --> 00:05:37.437 5 pi over 4 which is 225 degrees, and that's this angle. 00:05:37.439 --> 00:05:50.201 And then the last one is going to be this one which is 7 pi over 4 or 315 degrees. 00:05:50.203 --> 00:05:57.956 Once again, there are an infinite number of angles that are in the pi over 4 family, 00:05:57.956 --> 00:06:00.962 but they are all coterminal with one of these four. 00:06:00.964 --> 00:06:17.327 Now we look at the pi over 6 family. OK, first angle, pi over 6. 00:06:17.329 --> 00:06:37.484 So this is pi over 6 which is 30 degrees. Now another pi over 6 would be 2 pi over 6, 00:06:37.486 --> 00:06:46.487 but that's pi over 3. And then we have 3 pi over 6 which is here, 00:06:46.489 --> 00:06:54.011 and then 4 pi over 6 which is here. And so the next angle 00:06:54.011 --> 00:07:03.011 we're actually going to identify as being part of this family is 5 pi over 6 00:07:03.013 --> 00:07:15.783 which is 150 degrees. Then we'd have 6 pi over 6 which we don't identify 00:07:15.785 --> 00:07:29.037 because again that's pi. Then this one is 7 pi over 6 which is 210 degrees. 00:07:29.039 --> 00:07:36.308 Then we have 8 pi over 6, 9 pi over 6, 10 pi over 6, and those are all 00:07:36.310 --> 00:07:54.299 already represented, and then the last one is 11 pi over 6 which is 330 degrees. 00:07:54.301 --> 00:08:01.082 You need to practice drawing these angles and get used to identifying 00:08:01.084 --> 00:08:05.580 which of these special angles fit into which family. 00:08:07.335 --> 00:08:14.084 Let's take a look at identifying the angle family that a given angle belongs to. 00:08:14.086 --> 00:08:20.847 We will sketch the given angle. We will find theta c, the coterminal angle 00:08:20.849 --> 00:08:26.599 which is the angle of least non-negative measure coterminal with the given angle. 00:08:26.601 --> 00:08:32.350 And then we will identify the special angle family. The first one we're going to look at 00:08:32.352 --> 00:08:39.350 is the given angle theta is 33 pi over 6. Hopefully the first thing you will notice about this 00:08:39.352 --> 00:08:45.626 is that 33 and 6 have a common factor of 3, and we need to simplify this fraction 00:08:45.628 --> 00:08:55.132 before we look at anything else. So we're going to divide the numerator and the 00:08:55.134 --> 00:09:11.133 denominator each by 3, and we will get 11 pi over 2. Now we will sketch theta. 00:09:11.135 --> 00:09:17.382 Now let's look at what we have. Remember that one rotation in terms of pi over 2 00:09:17.384 --> 00:09:25.633 would be 4 pi over 2. OK, so what we're thinking is that 2 pi is 4 pi over 2. 00:09:25.635 --> 00:09:38.132 So to find theta c for 11 pi over 2 we'd have 4 pi over 2, 8 pi over 2, and then 00:09:38.134 --> 00:09:46.184 9, 10, 11 pi over 2. You could do this algebraically by just repeatedly subtracting 00:09:46.186 --> 00:09:53.934 4 pi over 2 from the given angle until you get an angle of least non-negative measure 00:09:53.936 --> 00:10:06.434 which would be in this case 3 pi over 2. So that is theta c. Once we look at this 00:10:06.436 --> 00:10:17.467 we realize that this is a special angle and it belongs to the quadrantal family, 00:10:17.469 --> 00:10:29.231 and so when we identify the angle family we would write the word quadrantal. 00:10:29.233 --> 00:10:34.485 Lets take a look at another example. So here we have theta equals negative 00:10:34.487 --> 00:10:47.232 840 degrees. First I'm going to draw this angle. Now because it's a negative angle, 00:10:47.234 --> 00:10:55.005 I start here going this direction, and each rotation is going to be negative 360 degrees. 00:10:55.007 --> 00:11:08.766 So this would be negative 360, negative 720, and then another negative 120. 00:11:08.768 --> 00:11:19.767 So this part is negative 120 degrees which means that this part would be 00:11:19.769 --> 00:11:26.519 positive 240 degrees. You can always check this algebraically by adding 00:11:26.521 --> 00:11:35.515 360 degrees repeatedly to this angle until you get the angle of least positive measure 00:11:35.517 --> 00:11:45.304 which is theta c. So here we have theta c is 240 degrees, and the angle family 00:11:45.306 --> 00:11:54.695 is pi over 3 or in this case we would say 60 degrees. 00:11:57.000 --> 00:12:03.321 In order to create the definitions of the trigonometric functions of general angles, 00:12:03.323 --> 00:12:09.577 we start by reviewing the right triangle definitions of trigonometric functions. 00:12:09.579 --> 00:12:14.334 So we have the sine of theta is opposite over hypotenuse, 00:12:14.336 --> 00:12:18.581 cosine of theta is adjacent over hypotenuse, 00:12:18.583 --> 00:12:24.334 and tangent of theta is opposite over adjacent. We know that the other three functions 00:12:24.336 --> 00:12:32.330 are reciprocals, each one of one of these. Notice here that theta is an acute angle. 