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Laplace transforms on cos^3 t

  1. Nov 9, 2011 #1
    I need to do a laplace transform on cos^3 t. I understand laplace but the trig is tripping me up.

    cos^3 t = Cos^2 t * Cos t = cos t * (cos 2t + 1)/2 (double angle formula)

    so i have (cos t)*(cos 2t)/2 + (cos t)/2.

    my book's solution says (cos t)*(cos 2t)/2 = (1/2)(cos (2t+1) + cost (2t-1))... how? I cant think of any formulas that give the above result. can someone please explain?

    thanks.
     
  2. jcsd
  3. Nov 9, 2011 #2

    AlephZero

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    Do you know the formulas for cos(a+b) and cos (a-b) in terms of cos and sin of a and b?

    Write them down, then eliminate the sin terms.
     
  4. Nov 9, 2011 #3

    lurflurf

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    The usual way is to use the triple angle identity
    Cos(3t)=4cos3(t)-3cos(t)
    It could also be done with any number of identities or the product formula for Laplace transforms.
     
    Last edited: Nov 9, 2011
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