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Math Help - Triple angle trig function

  1. #1
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    Post Triple angle trig function

    Solve:
    sin3xcosx = cos3xsinx


    Here's what I've done so far:

    sin3xcosx = cos3xsinx
    sin3xcosx - cos3xsinx = 0
    sin(2x+x)cosx - cos(2x+x)sinx = 0
    (sin2xcosx + cos2xsinx)cosx - (cos2xcosx - sin2xsinx)sinx = 0

    Here's where I'm stuck. Should I distribute the cosx and sinx on either side? Or plug in the sin2x/cos2x double angle identities?
    Any help is appreciated!
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  2. #2
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    Quote Originally Posted by live_laugh_luv27 View Post
    Solve:
    sin3xcosx = cos3xsinx
    I would write this as \sin3x\cos x - \cos3x\sin x = 0 and use the formula \sin A\cos B - \cos A\sin B = \sin(A-B).
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  3. #3
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    So, what should I do after I simplify the equation to sin(3x - x) = 0?
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  4. #4
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    Quote Originally Posted by live_laugh_luv27 View Post
    So, what should I do after I simplify the equation to sin(3x - x) = 0?
    3x x = 2x, so your equation is sin(2x) = 0.
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