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Trigonometric Identity Problem

  1. Nov 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove the Identity

    sinθ/(1+cosθ) = 1-cos(θ)/sinθ



    2. Relevant equations

    sinθ/cosθ = tanθ

    sin^2θ + cos^2θ = 1

    3. The attempt at a solution


    sinθ/(1 + cosθ) = LS

    cosθtanθ/(1+cosθ) = LS

    cosθtanθ/(sin^2θ + cos^2θ + cosθ) = LS

    cosθtanθ/(tan^2θcos^2θ + cos^2θ + cosθ) = LS

    (cosθ/cosθ) (tanθ/[tan^2θcosθ + cosθ + 1]) = LS

    sinθ/(cosθ[tan^2θcosθ + cosθ + 1]) = LS

    sinθ/(sin^2θ + cos^2θ +1) = LS

    sinθ/2 = LS
     
    Last edited by a moderator: Nov 4, 2012
  2. jcsd
  3. Nov 4, 2012 #2

    SammyS

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    The above line should be: [itex]\displaystyle \frac{\cos(\theta)}{\cos(\theta)}\left(
    \frac{\sin(\theta)}{\tan^2(\theta)\cos(\theta) + \cos(\theta) + 1}\right)[/itex]
    What is  1 - cos2(θ) ?

    How can you make the denominator of LHS or numerator of RHS equal to  1 - cos2(θ) ?
     
    Last edited by a moderator: Nov 4, 2012
  4. Nov 4, 2012 #3
    why should sinθ be in the numerator?

    I factored out (cosθ/cosθ), where the numerator was initially

    cosθtanθ

    thus leaving me with tanθ instead of sinθ in the numerator.
     
  5. Nov 4, 2012 #4

    SammyS

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    [itex]\displaystyle \cos(\theta)\tan(\theta)=\cos(\theta)\frac{\sin( \theta)}{\cos(\theta)}[/itex]
    [itex]\displaystyle =\frac{\cos(\theta)\,\sin(\theta)}{\cos(\theta)}[/itex]

    [itex]\displaystyle =\sin(\theta)[/itex]​
     
  6. Nov 4, 2012 #5
    In those steps you didn't distribute the cos back to the 1.
     
    Last edited by a moderator: Nov 4, 2012
  7. Nov 4, 2012 #6

    ehild

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    It is not an identity. sinθ/(1+cosθ) = (1-cos(θ))/sinθ is.

    ehild
     
  8. Nov 5, 2012 #7

    SammyS

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    Good point !
     
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