Proving a Trigonometry Identity
Date: 05/19/2003 at 18:37:18 From: Stephanie Subject: Proving Trigonometry Identities Here is my problem. Prove the following identity: cos 3x = 4 cos^3x - 3cosx Here is what I did and where I'm stuck: cos 3x = cos (2x+x) = (cos 2x)(cos x) - (sin 2x)(sin x) = (2 cos^2x-1)(cos x) - [(2 sinx)(cosx)](sinx) I'm not sure how you distribute this. What next?
Date: 05/19/2003 at 18:46:05 From: Doctor Schwa Subject: Re: Proving Trigonometry Identities Great start! Now, distributing is exactly right. With s = sin x and c = cos x, for short, you have (2c^2 - 1)(c) - (2sc)s so distributing gives you 2c^3 - c - 2s^2 c and finally substituting s^2 = 1 - c^2 will let you finish the problem. Does that make sense? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
Date: 05/19/2003 at 19:25:01 From: Stephanie Subject: Thank you (Proving Trigonometry Identities) Thanks so much. All I needed was that push in the right direction. I'll keep in mind to turn the sinx and cosx into s and c. It makes it much easier to see the next step to take. Thank you.
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum