# Stuck on Laplace Transform of Odd Trig Function

1. Mar 30, 2013

### Underhill

Hey guys!

I'm stuck on a Laplace transform. Following is the problematic function:

[cos(t)]^3

Seems simple, but I'm having issues doing the Laplace transform on odd trigonometric functions. When I use the half-angle formula, I get this, which I can't seem to solve:

1/2cos(t) + 1/2cos(t)*cos(2t)

How do I get this into a form on which I can perform a Laplace transform?

2. Mar 30, 2013

### SteamKing

Staff Emeritus
You can use the cosine addition formula with A = 2t and B = t to obtain an identity which involves (cos t)^3

3. Mar 30, 2013

### HallsofIvy

$$cos(3t)= cos(2t)cos(t)- sin(2t)sin(t)= (cos^2(t)- sin^2(t))cos(t)- (2sin(t)cos(t))sin(t)$$
$$= cos^3(t)- sin^2(t)cos(t)- 2sin^2(t)cos(t)= cos^3(t)- 3(1- cos^2(t))cos(t)$$
$$= 4cos^3(t)- 3cos(t)$$

So $cos^3(t)= cos(3t)/4+ 3cos(t)/4$.

4. Apr 2, 2013

### Underhill

Thanks both of you for your help.

I eventually solved it by using a half-angle formula on [cos(t)]^2, and then using a trig product formula on the resulting expression. I got the same answer as HallsofIvy.

Thanks again, guys!