00:12:32.330 --> 00:12:38.831 To extend the right triangle definitions of the trigonometric functions to include all 00:12:38.833 --> 00:12:44.083 general angles, that means angles with positive measure, negative measure, zero 00:12:44.085 --> 00:12:57.880 measure, and no limit in size, we take a look at a point on the terminal side of the angle. 00:12:57.882 --> 00:13:08.127 So here's theta and I pick a point on the terminal side, and that point has coordinates x, y, 00:13:08.129 --> 00:13:27.128 and I look at the length of this line segment. I choose to call it r, and looking at this 00:13:27.130 --> 00:13:38.166 right triangle, I can see that r is equal to the square root of x squared plus y squared. 00:13:38.168 --> 00:13:46.410 Now using the right triangle definitions for theta, and let's move this label for theta over 00:13:46.412 --> 00:13:55.433 to here. Much better. Then I can see that the sine of theta would be opposite, but 00:13:55.435 --> 00:14:03.436 if you look at this as the coordinate, this is actually the y-coordinate, and then of course 00:14:03.438 --> 00:14:10.935 this is the x-coordinate. So let's go back. The sine of theta is opposite over hypotenuse. 00:14:10.937 --> 00:14:21.962 OK, so the sine of theta is opposite over hypotenuse, and so what we use here 00:14:21.962 --> 00:14:31.462 are the coordinates of the point. By the same process, I can see that the cosine of theta, 00:14:31.464 --> 00:14:36.710 adjacent over hypotenuse would be x over r, and then 00:14:36.712 --> 00:14:47.711 the tangent of theta would be y over x, which is opposite over adjacent. 00:14:47.713 --> 00:14:56.565 Alright, now let's take a look at an angle that has a terminal side that is not in quadrant 1. 00:14:56.567 --> 00:15:11.210 So let's try quadrant 2. So here's theta. Here is a point on the terminal side of theta. 00:15:11.212 --> 00:15:21.764 The length of this line segment, r, is again equal to the square root of x squared 00:15:21.766 --> 00:15:31.021 plus y squared, and we can see now that we can use this same representation, 00:15:31.023 --> 00:15:42.291 but this time using the coordinate of the points. And we can find the representations 00:15:42.293 --> 00:15:48.539 of the other three trig functions using reciprocals. So let's write this all down. 00:15:48.541 --> 00:16:00.060 So the sine of theta for any angle is y over r, the cosine is x over r, 00:16:00.062 --> 00:16:09.563 and the tangent of theta is y over x, where x and y are the coordinates of the point 00:16:09.565 --> 00:16:15.060 on the terminal side of theta. Notice here that x cannot equal 0. 00:16:15.062 --> 00:16:24.559 Now the cosecant of theta, using the reciprocal identity, is r over y where 00:16:24.561 --> 00:16:38.563 y does not equal 0. The secant of theta is r over x where x does not equal 0, 00:16:38.565 --> 00:16:48.312 and finally, the cotangent of theta is x over y where y does not equal 0. 00:16:48.314 --> 00:17:00.000 And so now we have the definition of the six trigonometric functions for general angles. 00:17:02.000 --> 00:17:09.128 Let's take a look at an example: the point P, with coordinates (12, -10). 00:17:09.130 --> 00:17:15.385 This point lies on the terminal side of theta. We want to find the exact values of the 00:17:15.387 --> 00:17:22.395 six trigonometric functions of theta, and here are the definitions of the trigonometric 00:17:22.397 --> 00:17:31.620 functions for general angles. So the first thing we're going to do is we will 00:17:31.622 --> 00:17:41.159 plot the point (12, -10), so 12 is, we'll say, out here somewhere. 00:17:41.161 --> 00:17:46.168 Negative 10 is about here, 00:17:46.170 --> 00:18:00.181 so the point is about here. Then that means that theta has a terminal side that lies 00:18:00.183 --> 00:18:08.441 in quadrant 4, represented by this positive angle. In this situation, we know 00:18:08.443 --> 00:18:17.447 that x, the x-coordinate, is 12, and the y-coordinate is negative 10. 00:18:17.449 --> 00:18:26.190 The piece we're missing is r. OK, so we're going to find r. We know that 00:18:26.192 --> 00:18:37.470 r is the square root of x squared plus y squared, so r is the square root of 00:18:37.472 --> 00:18:52.466 12 squared plus negative 10 squared. So r is the square root of 144 plus 100, 00:18:52.468 --> 00:18:59.465 so that's 244. Now we must simplify this radical which means we have to look 00:18:59.467 --> 00:19:08.246 to see if there is a factor here of 244 that is a perfect square. It's pretty obvious, I think, 00:19:08.248 --> 00:19:27.370 that 4 times 61 is 244. The square root of 4 is 2, and so this is what I'm going to use for r. 00:19:27.372 --> 00:19:48.037 OK. Now all I have to do is fill in. So the sine of theta is y over r, and as soon as we look 00:19:48.039 --> 00:19:54.297 at this, we see that we have a common factor of 2, so we're going to simplify this 00:19:54.299 --> 00:20:04.547 expression and get [negative] 5 over the square root of 61. Then we have the cosine of theta is 00:20:04.549 --> 00:20:12.563 x over r, so that's 12 over 2 times the square root of 61, 00:20:12.565 --> 00:20:25.813 which is 6 over the square root of 61. And then the tangent of theta is y over x, 00:20:25.815 --> 00:20:39.062 so this is y over x and when I reduce this I get [negative] 5/6. So here's the sine, the cosine, 00:20:39.064 --> 00:20:49.096 and the tangent for theta. And now I can find the other three values by taking 00:20:49.098 --> 00:20:59.097 the reciprocals. So the cosecant of theta would be negative square root 61 over 5. 00:20:59.099 --> 00:21:13.348 The secant of theta would be square root of 61 over 6, and the cotangent of theta 00:21:13.348 --> 00:21:24.848 would be negative 6 over 5. And so now we have found the exact values of the six 00:21:24.850 --> 00:21:32.847 trigonometric functions of theta having a point on the terminal side (12, -10). 00:21:35.545 --> 00:21:41.097 Let's find the values of the trigonometric functions for quadrantal angles. 00:21:41.099 --> 00:21:51.598 Let's start with the first quadrantal angle. This would be an angle with the terminal side 00:21:51.600 --> 00:22:01.920 on the positive x-axis, and what we will do is we will pick any point on this terminal side. 00:22:01.922 --> 00:22:08.423 And you can see that a point here will have a coordinate of x, and then 00:22:08.425 --> 00:22:15.183 the y-coordinate would be 0. So this would be our point P. You can also see that the length 00:22:15.185 --> 00:22:21.430 of this line segment, which we usually call r, in this case is actually going to have 00:22:21.432 --> 00:22:32.182 the value of x. So from this, using our definitions, we can find the sine of 0, the cosine of 0, 00:22:32.184 --> 00:22:44.936 the tangent of 0, and then using the reciprocal property, we can find the other three values. 00:22:44.938 --> 00:22:55.729 So let's see. The sine of 0, from the definition, is y over r, so it would be 0 over x, 00:22:55.731 --> 00:23:13.228 so the sine of 0 is 0. The cosine of 0 is x over r, and so that would be x over x which is 1. 00:23:13.230 --> 00:23:30.722 Then the tangent of 0 would be, from the definition, y over x. y is 0, x is x, and we get 0. 00:23:30.724 --> 00:23:39.020 OK, what we can do now is find the other three using the reciprocal property. 00:23:39.022 --> 00:23:45.528 Well, that gets pretty interesting with the 0. So let's take a look at the definition. 00:23:45.530 --> 00:23:59.777 The cosecant of 0 would be r over y which is going to be x over 0. So what we see here 00:23:59.779 --> 00:24:10.027 is that the cosecant of 0 is undefined. 00:24:10.029 --> 00:24:23.526 The secant of 0 would be r over x, and that's x over x which is 1, and that's no surprise 00:24:23.528 --> 00:24:37.527 because the reciprocal of 1 is 1. And then the cotangent of 0 is x over y, so that's x over 0, 00:24:37.529 --> 00:24:47.830 and we have the same situation here, and so the cotangent of 0 is undefined. Now what we 00:24:47.832 --> 00:25:00.000 can do is we can find the trig functions for pi over 2, pi, 3 pi over 2 by the same method. 00:25:01.025 --> 00:25:14.327 Let's find the values of the trigonometric functions for pi. 00:25:14.329 --> 00:25:25.124 So here theta is pi or 180 degrees. A point on the terminal side of pi 00:25:25.126 --> 00:25:32.123 would have coordinates x, which is going to be a negative value, and then the y-coordinate 00:25:32.125 --> 00:25:42.120 will be 0. Now the length of this line segment is r, and r is a positive number. 00:25:42.122 --> 00:25:54.624 But when we identify the values of x, y, and r for this angle we know that x is equal to x, 00:25:54.626 --> 00:26:02.871 y is equal to 0, but r has to be positive and we know that x is negative. 00:26:02.873 --> 00:26:09.918 So in order to make this positive, we realize that r has to be the opposite of x. 00:26:09.920 --> 00:26:29.688 Now we use the definition, so the sine of pi is equal to y over r which is 0. The cosine of 00:26:29.690 --> 00:26:53.936 pi is x over r, and the tangent of pi is y over x which is 0. OK, now let's get the other three. 00:26:53.938 --> 00:27:23.746 The cosecant of pi is going to be r over y and this is undefined. The secant of pi is r over x, 00:27:23.748 --> 00:27:45.510 and the cotangent of pi is x over y, so that's x over y which is undefined. 00:27:47.500 --> 00:27:52.021 By continuing in a similar manner, we are able to find the values of the trigonometric 00:27:52.023 --> 00:27:57.779 functions for all of the quadrantal angles. Here they are in a table. 