WEBVTT
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In preparation for evaluating trigonometric functions of general angles,
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it is important to group the special angles into families.
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The first family we will look at is the quadrantal family.
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The first angle in the quadrantal family has a terminal side along the positive x-axis.
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This angle has a measure of 0 radians, also 0 degrees.
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The next quadrantal angle has a terminal side here, and it has a measure of
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pi over 2 radians or 90 degrees.
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The next angle has a terminal side here, and it has a measure of
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pi radians or 180 degrees.
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And then the fourth quadrantal angle we will look at has a terminal side here,
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and it has a measure of 3 pi over 2 radians or 270 degrees.
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Now there are infinitely many quadrantal angles, but they are all coterminal
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with one of these four angles.
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The next family we will look at is the pi over 3 family.
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Alright, the first angle is pi over 3. So pi over 3 sits about here, has a
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terminal side in quadrant 1, and has a measure of pi over 3 radians or 60 degrees.
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The next angle in the pi over 3 family, let's draw it, has a terminal side about here,
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and it has a measure of 2 pi over 3 radians or 120 degrees.
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Now watch what happens next. If we add another pi over 3, we would
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get an angle that has a terminal side on this axis, but we can see that that angle
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already belongs to the quadrantal family. So we skip 3 pi over 3,
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and we move to this angle which has a measure of 4 pi over 3,
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which is equal to 240 degrees. Now we add another pi over 3. This angle
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here has the measure of 5 pi over 3 or 300 degrees.
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And we can check this because if we add another pi over 3 we get right back to
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one rotation, which is 2 pi. So now we have the four major angles in
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the pi over 3 family, and then there of course are infinitely many others,
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but they are all coterminal with one of these four. OK.
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Next, we look at the pi over 4 family, first angle pi over 4 or 45 degrees.
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Now if you add another pi over 4 you would get to 2 pi over 4, but when you
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reduce that, you get pi over 2, and once again that angle is already a quadrantal angle.
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So now we have 1 pi over 4, 2 pi over 4,
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so the next angle in this family would be 3 pi over 4 which is 135 degrees.
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OK, now we add another pi over 4 and you see what's going to happen again.
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We get an angle that's coterminal with any angle on this axis, and of course
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once again that's here. So we add another pi over 4 and you see the routine.
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So we have pi over 4, 2 pi over 4, 3 pi over 4, 4 pi over 4, and now this is
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5 pi over 4 which is 225 degrees, and that's this angle.
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And then the last one is going to be this one which is 7 pi over 4 or 315 degrees.
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Once again, there are an infinite number of angles that are in the pi over 4 family,
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but they are all coterminal with one of these four.
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Now we look at the pi over 6 family. OK, first angle, pi over 6.
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So this is pi over 6 which is 30 degrees. Now another pi over 6 would be 2 pi over 6,
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but that's pi over 3. And then we have 3 pi over 6 which is here,
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and then 4 pi over 6 which is here. And so the next angle
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we're actually going to identify as being part of this family is 5 pi over 6
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which is 150 degrees. Then we'd have 6 pi over 6 which we don't identify
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because again that's pi. Then this one is 7 pi over 6 which is 210 degrees.
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Then we have 8 pi over 6, 9 pi over 6, 10 pi over 6, and those are all
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already represented, and then the last one is 11 pi over 6 which is 330 degrees.
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You need to practice drawing these angles and get used to identifying
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which of these special angles fit into which family.
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Let's take a look at identifying the angle family that a given angle belongs to.
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We will sketch the given angle. We will find theta c, the coterminal angle
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which is the angle of least non-negative measure coterminal with the given angle.
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And then we will identify the special angle family. The first one we're going to look at
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is the given angle theta is 33 pi over 6. Hopefully the first thing you will notice about this
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is that 33 and 6 have a common factor of 3, and we need to simplify this fraction
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before we look at anything else. So we're going to divide the numerator and the
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denominator each by 3, and we will get 11 pi over 2. Now we will sketch theta.
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Now let's look at what we have. Remember that one rotation in terms of pi over 2
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would be 4 pi over 2. OK, so what we're thinking is that 2 pi is 4 pi over 2.
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So to find theta c for 11 pi over 2 we'd have 4 pi over 2, 8 pi over 2, and then
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9, 10, 11 pi over 2. You could do this algebraically by just repeatedly subtracting
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4 pi over 2 from the given angle until you get an angle of least non-negative measure
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which would be in this case 3 pi over 2. So that is theta c. Once we look at this
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we realize that this is a special angle and it belongs to the quadrantal family,
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and so when we identify the angle family we would write the word quadrantal.
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Lets take a look at another example. So here we have theta equals negative
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840 degrees. First I'm going to draw this angle. Now because it's a negative angle,
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I start here going this direction, and each rotation is going to be negative 360 degrees.
