How can I use Laplace transforms to solve for cos^3 t?

[wown]

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 can't think of any formulas that give the above result. can someone please explain?

thanks.

 

[AlephZero]

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.

 

[lurflurf]

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.