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Homework Help: Trig manipulation troubles

  1. Oct 21, 2008 #1
    I'm following through a solution to a problem I couldn't do and there's a bit of trig manipulation I can't get my head around:

    2cos^{2}x - sin^{2}x = 2 - 3sin^{2}x

    I tried using this identity to get it to work:

    cos^{2}x - sin^{2}x = 1 - 2sin^{2}x

    but alas, I could not. And after that I have no more ideas.

    If someone could point out how this transforms that'd be great. Evil, evil trig...
  2. jcsd
  3. Oct 21, 2008 #2
    What about just cos2x = 1 - sin2x?
  4. Oct 21, 2008 #3
    Certainly hear you:)

    How about cos2x =0.5(1+cos2x) and sin2x =0.5(1-cos2x)
    Last edited: Oct 21, 2008
  5. Oct 22, 2008 #4
    Doesn't that work though?

    2(1-sin^2(x)) - sin^2(x) = 2 -2 sin^2(x) - sin^2(x) = 2 - 3sin^2(x)
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