00:28:00.000 --> 00:28:07.286 Let's take a look at an example. Find the exact value of the tangent of negative 7 pi over 2 00:28:07.288 --> 00:28:19.052 without a calculator. So first of all, we have to find the terminal side of negative 7 pi over 2. 00:28:19.054 --> 00:28:28.309 We know that one rotation in this direction is negative 4 pi over 2, and then this would be 00:28:28.311 --> 00:28:36.058 negative 5, negative 6, negative 7 pi over 2. So we see that the terminal side 00:28:36.060 --> 00:28:42.309 of this angle lies on the positive y-axis. This is a quadrantal angle. We will pick 00:28:42.311 --> 00:28:57.568 a point on the terminal side here, and we will label this point (0,y). And now we will also 00:28:58.570 --> 00:29:06.336 look at the length of this line, which is also, we call it r, but in this case you can see that it is y. 00:29:06.338 --> 00:29:18.835 So I'm going to identify x is 0, y is actually the value of y, and r is also y. 00:29:18.837 --> 00:29:34.334 I want to find the tangent of negative 7 pi over 2, and so I use the definition y over x and I get 00:29:34.336 --> 00:29:47.383 y over 0 which we know is undefined. So I can't really find the exact value of this 00:29:47.385 --> 00:29:52.887 expression, but I can say that it is undefined. 00:29:54.500 --> 00:30:04.392 Let's find the signs, that's the s-i-g-n-s signs, of the trigonometric functions by quadrant. 00:30:04.394 --> 00:30:13.892 Well, since r is always positive, the sign, the s-i-g-n sign, is dependent completely on 00:30:13.894 --> 00:30:22.420 the sign of x and the sign of y, based on the point on the terminal side of the angle. 00:30:22.422 --> 00:30:38.169 So if we look at the four quadrants, one, two, three, and four, let's think about the values 00:30:38.171 --> 00:30:46.441 of x and y. In this quadrant, x and y are both positive. 00:30:46.443 --> 00:30:51.698 If the angle has a terminal side in this quadrant, think about what you would get. 00:30:51.700 --> 00:30:58.454 Well, you would get a negative value for x and a positive value for y. 00:30:58.456 --> 00:31:04.205 If the angle has a terminal side in this quadrant, both x and y will be negative numbers, 00:31:04.207 --> 00:31:14.964 and here x will be a positive value and y will be negative. Let's take a look at each quadrant, 00:31:14.966 --> 00:31:25.714 one at a time. Quadrant one: In quadrant one, x, y, and r are all positive, and therefore, 00:31:25.716 --> 00:31:33.214 for an angle that has a terminal side in quadrant one all of the trig functions will be positive. 00:31:33.216 --> 00:31:46.504 In quadrant two: If an angle has a terminal side in quadrant two, x will be a negative number, 00:31:46.506 --> 00:31:59.264 y will be a positive number, and r, as always, will be a positive number. So the sine 00:31:59.266 --> 00:32:09.522 of an angle with a terminal side in quadrant two would be y over r which in this case would 00:32:09.524 --> 00:32:16.274 be represented by a positive number over a positive number which would yield a 00:32:16.276 --> 00:32:26.527 positive number. The cosine of an angle that has a terminal side in quadrant two would be 00:32:26.529 --> 00:32:38.054 represented by x over r where x is a negative number and r is a positive number, 00:32:38.056 --> 00:32:43.305 and a negative number divided by a positive number would yield a negative number. 00:32:43.307 --> 00:32:56.054 OK, the tangent of theta where theta has a terminal side in quadrant two would be y over x 00:32:56.056 --> 00:33:03.554 where y is represented by a positive number and x is represented by a negative number, 00:33:03.556 --> 00:33:09.838 and the result would produce a negative number. So what we see here is that 00:33:09.840 --> 00:33:18.092 if an angle has a terminal side in quadrant two, the sign, the s-i-g-n sign, would be for the 00:33:18.094 --> 00:33:24.093 sine function positive, for the cosine function negative, and the tangent function negative. 00:33:24.095 --> 00:33:33.093 So what that means is that the cosecant for an angle that has a terminal side in quadrant two 00:33:33.095 --> 00:33:41.592 would also be positive because these are reciprocal functions. For the secant negative, 00:33:41.594 --> 00:33:49.845 and the cotangent negative. We can continue 00:33:49.847 --> 00:33:57.843 in a like manner for angles that have terminal sides in quadrants three and four. 00:33:59.880 --> 00:34:05.142 After finding the signs of the trigonometric functions in the remaining quadrants, 00:34:05.144 --> 00:34:13.641 this is the result. We see that if an angle has a terminal side in quadrant one, all six trig 00:34:13.643 --> 00:34:20.654 functions will have a positive value. If an angle has a terminal side in quadrant two we see 00:34:20.656 --> 00:34:25.662 that both the sine function and the cosecant function will be positive and the other four 00:34:25.664 --> 00:34:32.660 functions will have a negative value. If an angle has a terminal side in quadrant three, both 00:34:32.662 --> 00:34:38.175 the tangent and the cotangent will have positive values. The other four will have negative 00:34:38.177 --> 00:34:46.682 values. And then in quadrant four, if an angle has a terminal side here, then the cosine and 00:34:46.684 --> 00:34:53.688 the secant will both be positive, and the other four trig functions will have negative values. 