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Proving a trig identity

  1. Sep 14, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that

    [tex]\frac{cos 3x}{cos x} = 2cos (2x) - 1[/tex]

    2. Relevant equations

    The ones I used:

    [tex] cos 2x = cos^2 x - sin^2 x[/tex]
    [tex]sin^2 x + cos^2 x = 1
    [/tex]

    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...

    [tex] 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 [/tex]
    And I'm not sure what do to from there, so I did another approach (from the original right hand side):

    [tex] 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. [/tex]

    And I'm stuck at this point. Any suggestions?
     
    Last edited: Sep 14, 2008
  2. jcsd
  3. Sep 14, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

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