# Proving a trig identity

1. Sep 14, 2008

### kirab

1. The problem statement, all variables and given/known data

Prove that

$$\frac{cos 3x}{cos x} = 2cos (2x) - 1$$

2. Relevant equations

The ones I used:

$$cos 2x = cos^2 x - sin^2 x$$
$$sin^2 x + cos^2 x = 1$$

3. The attempt at a solution

I *think* that the left hand side cannot be manipulated so I only fooled around with the right hand side...

$$2cos 2x - 1 = 2 (cos^2 x - sin^2 x) - 1 = 2 (cos^2 x - sin^2 x) - (sin^2 x + cos^2 x) = 2cos^2 x - 2sin^2 x - sin^2 x - cos^2 x = cos^2 x - 3sin^2 x$$
And I'm not sure what do to from there, so I did another approach (from the original right hand side):

$$2cos 2x - 1 = 2 (cos2x - 1/2) = 2 (cos 2x - ((sin^2 x + cos^2 x)/2)) = 2 ((2cos2x - sin^2 x - cos^2 x)/2) = 2cos2x - sin^2 x - cos^2 x.$$

And I'm stuck at this point. Any suggestions?

Last edited: Sep 14, 2008
2. Sep 14, 2008

### rock.freak667

Write cos(3x) as cos(2x+x) and then expand it.

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