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General solution of trig functions

  1. Jan 2, 2014 #1
    Hello,

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

    find the general solution to cos3θ = sin2θ

    2. Relevant equations



    3. The attempt at a solution

    I know that sinθ = cos(π/2 - θ) but I am unsure of how to apply this when I have sin2θ.

    Do I say that sin2θ = cos2(π/2 - θ)?

    I think not because when I do this my answer does not agree with what my book says.

    Thanks!
     
  2. jcsd
  3. Jan 2, 2014 #2
    I thought i'd resolved my problem, but it still does not agree with the book.


    sinθ = cos(π/2 - θ)

    sin2θ = cos(π/2 - 2θ)

    cos3θ = cos(π/2 - 2θ)

    3θ = nπ +/- π/2 - 2θ

    θ = (nπ +/- π/2 - 2θ)/3
     
  4. Jan 2, 2014 #3
    I feel like a crazy person talking to myself, but just to save anyone the trouble of replying, I have figured it out.

    3θ = nπ +/- π/2 - 2θ

    either 3θ = nπ - π/2 - 2θ

    or

    3θ = nπ + π/2 - 2θ

    It works out that θ = π/10(4n+1)

    or

    θ = π/2(4n-1)

    Which is precisely what I wanted :)

    Perhaps a mod can delete this thread.
     
  5. Jan 2, 2014 #4

    Ray Vickson

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    I hope a moderator does not delete this thread, because you have certainly not found the general solution.

    Applying ##\sin(\phi) = \cos(\pi/2 \: - \: \phi)## to ##\phi = 2 \theta##, we have
    [tex] \sin(2 \theta) = \cos\left( \frac{\pi}{2} - 2 \theta \right)[/tex]
    for what it's worth (which is not much in this problem).
     
  6. Jan 2, 2014 #5
    My answer agrees with the textbook... Is what i've done not what you'd call the general solution to the original equation, or are you saying I have made a mistake?

    Thanks.
     
  7. Jan 2, 2014 #6

    Ray Vickson

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    No, I'm saying I made a dumb mistake, and your solution is OK. Sorry.
     
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