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About this Lesson
 Type: Video Tutorial
 Length: 6:46
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 72 MB
 Posted: 01/22/2009
This lesson is part of the following series:
Trigonometry: Full Course (152 lessons, $148.50)
PreCalculus Review (31 lessons, $61.38)
Trigonometry: Trigonometric Identities (23 lessons, $26.73)
Trigonometry: DoubleAngle Identities (3 lessons, $4.95)
Doubleangle identities allow you to simplify trigonometric equations with a 2 as the coefficient. (similar formulae exist for trig functions with 1/2 or 3 as the coefficient). In this lesson, Professor Burger uses the equation cos2x = sinx as an example. If this equation were simply cos x = sinx, we could divide to rewrite the formula as sinx/cosx = tan x = 0, but in this case, we have a coefficient in advance of one of the arguments, which is why we need to use the doubleangle formulas. After using the doubleangle formulas in the provided example to simplify, you can further simplify these equations using trig identities (like the Pythagorean identity) and factoring. These tools will help you to solve many trig equations. The duble angle identities for sine, cosine, tangent and cotangent are: sin2x = 2sinxcosx, cos2x = cos^2xsin^2x, tan 2x = 2tanx/(1tan^2x), and cot2x = (cot^2x1)/2cotx.
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, Precalculus. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/precalculus. The full course covers angles in degrees and radians, trigonometric functions, trigonometric expressions, trigonometric equations, vectors, complex numbers, 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
 Thinkwell
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11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
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Using DoubleAngle Identities
All right, let’s take a look at how we could actually use this doubleangle formula business to actually start solving even more exotic trig equations. I know that’s what you live for, solving these trig equations. Suppose I want to solve this one: cos2sinxx=, and I only want to find the solutions for x inside of zero to 2?. So that means I just want the angles between zero and 2? and nothing else. Well, you see, this one is a little bit annoying. Your first instinct maybe is to divide by the cosine, so I get sincos, which is tangent. But see, they’re not the same angle now. So sincos2xx is not tangent of anything. It’s just a big mess. So this is actually pretty scary. How would you handle it?
Well, now we’re armed with a formula for doubleangles for cosine, so I can actually take cos2x and convert it into just cosines and sines of x’s. So let’s do that. So using the doubleangle formula for cosine, I see this actually equals 2cossinxx?, and now that equals, by this thing, sinx. Well, now we’re in pretty good shape because now everything is just x’s. There are no 2x’s here. I’m a little bit troubled and disturbed by the fact that this is cosine stuff and everything else seems to be sine. I would love everything to be uniform, either sine or cosine.
Well, I can’t change that to cosine. That seems almost impossible. Could I change this to sine? Well, yes, because now I remember the fundamental Pythagorean identity, which is equivalent to saying that 22sincos1xx+. So what does cosine squared equal? Well, if I bring this term over to this side, I see that 2cos1sinxx=?. So in place of cosine squared here, I can replace it by . So let’s do that. If I do that, I now see, in place of this I write . And don’t forget that term, 21sin?21sin?2sinx?. And that’s still equals sinx. So notice that now everything just has x’s, no 2x’s, and everything happens to the sine. So in fact, that’s a good sign.
All right. Now, you may think these cancel out but be careful, they don’t because they have a and a , and that’s . So in fact those squared terms remain. Let’s bring everything over to this side. So I’ll bring the 2sin?2sin?22sin?22sinx? to this side and it becomes 22sinx+, so I have a 22sinx+. I have that term, sinx. I’ll bring everything over to this side. I bring that over. I see a 1. And what does that equal? Well, that would equal zero.
Well, now what you might want to do is just call this “something” in order not to confuse the issue because I’m going to try to factor now. Or you may just begin to start to think about the “something” at the same time as actually factoring. So I’m going to put a sinx here and a sinx here. I’ll put the 2 here. That product is 22sinx. Now I want the terms I multiply at the end to give 1, so I’ll put in a 1 and a 1. And the sines should be opposite. But how should I put them in so that when I do this inside and outside, I get a +1? Well, I think if I put a plus sign here and minus sign here, it’s going to work well. Because here I see 2sinx+, minus a sinx gives me a sinx+. So in fact this actually works out great.
So I factored this in one fell swoop, and so what does that mean? Well, it means that now what I can do is set up two equations. One is that this thing equals zero, and the other is that this equals zero. So let’s just rewrite that up here. So I see (. That’s what I’m left with now. So either this equals zero—if 2sin1)(sin1)0xx?+2sin1x? equals zero, that means that sinx must equal 12, right? Because I bring this over to the other side with a +1 and divide both sides by 2, so I’d see this. Or the other possibility is that this equals zero, which means sin. 1x=?
Well, now we can find the solutions. And remember, we’re only looking for solutions between zero and 2?, so you have to look only inside here. Well 1 is an easy one. There’s only one answer. It equals 1 right here at 32?, so that’s easy. Here 32x?=.
Well, what about this one? This one is a teeny bit more work but not too much more work. At 12, it crosses the half here but also here, so there will be two answers to this one and they’re going to be both symmetric to this thing, so whatever I have here, I’ll take ? minus that here. And what’s the answer here? Well, the sine of what angle is 12? Well, I happen to know that’s actually going to be 30° or 6? radians. So I see that 6x?= radians. That’s this answer right here gives me 12. And how do I get the one that’s sort of symmetric on this side? Well, l take this length, which is 6?, and subtract it from ?. I move off that much this way. So I take ? and I subtract 6?, and ? is actually 66?; that’s just ?. But minus 6? gives me 56?. So the other answer is 56x?=.
So this equation actually has three solutions: 6x?=, 56x?=, and 32x?=. Those are all solutions to this original question of where the x is found in this region. So notice what I did. I had this thing that had cosines and sines but, more fatal, it has a 2x, and an x. So I got rid of that by using the doubleangle formula. Then I didn’t like the cosine so I got rid of that using the Pythagorean theorem formula. And then I set up this nice thing, which turned out to be just a factorable quadratic. This had two solutions; this had one solution; and they’re all the solutions in that range.
Okay, great. So you can actually solve these kinds of exotic equations now using the doubleangle formula.
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