00:34:53.690 --> 00:34:58.940 Now let me show you how you will more often than not use this information. 00:34:58.942 --> 00:35:10.440 When evaluating the sine of theta if theta has a terminal side in quadrant one, then the sine 00:35:10.442 --> 00:35:19.216 of theta will be positive. If theta has a terminal side in quadrant two, the sine will be positive. 00:35:19.218 --> 00:35:26.466 If theta has a terminal side in quadrant three, the sine will not be positive. It will be negative, 00:35:26.468 --> 00:35:37.002 and the same thing if theta has a terminal side in quadrant four. The sine of theta will be negative. 00:35:37.004 --> 00:35:47.013 If we're evaluating the cosine of theta, then if theta has a terminal side in quadrant one, once again 00:35:47.015 --> 00:35:56.022 we know the cosine of theta will be positive. In quadrant two, when the terminal side of theta 00:35:56.024 --> 00:36:08.022 lies here, the cosine will be negative. If the terminal side of theta lies in quadrant three, the cosine 00:36:08.024 --> 00:36:20.520 will be negative, and if the terminal side of theta lies in quadrant four, the cosine will be positive. 00:36:20.522 --> 00:36:33.557 And then we will look at the tangent function. Again, if the terminal side of theta lies in 00:36:33.559 --> 00:36:42.058 quadrant one, the tangent of theta will be positive. If the terminal side of theta lies in quadrant two, 00:36:42.060 --> 00:36:50.309 the tangent will be negative. If the terminal side of theta lies in quadrant three, the tangent of theta 00:36:50.311 --> 00:37:04.058 will be positive, and if the terminal side of theta lies in quadrant four, the tangent will be negative. 00:37:04.060 --> 00:37:14.000 These three diagrams will be particularly useful as you further your knowledge and study of trigonometry. 00:37:16.075 --> 00:37:27.590 Let's look at an example. If the tangent of theta is greater than zero, positive, and the sine of theta is 00:37:27.592 --> 00:37:37.104 less than zero, negative, then theta has a terminal side that lies in what quadrant? Well, let's take a look. 00:37:37.106 --> 00:37:48.361 If the tangent of theta is greater than zero that means the tangent is positive, so we look here and 00:37:48.363 --> 00:37:53.611 we see that the tangent of theta would have a positive value if the terminal side of theta lies in 00:37:53.613 --> 00:38:10.388 quadrant one or quadrant three. So this is true for these two quadrants. Now the sine of theta is negative 00:38:10.390 --> 00:38:24.150 means that the terminal side of theta lies in quadrant three or quadrant four, and because we have here 00:38:24.152 --> 00:38:33.400 "and" we're looking for the quadrant that applies in both cases, and we see that quadrant three 00:38:33.402 --> 00:38:41.918 is the only possible quadrant that the terminal side of theta could lie in that meets these two conditions. 00:38:43.688 --> 00:38:53.168 Let's look at another example. If the tangent of theta equals negative 3/7 and the cosecant of theta is 00:38:53.170 --> 00:39:02.418 positive, then find the sine of theta. Well the first thing we have to do is determine where the terminal 00:39:02.420 --> 00:39:15.667 side of theta lies, and we will get that information from the fact that the tangent of theta is negative and 00:39:15.669 --> 00:39:25.461 the cosecant of theta is positive, so let's find the quadrants. The tangent is negative when theta has a 00:39:25.463 --> 00:39:41.213 terminal side in quadrant two or four. The cosecant is positive which means that the sine is positive 00:39:41.215 --> 00:39:49.463 because they're reciprocal functions. So the sine is positive if theta has a terminal side in quadrant one 00:39:49.465 --> 00:39:59.746 or quadrant two, and the only quadrant here that these two conditions have in common is quadrant two, 00:39:59.748 --> 00:40:12.496 so that is where the terminal side of theta lies. So let's draw theta and see what we can determine. 00:40:12.498 --> 00:40:27.245 So here's theta, and we know from what was given that we can find x, y, and r. Now be careful here. 00:40:27.247 --> 00:40:40.996 In this quadrant, a point on the terminal side of theta will have a negative x-value and a positive y-value. 