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A little push on this trig identity

  1. Feb 26, 2005 #1
    A little push on this trig identity plz

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

    I need a little push I know from my other post that [tex] \cos(2x)=\cos^2 (x) - \sin^2 (x) [/tex] (can someone explain why?)
  2. jcsd
  3. Feb 26, 2005 #2
    what is your questioin? proving the tan identity or the cos one?
  4. Feb 26, 2005 #3
    I have to make the left hand side equal the right hand side I dont think it matters if you use cos or tan which ever one is easier.
  5. Feb 26, 2005 #4
    It would be easier to work with the right side.
    So it would be like:

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

    [tex]\frac {1 - (1 - 2\sin^2x)}{1 + 2\cos^2x - 1}[/tex] Double angle identities

    [tex]\frac{2\sin^2x}{2\cos^2x}[/tex] 2's cancel out

  6. Feb 26, 2005 #5
    Thanks sooo much BLUE SODA Im not good with the double angle identity thanks again :smile:
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