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About this Lesson
 Type: Video Tutorial
 Length: 11:19
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 122 MB
 Posted: 07/01/2009
This lesson is part of the following series:
Trigonometry: Full Course (152 lessons, $148.50)
Trigonometry: Trigonometric Functions (28 lessons, $26.73)
Trigonometry: Trig Functions (4 lessons, $6.93)
Taught by Professor Edward Burger, this lesson comes from a comprehensive course, Trigonometry. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/trigonometry. The full course covers trigonometric functions, trigonometric identities, application of trig, complex numbers, polar coordinates, exponential functions, logarithmic functions, conic sections, and more.
Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.
He has also taught at UTAustin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".
Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, padic analysis, the geometry of numbers, and the theory of continued fractions.
Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.
About this Author
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11/13/2008
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Evaluating Trigonometric Functions Using the Reference Angle
You know, once you get down where the positive terms are and the negative terms of the trig functions are, if you remember this little thing about "all students take calculus", it's just a silly little thing, but then "all students take calculus" tells you that all the trig functions here are actually going to be positive, here only the sine and it's flip which is cosecant, are positive. Here, "t" take, tangent and cotangent are positive. And here, cosine and secant will be positive. And everyone else will be negative.
Just knowing those things, in fact, allows us to find the trig values of all sorts of angles that are huge, just by using that fact and this notion of a reference angle. Now, a reference angle is basically what we've already been thinking about, but now let's make it sort of more precise. So, let's first of all find the cosine of 120. Well, where's 120? Well, 120 is going to be somewhere that's bigger than 90, but less than 180, so it sort of looks like this. So, this angle here, is 120. And I want to find the cosine of it. Well, how can I do that? Well, what I do is I look at the reference angle, and the reference angle is just the smaller angle that is made with the axis and this terminal ray. So, this is 120, this whole thing, we know, is 180, so what does that force this thing to be? It forces that to be 60. So, in this case, the reference angle is actually 60. So, here the reference angle is 60.
So, to find the cosine of this, remember now I'm just looking at this triangle so the cosine of this will actually be the cosine of 60. Because I'm just looking at this triangle. However, I've got to be careful of which quadrant I'm in, so I have to see whether it should be positive or negative. Well, let's think about it  "all students take calculus." So, the only things that are positive here is actually sine. This is cosine, so, in fact, the answer should be negative. And you can see that. Remember that cosine is adjacent over hypotenuse and adjacent here is negative. I'm to the left of the origin. So, this is negative. In fact, what this equals is negative cosine of the reference angle. So, in fact, I can actually start computing the cosines of large things by first finding the reference angle. That's the angle  and in fact these angles you'll notice, by the way, are supplemental, right? They add up to 180. And I take the small one and I can sort of now look at the right triangle formed there, and this reference angle, but I've got to be careful of the sine, depending upon which quadrant I'm in, in this case, it's going to be negative because I'm in the second quadrant, cosine is negative there. Well, now, what's cosine of 60. If you're not quite happy with that, that's okay, too. Because there's other ways of thinking about it. What we can do is think about it as follows. We can say, well, if it's 60, what kind of triangle do we have here? This is 60, this is 90, that forces this to be 30. And you can recognize this and say, "Oh, wait a minute, okay," so if this one of those ones where if I continue, this is going to be 60, 60 and 60, so, in fact, if this length were to be two, this would be half of that length, which is a 1, and then I can use Pythagorean Theorem and see this is the square root of 3, and then I can just read off  I want the cosine. So, cosine would be adjacent over hypotenuse, so it would be , but there's a negative sign in front, so . Cosine of 120 is .