00:40:40.998 --> 00:40:56.053 So what that means here is that because the tangent is y over x, it means that the negative sign actually 00:40:56.055 --> 00:41:06.553 goes with the 7, so what I'm going to put is the 7 here and then 3 here. Next I have to find r. 00:41:06.555 --> 00:41:29.586 So r is the square root of x squared plus y squared, so r is [the square root of] 49 plus 9 which is the square root of 58. 00:41:29.588 --> 00:41:40.336 Now we check to see if we can simplify this radical, and we look for a factor of 58 that is a perfect square. 00:41:40.338 --> 00:41:50.605 OK. We see that there aren't any, and so we use the square root of 58. And then the last thing we have 00:41:50.607 --> 00:42:08.125 to do is find the sine of theta. So the sine of theta is y over r, and in this case, y is 3 and 00:42:08.127 --> 00:42:22.375 r is the square root of 58, and that is the sine of theta for these conditions. 00:42:24.100 --> 00:42:36.650 Let's take a look at determining reference angles. The reference angle, denoted by theta sub r, is the 00:42:36.652 --> 00:42:44.160 positive acute angle associated with a given angle theta. Now all angles, except quadrantal angles, 00:42:44.162 --> 00:42:50.916 have an associated reference angle. You can find this angle, this reference angle, because it is the angle 00:42:50.918 --> 00:42:59.166 between the "nearest" x-axis and the terminal side of theta c. And we put "nearest" in quotation marks 00:42:59.168 --> 00:43:09.417 because we really mean the nearest half of the x-axis. So let's take a look at what this means. 00:43:09.419 --> 00:43:28.167 So if I have an angle theta and I find theta c, the coterminal angle, and that angle has a terminal side 00:43:28.169 --> 00:43:42.167 that lies in quadrant one, then the reference angle is equal to theta c, and that's the simplest case. 00:43:42.169 --> 00:44:02.668 Alright, let's look at the situation where theta c has a terminal side that lies in quadrant two. 00:44:02.670 --> 00:44:16.748 The reference angle is this angle because this is the nearest half of the x-axis. We find the measure of 00:44:16.750 --> 00:44:26.001 the reference angle by starting with pi which would be the angle from here to here, and then we subtract 00:44:26.003 --> 00:44:36.021 theta c, and the remainder is then theta r. 00:44:36.023 --> 00:44:53.787 For the case where the terminal side of theta c lies in quadrant three, so this is theta c, this angle, then 00:44:53.789 --> 00:45:03.535 this is the reference angle because again it is the angle between the nearest half of the x-axis and the 00:45:03.537 --> 00:45:15.560 terminal side of theta c. So to find the reference angle in this case, we would take theta c, 00:45:15.562 --> 00:45:26.820 and then we would subtract pi, and what is left would be the reference angle. 00:45:26.822 --> 00:45:44.842 The fourth case would be where the terminal side of theta c lies in quadrant four, and this would be 00:45:44.844 --> 00:46:01.359 the reference angle, and we would find that angle by starting with 2 pi and subtracting theta c. 00:46:01.361 --> 00:46:08.110 And that is how we would find the reference angle in each of the four quadrants. 00:46:10.000 --> 00:46:20.625 Let's take a look at some examples of determining the reference angle. We want to find theta c, then 00:46:20.627 --> 00:46:28.889 we want to find the quadrant in which the terminal side of theta c lies, and then we will find the reference angle. 00:46:28.891 --> 00:46:38.637 So let's look at the first theta. We have negative 41 pi over 6. We have to find theta c, and so one 00:46:38.639 --> 00:46:47.136 thing to do is to realize that one rotation would be 12 pi over 6, and so if you think in multiples of 12, 00:46:47.138 --> 00:46:57.420 you can add to the fraction multiples of 12 until you get a positive number. So what I'm thinking here is 00:46:57.422 --> 00:47:07.428 I could add 12 pi over 6 or 24 pi over 6 or 36 pi over 6 or 48 pi over 6, and as soon as I get to 48, 00:47:07.430 --> 00:47:25.947 I realize that the theta c would be this angle plus 48 pi over 6 because that will give me a positive number, 00:47:25.949 --> 00:47:37.946 and so this would be 7 pi over 6, and that would be theta c. Now I know that 7 pi over 6 has a terminal 00:47:37.948 --> 00:47:53.471 side in quadrant three, so here's my theta c, terminal side in quadrant three, and now I have to find the 00:47:53.473 --> 00:48:09.220 reference angle. Well, this is the reference angle. I can find it algebraically by taking the value of theta c 00:48:09.222 --> 00:48:30.470 and then subtracting pi, so that would be 7 pi over 6 minus pi, which would be 6 pi over 6, and so what 00:48:30.472 --> 00:48:43.770 I get is that the reference angle is pi over 6. As you practice doing exercises of this type, you will come 00:48:43.772 --> 00:48:51.016 to discover that if you have a special angle that's reduced to lowest terms then the reference angle 00:48:51.018 --> 00:48:59.768 is going to be pi divided by whatever the denominator is, which in this case is pi over 6. 