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So this would be negative 360, negative 720, and then another negative 120.
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So this part is negative 120 degrees which means that this part would be
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positive 240 degrees. You can always check this algebraically by adding
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360 degrees repeatedly to this angle until you get the angle of least positive measure
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which is theta c. So here we have theta c is 240 degrees, and the angle family
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is pi over 3 or in this case we would say 60 degrees.
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In order to create the definitions of the trigonometric functions of general angles,
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we start by reviewing the right triangle definitions of trigonometric functions.
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So we have the sine of theta is opposite over hypotenuse,
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cosine of theta is adjacent over hypotenuse,
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and tangent of theta is opposite over adjacent. We know that the other three functions
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are reciprocals, each one of one of these. Notice here that theta is an acute angle.
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To extend the right triangle definitions of the trigonometric functions to include all
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general angles, that means angles with positive measure, negative measure, zero
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measure, and no limit in size, we take a look at a point on the terminal side of the angle.
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So here's theta and I pick a point on the terminal side, and that point has coordinates x, y,
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and I look at the length of this line segment. I choose to call it r, and looking at this
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right triangle, I can see that r is equal to the square root of x squared plus y squared.
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Now using the right triangle definitions for theta, and let's move this label for theta over
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to here. Much better. Then I can see that the sine of theta would be opposite, but
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if you look at this as the coordinate, this is actually the y-coordinate, and then of course
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this is the x-coordinate. So let's go back. The sine of theta is opposite over hypotenuse.
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OK, so the sine of theta is opposite over hypotenuse, and so what we use here
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are the coordinates of the point. By the same process, I can see that the cosine of theta,
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adjacent over hypotenuse would be x over r, and then
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the tangent of theta would be y over x, which is opposite over adjacent.
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Alright, now let's take a look at an angle that has a terminal side that is not in quadrant 1.
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So let's try quadrant 2. So here's theta. Here is a point on the terminal side of theta.
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The length of this line segment, r, is again equal to the square root of x squared
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plus y squared, and we can see now that we can use this same representation,
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but this time using the coordinate of the points. And we can find the representations
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of the other three trig functions using reciprocals. So let's write this all down.
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So the sine of theta for any angle is y over r, the cosine is x over r,
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and the tangent of theta is y over x, where x and y are the coordinates of the point
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on the terminal side of theta. Notice here that x cannot equal 0.
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Now the cosecant of theta, using the reciprocal identity, is r over y where
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y does not equal 0. The secant of theta is r over x where x does not equal 0,
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and finally, the cotangent of theta is x over y where y does not equal 0.
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And so now we have the definition of the six trigonometric functions for general angles.
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Let's take a look at an example: the point P, with coordinates (12, -10).
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This point lies on the terminal side of theta. We want to find the exact values of the
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six trigonometric functions of theta, and here are the definitions of the trigonometric
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functions for general angles. So the first thing we're going to do is we will
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plot the point (12, -10), so 12 is, we'll say, out here somewhere.
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Negative 10 is about here,
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so the point is about here. Then that means that theta has a terminal side that lies
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in quadrant 4, represented by this positive angle. In this situation, we know
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that x, the x-coordinate, is 12, and the y-coordinate is negative 10.
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The piece we're missing is r. OK, so we're going to find r. We know that
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r is the square root of x squared plus y squared, so r is the square root of
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12 squared plus negative 10 squared. So r is the square root of 144 plus 100,
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so that's 244. Now we must simplify this radical which means we have to look
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to see if there is a factor here of 244 that is a perfect square. It's pretty obvious, I think,
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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.
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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
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at this, we see that we have a common factor of 2, so we're going to simplify this
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expression and get [negative] 5 over the square root of 61. Then we have the cosine of theta is
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x over r, so that's 12 over 2 times the square root of 61,
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which is 6 over the square root of 61. And then the tangent of theta is y over x,
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so this is y over x and when I reduce this I get [negative] 5/6. So here's the sine, the cosine,
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and the tangent for theta. And now I can find the other three values by taking
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the reciprocals. So the cosecant of theta would be negative square root 61 over 5.
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The secant of theta would be square root of 61 over 6, and the cotangent of theta
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would be negative 6 over 5. And so now we have found the exact values of the six
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trigonometric functions of theta having a point on the terminal side (12, -10).
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Let's find the values of the trigonometric functions for quadrantal angles.
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Let's start with the first quadrantal angle. This would be an angle with the terminal side
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on the positive x-axis, and what we will do is we will pick any point on this terminal side.
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And you can see that a point here will have a coordinate of x, and then
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the y-coordinate would be 0. So this would be our point P. You can also see that the length
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of this line segment, which we usually call r, in this case is actually going to have
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the value of x. So from this, using our definitions, we can find the sine of 0, the cosine of 0,
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the tangent of 0, and then using the reciprocal property, we can find the other three values.