Let's try another one. How about tangent of 300? So let's find the tangent of 300. That's a lot of degrees. Let's draw a picture. Remember this is zero degrees, this is 90 degrees, this is 180 degrees, this is 270 degrees, and this is 360 degrees. So, it's going to be somewhere in here. What do I do? Well, there's the angle, 300. And so, what do I want to do? Well, what I'll do is I'll look for the reference angle. What's the reference angle? It's this little angle right in here and that angle is going to be what? This whole thing is 360, this is 300, so this must be 60. So, in fact, this reference angle here is 60. So I can now think about this triangle, this right triangle, that has an angle of 60. And in fact it's exactly what we just had. What do I want? I want tangent. So what's tangent. Well, the tangent of 300  first of all, in the fourth quadrant, will tangent be positive or negative? "All students take calculus." So, only the cosine and secant are positive here. So, therefore, tangent will be negative. So, it's negative and then the value of tangent at the reference angle, which is . So, do you see how I can stick in the reference angle, but I have to be careful of the sign. I've got to think about where I'm landing. And now, with the tangent of this 60  tangent of 60 is going to be what? It's going to be opposite over adjacent, and so what do I see here? It's . But don't forget the negative sign because I'm in the fourth quadrant, and so, in fact, the answer is . That is the tangent of 300.
Look what I did. What I first did was I found the reference angle  that smaller angle that takes me up to or down to the axis, and then I have a little triangle here, and now I just can find the tangent of that angle, and put on the appropriate sign.
One last one. How about this one? Cosecant  and now I'm going to give you radian measure just to sort of get you in the mood, . Cosecant of . What's the first thing I think of? Well, first of all, I want to get rid of the cosecant stuff. So, that equals just one over sine of . That got rid of the cosecant. Now I've got to find the sine of that. Now that angle may be a little bit disturbing to you, but let's get in the habit of thinking about things in terms now of radians. So this is zero radians. This angle here is radians. Halfway around is radians. This is radians, and this is 2 radians. So, where is ? Well, you could actually sort of count it out if you wanted to, because you could think about it as follows. You could say, well, let's see . Well, this is right here, because it's just half of , 45, so it's right there, and then this is , and therefore, this would be . So, looks just like this. This angle here is . What would the reference angle be? Well, this whole thing is . This is , so how many 's is this? Well, one . + is . So, in fact, this is just right here. The reference angle is . What is ? That's 45 and you convert it to 45. So, what I see here is the reference angle is , so what do I see over here? Well, here what I see, continuing this, is one over  now, sine, is sine positive or negative in the second quadrant? "All students take calculus." Well, students, "s", in fact, the sine is positive here. So, even though it's way over here, the sine will be positive, so in fact I can just now put in sine of the reference angle, . Because I know it's positive, so these two things will actually be the exact same value. There's no negative difference. I just compute the reference angle there, and I'm set. Why? Because sine in this case is going to be opposite over hypotenuse, and opposite is positive here. So, it would be the exact same as if we were in the first quadrant. Now, what is sine of ? Well, it's going to be the sine of 45 and you can do that in a variety of ways. You can just think about it, 45, that means that if this is 45, or let me start using , then this would be , too. So, these two things would be the same. So this has length 1, this has length 1, and this has length the , so what is the sine? The sine would be opposite over hypotenuse, so this would equal one over opposite, one over hypotenuse, . But don't forget the big one over, over here. So, I've got to invert this, and multiply. So that would give me now the square root of two on the top and so what I see is the cosecant of is actually the . How did I find it? I first graphed this thing roughly. I found the reference angle, and I know that the cosecant of will equal the cosecant of the reference angle, except there may be a sine in front of it. And I determined that by thinking about what's positive here, but here I realize that both sine and therefore, cosecant, are both positive here, so in fact, there's no sign difference, and therefore I just compute away and I've got it. If that angle were way over here, but still 45 from this point, then the number would be the same, but now I'd have a negative sign, because here the cosecant and the sine would be negative. So you just compute the reference angle and you find the trigonometric function of the reference angle but then put in a plus or minus sign in front, depending upon which quadrant you're in and depending upon which trig function you're looking at.
Okay, so that gives you a sense of actually how to compute trig functions of any size angle at all. You just find the quadrant, see if you're positive or negative, and then find the trig function of a simple reference angle, using the old methods. So, you can see the importance of knowing those special trig functions for the acute angles.
All right, well try these and see how you make out and just take your time, take it easy. And enjoy.
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Clear explanation: Using Reference Angles & CAST rule (or mnemonic "Every Student Takes Calculus") to find trig function values in 2, 3 & 4 quadrants!