00:48:59.770 --> 00:49:11.302 OK, let's look at this example. So I have theta is 24 pi over 9. What's the first thing you notice about this? 00:49:11.304 --> 00:49:17.555 Well, you should notice that both of these numbers are both divisible by 3. So what I'm going to do is 00:49:17.557 --> 00:49:26.565 I'm going to rewrite this angle as 8 pi over 3 by dividing the numerator and the denominator each by 3. 00:49:26.567 --> 00:49:39.329 And now I'm going to find theta c. So one rotation in terms of pi over 3 would be 6 pi over 3, 00:49:39.331 --> 00:49:52.840 that's equal to 2 pi, so I'm going to subtract that out and get 2 pi over 3, and that's theta c. OK. 00:49:52.842 --> 00:50:11.111 2 pi over 3 has a terminal side in quadrant two, and so the reference angle, and from what you learned 00:50:11.113 --> 00:50:17.360 on this example you probably already know that the reference angle is pi over 3, but let's work it out. 00:50:17.362 --> 00:50:31.612 So here, theta r is pi minus theta c, so pi, and we want a denominator of 3 so we're going to rewrite pi 00:50:31.614 --> 00:50:49.148 3 pi over 3 minus theta c, and we are not at all surprised to see that the reference angle is pi over 3. 00:50:49.150 --> 00:50:57.146 On the next example the first thing you might notice is that the denominator is 10, so this is not going to 00:50:57.148 --> 00:51:11.148 be one of the special angles, but we will use the same procedure. So theta c is going to be 37 pi over 10, 00:51:11.150 --> 00:51:27.933 and then we will subtract 2 pi in the form of 20 pi over 10, and we will get 17 pi over 10, and that is theta c. 00:51:27.935 --> 00:51:35.686 Now we don't just know exactly where the terminal side of this lies like we do with special angles so let's 00:51:35.688 --> 00:51:47.185 count. One rotation would be 20 pi over 10, so each quarter of a rotation would be 5 pi over 10. So this 00:51:47.187 --> 00:52:00.967 would be 5, 10, 15, and all the way around would be 20, so somewhere in about here is 17 pi over 10. 00:52:00.969 --> 00:52:18.220 And so the reference angle is here, and we would find the value of it by writing 2 pi and then subtracting 00:52:18.222 --> 00:52:30.753 theta c. Now since our denominator is 10, we're going to write 2 pi again as 20 pi over 10, and then we're 00:52:30.755 --> 00:52:44.750 going to subtract 17 pi over 10 and what we get is the reference angle is 3 pi over 10, and at this point 00:52:44.752 --> 00:52:51.497 you realize that your reference angle is not always going to just be pi over the denominator because in 00:52:51.499 --> 00:53:00.785 this case it is not. Let's take a look at an example when theta is measured in degrees. Same process. 00:53:00.787 --> 00:53:13.776 We look for a coterminal angle. We're going to subtract in this case 360 degrees, and when we 00:53:13.778 --> 00:53:28.306 subtract we get 60 degrees, and that's theta c. Well we know that 60 degrees has a terminal side 00:53:28.308 --> 00:53:42.804 in quadrant one, and in this case the reference angle is the coterminal angle which is 60 degrees. 00:53:44.898 --> 00:53:52.077 Before we can evaluate trigonometric functions of angles belonging to the pi over 6, pi over 4, or pi over 3 00:53:52.079 --> 00:54:04.341 families we need to make an observation. Notice here that all four of these angles are the angles that 00:54:04.343 --> 00:54:17.106 belong to the pi over 6 family of special angles. Notice that they all have the same reference angle, 00:54:17.108 --> 00:54:31.866 that being pi over 6. Notice that because the tangent is defined as y over x we can label a point on the 00:54:31.868 --> 00:54:41.880 terminal side of pi over 6 as square root of 3 comma 1. By that same token, we can label points on the 00:54:41.882 --> 00:54:50.634 terminal sides of 5 pi over 6, 7 pi over 6, and 11 pi over 6, and from there we can find the trig functions, 00:54:50.636 --> 00:54:58.146 in particular the tangent, of those three angles. Now what do we observe? 00:54:58.148 --> 00:55:07.156 We observe that the only thing that differs here is the plus or minus sign. So in order to find 00:55:07.158 --> 00:55:13.654 the trigonometric functions of angles belonging to special angle families, all we really need to know 00:55:13.656 --> 00:55:20.654 is the reference angle, and then we need to know the quadrant that the terminal side of the angle lies in 00:55:20.656 --> 00:55:26.652 because that will help us to determine whether the trig function is positive or negative. 