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Double-angle Formulae

  1. Nov 4, 2012 #1
    Having some trouble with this one. Can anyone help me out?

    Many thanks.

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

    Prove that [itex]\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan\theta}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    [itex]\frac{3\sin\theta}{\cos\theta}-\frac{\sin^3\theta}{\cos^3\theta}/1-\frac{3\sin^2\theta}{\cos^2\theta}[/itex]
    [itex]\frac{3\sin\theta\cos\theta-\sin^3\theta}{\cos^2\theta-3\sin^2\theta}[/itex]
    [itex]\frac{\sin\theta(3\cos\theta-\sin^2\theta)}{\cos^2\theta-3\sin^2\theta}[/itex]
    ...
     
  2. jcsd
  3. Nov 4, 2012 #2

    rock.freak667

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    It will be easier to use the left side to prove the right side.


    tan3θ = tan(2θ+θ)

    Expand out tan(2θ+θ), what do you get?
     
  4. Nov 4, 2012 #3

    vela

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    The tangent in the denominator should be squared.

    Use parentheses! That should be
    [tex]\left( \frac{3\sin\theta}{\cos\theta} - \frac{\sin^3\theta}{\cos^3\theta}\right) / \left(1-\frac{3\sin^2\theta}{\cos^2\theta}\right)[/tex]

    Check your algebra. That line doesn't follow from the one above.
     
  5. Nov 4, 2012 #4
    Ok, thank you. I'll give that a 2nd look.
     
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