Laplace transform of sin(2t)cos(2t)

Click For Summary

Homework Help Overview

The discussion revolves around finding the Laplace transform of the function f(t) = sin(2t)cos(2t) using a trigonometric identity. The subject area is focused on Laplace transforms and trigonometric identities in calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the use of trigonometric identities, particularly the double-angle formula, to simplify the expression for the Laplace transform. Questions arise about how to proceed with the integration involving the exponential function.

Discussion Status

Some participants have provided insights into using trigonometric identities to express sin(2t)cos(2t) in terms of sin(4t). There is an ongoing exploration of how to handle the integral involving the exponential term, with no explicit consensus reached on the next steps.

Contextual Notes

Participants are working under the constraints of deriving the Laplace transform and are considering the implications of integrating with respect to t while treating s as a constant. There is a mention of known transforms which may influence the approach taken.

JJBladester
Gold Member
Messages
281
Reaction score
2

Homework Statement



Find the Laplace transform of

f(t) = sin(2t)cos(2t)

using a trig identity.


Homework Equations



N/A.

The Attempt at a Solution



I know the double-angle formula sin(2t) = 2sin(t)cos(t) but that's not helping much. Can you give me some advice on how to proceed.
 
Physics news on Phys.org
Consider that

[tex]\sin{(4t)}=2\sin{(2t)}\cos{(2t)}[/tex]
 
jeffreydk said:
Consider that

[tex]\sin{(4t)}=2\sin{(2t)}\cos{(2t)}[/tex]

Okay, then [tex]\sin{(2t)}\cos{(2t)} = 1/2\sin{(t)}\cos{(t)}[/tex]

The Laplace integral is then:

[tex]1/2\int{e^{-st}\sin{(t)}\cos{(t)}}dt[/tex]

So where do I go from here to solve this? The [tex]e^{-st}[/tex] is throwing me off, otherwise I'd just do a u substitution with [tex]u = \sin{(t)}[/tex] and [tex]du = \cos{(t)}dt[/tex].
 
Well if you have that

[tex]\sin{(4t)}=2\sin{(2t)}\cos{(2t)}[/tex]

then

[tex]\sin{(2t)}\cos{(2t)}=\frac{1}{2}\sin{(4t)}[/tex]

So then if you want to do it from the integral you just need to integrate

[tex]\int_0^{\infty}\frac{1}{2}\sin{(4t)}e^{-st}dt[/tex]

where the [itex]s[/itex] you treat as a constant (you're integrating with respect to [itex]t[/itex]). But you could just use the known laplace transform for [itex]\sin{(at)}[/itex] unless you are meaning to derive it.
 
Ahhhh lights coming on. Thanks for the help!
 
Ah good--glad to help.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K