Trig Identities and Double AnglesDate: 01/04/98 at 19:30:45 From: Monique Sandiford Subject: Trig Identities- Double Angles Prove that cosA + sinA sec2A + tan2A = ------------ cosA - sinA Date: 01/05/98 at 11:41:52 From: Doctor Wilkinson Subject: Re: Trig Identities- Double Angles My first step in a problem like this would be to express everything in terms of sines and cosines, since you have sines and cosines on the right side. Also I know the identities for sines and cosines pretty well, but I tend to forget them for the other functions. So sec2A + tan2A = 1 + sin2A = 1 + sin2A ----- ----- --------- cos2A cos2A cos2A Now apply the double angle formulas for the sin and cosine to get 1 + 2sinA cosA ------------------- cos^2(A) - sin^2(A) Now it's time to get clever, because there are a number of directions we might consider going. The denominator factors as a difference of squares: cos^2(A) - sin^2(A) = (cosA + sinA)(cosA - sinA) This is kind of promising, because one of the factors is cosA - sinA, which is what we want to get in the denominator. This tells us we probably don't want to substitute cos^2(A) = 1 - sin^2(A) or sin^2(A) = 1 - cos^2(A) in the denominator. We already have the form we want in the denominator; let's look at what we can do with the numerator. At first glance things don't look so good. We could substitute for sin in terms of cosine or cosine in terms of sine, but that would bring square roots in, and would complicate things rather than simplify them. A useful trick in trig identities is sometimes to make a substitution that looks crazy until you see how it works. Let's substitute sin^2(A) + cos^2(A) for 1 This turns the numerator into sin^2(A) + cos^2(A) + 2 sinAcosA which we can arrange to sin^2(A) + 2 sinAcosA + cos^2(A) which factors into (sin(A) + cos(A))^2 = (cosA + sinA)^2 and the rest is easy. -Doctor Wilkinson, 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- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/