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.