Simplifying Trigonometric EquationsDate: 06/02/98 at 10:13:08 From: Chris Keenan Subject: Trigonometric equations I am a student at Portsmouth high school in NH and our math teacher asked us to answer the following problems with help from the Internet. We found you, so here goes. (He didn't want us to try it ourselves.) Problem 1: cos 2x cos 2x --------------- - --------------- - csc 2x sec 2x - tan 2x sec 2x + tan 2x Problem 2: sin x tan(x/2) = --------- where 0 < x < (Pi/2) 1 + cos x I hope you can help. Thanks. Chris Date: 06/02/98 at 12:57:16 From: Doctor Barrus Subject: Re: Trigonometric equations Hi, Chris! Well, these look like some interesting problems. Since it looks like none of the other math doctors has answered you yet, I guess I'll give it a shot. Problem 1: I'm assuming that the problem is to simplify this as much as possible. The first thing I usually do in problems like this is to write everything in terms of sines and cosines. I do this because I know and can derive more identities involving sines and cosines, and textbooks give more identities for sines and cosines than for other trigonometric functions. So let's do that: cos 2x cos 2x 1 ------------------- - --------------------- - -------- 1 sin 2x 1 sin 2x sin 2x ------ - -------- -------- + -------- cos 2x cos 2x cos 2x cos 2x Now I'll simplify the denominators: cos 2x cos 2x 1 ---------------- - ---------------- - -------- 1 - sin 2x 1 + sin 2x sin 2x ------------ ------------ cos 2x cos 2x and simplify the fraction division: (cos 2x)^2 (cos 2x)^2 1 (^2 means raised to ------------ - ------------ - -------- the second power) 1 - sin 2x 1 + sin 2x sin 2x Next I'll get a common denominator for the first two terms, and add: (1 + sin 2x)(cos 2x)^2 - (1 - sin 2x)(cos 2x)^2 1 ------------------------------------------------- - -------- (1 - sin 2x)(1 + sin 2x) sin 2x Then I'll simplify the expressions in the numerator and denominator of the first term a bit: [1 + sin 2x - (1 - sin 2x)](cos 2x)^2 1 --------------------------------------- - -------- 1 - (sin 2x)^2 sin 2x and then a little more: (2 sin 2x)(cos 2x)^2 1 (Here I used the identity ---------------------- - -------- (sin x)^2 + (cos x)^2 = 1) (cos 2x)^2 sin 2x and some more 1 2 sin 2x - -------- sin 2x Now I have something that really looks good. I'll put everything over a common denominator and subtract: 2(sin 2x)^2 - 1 ----------------- sin 2x Looking at this, I'd say that this looks similar to something in the identity: cos 2k = 1 - 2(sin k)^2 If I put 2x in everywhere there's an k, and multiply by -1, here's what I get: -cos 4x = 2(sin 2x)^2 - 1 So now I can reduce what I have to: cos 4x - -------- sin 2x This looks a ton better than what we started out with, doesn't it? At this point I would say that this is the simplest form I can write this in. Of course, as with most trigonometric expressions, there are lots of ways to write this in other forms. For example, we could write this particular expression as -(cos 4x)(csc 2x), so the exact form of your answer is going to be somewhat a matter of personal taste. Since I like sines and cosines, I'm going to keep my answer as it is. So cos 2x cos 2x cos 4x --------------- - --------------- - csc 2x = - -------- sec 2x - tan 2x sec 2x + tan 2x sin 2x Okay. Now on to the next one: Problem 2: I'm going to start with three identities from my trigonometry textbook: sin(x/2) = sqrt[(1 - cos x)/2] cos(x/2) = sqrt[(1 + cos x)/2] tan x = (sin x)/(cos x) where sqrt means the square root. Then: sin(x/2) tan(x/2) = ---------- cos(x/2) sqrt[(1 - cos x)/2] = --------------------- sqrt[(1 + cos x)/2] sqrt(1 - cos x)/sqrt(2) (A property of square = ------------------------- roots of real numbers.) sqrt(1 + cos x)/sqrt(2) sqrt(1 - cos x) = ---------------- (Simplifying the fraction) sqrt(1 + cos x) [sqrt(1 - cos x)][sqrt(1 + cos x)] = ------------------------------------ [sqrt(1 + cos x)][sqrt(1 + cos x)] because multiplying the top and bottom of a fraction by the same nonzero number does not change the fraction value (and we know that the value is nonzero, because x < pi). sqrt[(1 - cos x)(1 + cos x)] = ------------------------------ (Properties of square 1 + cos x roots.) sqrt[1 - (cos x)^2] = --------------------- 1 + cos x sqrt[(sin x)^2] (because (sin x)^2 + (cos x)^2 = 1) = ----------------- 1 + cos x sin x = ----------- 1 + cos x as desired. Please note that it is important to keep 0 < x < pi. When x is in this range, both sin x and cos x have positive values, so the rules I have used for square roots work. Well, this has been a long answer, but I hope it helps. Good luck! -Doctor Barrus, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 06/02/98 at 13:41:39 From: Doctor Barrus Subject: Re: Trigonometric equations By the way, Chris, a lot of people have been mailing this question in (I assume they're your classmates). Their questions have been a little different, though. Instead of having the first thing as cos 2x, they have cot 2x. You might want to check the assignment. Just in case you mistyped it, I've modified my answer to fit the other problem. Here it is: cot 2x cos 2x --------------- - --------------- - csc 2x sec 2x - tan 2x sec 2x + tan 2x I'm assuming that the problem is to simplify this as much as possible. The first thing I usually do in problems like this is to write everything in terms of sines and cosines. I do this because I know and can derive more identities involving sines and cosines, and textbooks give more identities for sines and cosines, than for other trigonometric functions. So let's do that: cos 2x -------- sin 2x cos 2x 1 ------------------- - --------------------- - -------- 1 sin 2x 1 sin 2x sin 2x ------ - -------- -------- + -------- cos 2x cos 2x cos 2x cos 2x Now I'll simplify the denominators: cos 2x -------- sin 2x cos 2x 1 ---------------- - ---------------- - -------- 1 - sin 2x 1 + sin 2x sin 2x ------------ ------------ cos 2x cos 2x and simplify the fraction division: (cos 2x)^2 (cos 2x)^2 1 -------------------- - ------------ - -------- (sin 2x)(1 - sin 2x) 1 + sin 2x sin 2x where ^2 means raised to the second power. Next I'll get a common denominator for everything, and add: (1 + sin 2x)(cos 2x)^2 - (sin 2x)(1 - sin 2x)(cos 2x)^2 - (1 - sin 2x)(1 + sin 2x) --------------------------------------------------------- (sin 2x)(1 - sin 2x)(1 + sin 2x) Then I'll simplify the expressions in the numerator and denominator of the first term a bit: [1 + sin2x - (sin2x)(1 - sin2x)](cos2x)^2 - [1 - (sin2x)^2] ----------------------------------------------------------- (sin 2x)(1 - (sin 2x)^2) and then a little more (using the identity (sin x)^2 + (cos x)^2 = 1): [1 + (sin 2x)^2](cos 2x)^2 - (cos 2x)^2 --------------------------------------- (sin 2x)(cos 2x)^2 and some more: (sin 2x)^2 ------------ sin 2x And finally divide: sin 2x This looks a lot better than what we started out with, doesn't it? At this point I would say that this is the simplest form I can write this in. Of course, as with most trigonometric expressions, there are many ways to write this in other forms. For example, we could write this particular expression as 2(sin x)(cos x), so the exact form of your answer is going to be somewhat a matter of personal taste. Since it's pretty compact, I'm going to keep my answer as it is. So cot 2x cos 2x --------------- - --------------- - csc 2x = sin 2x sec 2x - tan 2x sec 2x + tan 2x Good luck! -Doctor Barrus, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/