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So let's see. The sine of 0, from the definition, is y over r, so it would be 0 over x,
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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.
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Then the tangent of 0 would be, from the definition, y over x. y is 0, x is x, and we get 0.
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OK, what we can do now is find the other three using the reciprocal property.
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Well, that gets pretty interesting with the 0. So let's take a look at the definition.
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The cosecant of 0 would be r over y which is going to be x over 0. So what we see here
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is that the cosecant of 0 is undefined.
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The secant of 0 would be r over x, and that's x over x which is 1, and that's no surprise
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because the reciprocal of 1 is 1. And then the cotangent of 0 is x over y, so that's x over 0,
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and we have the same situation here, and so the cotangent of 0 is undefined. Now what we
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can do is we can find the trig functions for pi over 2, pi, 3 pi over 2 by the same method.
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Let's find the values of the trigonometric functions for pi.
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So here theta is pi or 180 degrees. A point on the terminal side of pi
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would have coordinates x, which is going to be a negative value, and then the y-coordinate
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will be 0. Now the length of this line segment is r, and r is a positive number.
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But when we identify the values of x, y, and r for this angle we know that x is equal to x,
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y is equal to 0, but r has to be positive and we know that x is negative.
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So in order to make this positive, we realize that r has to be the opposite of x.
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Now we use the definition, so the sine of pi is equal to y over r which is 0. The cosine of
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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.
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The cosecant of pi is going to be r over y and this is undefined. The secant of pi is r over x,
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and the cotangent of pi is x over y, so that's x over y which is undefined.
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By continuing in a similar manner, we are able to find the values of the trigonometric
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functions for all of the quadrantal angles. Here they are in a table.
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Let's take a look at an example. Find the exact value of the tangent of negative 7 pi over 2
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without a calculator. So first of all, we have to find the terminal side of negative 7 pi over 2.
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We know that one rotation in this direction is negative 4 pi over 2, and then this would be
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negative 5, negative 6, negative 7 pi over 2. So we see that the terminal side
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of this angle lies on the positive y-axis. This is a quadrantal angle. We will pick
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a point on the terminal side here, and we will label this point (0,y). And now we will also
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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.
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So I'm going to identify x is 0, y is actually the value of y, and r is also y.
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I want to find the tangent of negative 7 pi over 2, and so I use the definition y over x and I get
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y over 0 which we know is undefined. So I can't really find the exact value of this
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expression, but I can say that it is undefined.
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Let's find the signs, that's the s-i-g-n-s signs, of the trigonometric functions by quadrant.
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Well, since r is always positive, the sign, the s-i-g-n sign, is dependent completely on
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the sign of x and the sign of y, based on the point on the terminal side of the angle.
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So if we look at the four quadrants, one, two, three, and four, let's think about the values
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of x and y. In this quadrant, x and y are both positive.
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If the angle has a terminal side in this quadrant, think about what you would get.
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Well, you would get a negative value for x and a positive value for y.
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If the angle has a terminal side in this quadrant, both x and y will be negative numbers,
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and here x will be a positive value and y will be negative. Let's take a look at each quadrant,
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one at a time. Quadrant one: In quadrant one, x, y, and r are all positive, and therefore,
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for an angle that has a terminal side in quadrant one all of the trig functions will be positive.
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In quadrant two: If an angle has a terminal side in quadrant two, x will be a negative number,
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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
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be represented by a positive number over a positive number which would yield a
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positive number. The cosine of an angle that has a terminal side in quadrant two would be
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represented by x over r where x is a negative number and r is a positive number,
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and a negative number divided by a positive number would yield a negative number.
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OK, the tangent of theta where theta has a terminal side in quadrant two would be y over x
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where y is represented by a positive number and x is represented by a negative number,
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and the result would produce a negative number. So what we see here is that
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if an angle has a terminal side in quadrant two, the sign, the s-i-g-n sign, would be for the
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sine function positive, for the cosine function negative, and the tangent function negative.
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So what that means is that the cosecant for an angle that has a terminal side in quadrant two
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would also be positive because these are reciprocal functions. For the secant negative,
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and the cotangent negative. We can continue
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in a like manner for angles that have terminal sides in quadrants three and four.
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After finding the signs of the trigonometric functions in the remaining quadrants,
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this is the result. We see that if an angle has a terminal side in quadrant one, all six trig
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functions will have a positive value. If an angle has a terminal side in quadrant two we see
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that both the sine function and the cosecant function will be positive and the other four
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functions will have a negative value. If an angle has a terminal side in quadrant three, both
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the tangent and the cotangent will have positive values. The other four will have negative
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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.