00:55:28.400 --> 00:55:35.685 In order to evaluate trigonometric functions of angles belonging to the pi over 6, pi over 4, or pi over 3 00:55:35.687 --> 00:55:43.937 families, you will need to use many of the things you have learned earlier in this section. First we're 00:55:43.939 --> 00:56:02.961 going to find the cosine of 7 pi over 6. I will sketch 7 pi over 6. I will see that the terminal side of 00:56:02.963 --> 00:56:12.718 7 pi over 6 lies in quadrant three. I know that the cosine for an angle that has a terminal side in 00:56:12.720 --> 00:56:35.491 quadrant three will be a negative number. I know that the reference angle will be pi over 6. Therefore, 00:56:35.493 --> 00:56:51.501 the cosine of 7 pi over 6 is equal to negative cosine pi over 6. I know that the cosine of pi over 6, 00:56:51.503 --> 00:57:01.502 from either the special angle, so this is pi over 6. I know the cosine is adjacent over hypotenuse, 00:57:01.504 --> 00:57:07.272 square root of 3 over 2, or if you prefer to use the table, the cosine of pi over 6 here 00:57:07.274 --> 00:57:14.275 is the square root of 3 over 2. So I substitute the square root of 3 over 2 here, and I get that 00:57:14.277 --> 00:57:28.043 the cosine of 7 pi over 6 is equal to negative square root of 3 over 2. Now let's find the cosecant 00:57:28.045 --> 00:57:47.315 of 2 pi over 3. 2 pi over 3 has a terminal side that lies in quadrant two. Now the cosecant is going to have 00:57:47.317 --> 00:58:04.814 the same sign, s-i-g-n sign, as the sine function has in quadrant two which is positive. The reference angle 00:58:04.816 --> 00:58:18.318 is pi over 3, and now we have to be careful because we are going to write the cosecant of 2 pi over 3 00:58:18.320 --> 00:58:37.110 is equal to the cosecant of the reference angle, and that is equal to 1 over the sine of pi over 3. I'm going 00:58:37.112 --> 00:58:45.615 to go to the table, and I'm going to read the sine of pi over 3 is the square root of 3 over 2. That goes in the 00:58:45.617 --> 00:59:01.616 denominator, and so what I have is 1 over the square root of 3 over 2. And so when I simplify this expression, 00:59:01.618 --> 00:59:21.860 I'm going to get 2 over the square root of 3. We now want to evaluate the cosine of negative 4 pi over 3. 00:59:21.862 --> 00:59:31.420 Negative 4 pi over 3, well this would be negative 3 pi over 3, so negative 4 pi over 3 would have 00:59:31.422 --> 00:59:41.680 a terminal side that lies in quadrant two. The cosine in quadrant two, for an angle that has a terminal side 00:59:41.682 --> 01:00:01.948 in quadrant two, is going to be a negative number. The reference angle will be pi over 3. Now the cosine 01:00:01.950 --> 01:00:13.964 of negative 4 pi over 3 is equal to the cosine of the reference angle, but I have to incorporate 01:00:13.966 --> 01:00:24.977 the negative here, so let's squeeze that in here, and then put pi over 3. So I get negative, and then 01:00:24.979 --> 01:00:33.700 the cosine of pi over 3 is 1/2. 01:00:35.000 --> 01:00:41.734 Here are two more examples of evaluating trigonometric functions of angles belonging to one of these 01:00:41.736 --> 01:00:54.250 families. We want to find the tangent of 13 pi over 4. We want to sketch 13 pi over 4 and determine the 01:00:54.252 --> 01:01:02.750 quadrant that the terminal side of the angle lies in. We know that one rotation is 8 pi over 4, so this would be 01:01:02.752 --> 01:01:06.764 8 pi over 4. And then what does that leave? 01:01:06.766 --> 01:01:20.016 Well 13 pi over 4 minus 8 pi over 4 leaves 5 pi over 4, and so the terminal side of this angle lies in 01:01:20.018 --> 01:01:35.514 quadrant three. And in quadrant three the tangent will be positive, and the reference angle will be 01:01:35.516 --> 01:02:02.570 pi over four. And so the tangent of 13 pi over 4 is equal to positive tangent of pi over 4. 01:02:02.572 --> 01:02:12.328 And according to the table, or from the triangle, the tangent of pi over 4 is equal to 1. 01:02:12.330 --> 01:02:24.094 We want to evaluate the cotangent of negative 19 pi over 6, so I'm going to draw negative 19 pi over 6. 01:02:24.096 --> 01:02:38.592 So this would be negative 12 pi over 6, and then another negative 7 pi over 6 would give us a terminal 01:02:38.594 --> 01:02:48.591 side in quadrant two. Another way to look at this would be to take [negative] 19 pi over 6, add 01:02:48.593 --> 01:03:01.126 12 pi over 6 twice, which is 24 pi over 6, and get 5 pi over 6, which is a good way to check to see if you've 01:03:01.128 --> 01:03:08.374 done this correctly. Once we have this and we know we want the cotangent, then the tangent in this 01:03:08.376 --> 01:03:26.629 quadrant would be negative, and the reference angle would be pi over 6. So the cotangent of 01:03:26.631 --> 01:03:45.424 negative 19 pi over 6 would equal negative cotangent of the reference angle, so this is equal to 01:03:45.426 --> 01:03:58.923 negative 1 over the tangent of pi over 6, where the tangent of pi over 6 is 1 over the square root of 3. 01:03:58.925 --> 01:04:12.900 So what we get is negative 1 over 1 over the square root of 3 which simplifies to this.