Stuck on Laplace Transform of Odd Trig Function

Underhill
Messages
17
Reaction score
0
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?
 
Physics news on Phys.org
You can use the cosine addition formula with A = 2t and B = t to obtain an identity which involves (cos t)^3
 
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.
 
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!
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...

Similar threads

Replies
7
Views
5K
Replies
5
Views
3K
Replies
17
Views
3K
Replies
2
Views
2K
Replies
1
Views
3K
Replies
1
Views
4K
Replies
5
Views
3K
Replies
1
Views
1K
Back
Top