Laplace transforms on cos^3 t

  • Thread starter wown
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  • #1
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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.
 

Answers and Replies

  • #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.
 
  • #3
lurflurf
Homework Helper
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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.
 
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