Laplace Transform of sin(3t)cos(3t) on t>=0 | Convolution Method

[crxyem]

Homework Statement

Use a table to fing L{f(t)} for the given function f(t) defined on the interval t>=0

Homework Equations

f(t)=sin(3t)cos(3t)

The Attempt at a Solution

Well this type of Laplace transform has got me stumped, I'm not to sure how to proceed. It doesn't seem to fit any of the forms in the table that is in my textbook. If it were sin(3t)+cos(3t) I could use the form sinwt = w/(s^2+w^2) and cos(3t)=2/(s^2+w^2) . I've read about convolution were L{f*g}=F(s)G(s), but I'm not convinced that I need to use this method.

I know the solution is supposed to be 3/(s^2+36), I've tried working my way to this solution. I'm sure my method is incorrect but I thought it might be possible to apply the Laplace transform of a derivative.

given that
L{f '' (t)} = s^2L{f(t)} - sf(0) - f ' (0) (a)

let f (t)= 3·COS(6·t)
let f'' (t)= - 18·SIN(6·t)
let f '' (t) = - 108·COS(6·t)

evaluate at zero

f (0)= 3*cos(3*0) = 3
f '' (0) = -18 sin(3*0) = 0

Substitute into (a)
L{f '' (t)} = s^2L{f(t)} - sf(0) - f ' (0)

L{- 108·COS(6·t) } = s^2L{3·COS(6·t) } - s(3) - 0

Rearrange terms
s^2L{3·COS(6·t) } - L{- 108·COS(6·t) } = 3s

L is a linear operator this equation becomes
3+s^2L{COS(6·t) } +108*L{COS(6·t) } = 3s

Collect like terms
(s^2 + 108/3) = 3s

Then the following is true.
L{3·COS(6·t)} = 3s/(s^2 + 36)

but there's an extra s in the numerator ??
do I need to start with
L{f ''' (t)} = s^3L{f(t)} - s^2L{f(0)} - sf ' (0) - f '' (0)

what might I have done wrong ?
have I attempted the proper approach ??
Is my identity of f(t)=sin(3t)cos(3t) correct ?

 

[TD]

Since sin(2x) = 2sin(x)cos(x); sin(3t)cos(3t) = sin(6t)/2.

 

[crxyem]

thanks I've always been terrible with trig identities.

works our flawlessly now