What is the Laplace transform of sin2tcos2t?

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Discussion Overview

The discussion revolves around finding the Laplace transform of the function f(t) = sin(2t)cos(2t). Participants explore various approaches to simplify the product of the two trigonometric functions and apply the Laplace transform.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about how to handle the product of two trigonometric functions.
  • Another suggests transforming f(t) to a different form to facilitate the Laplace transform.
  • A participant proposes using the identity sin(2x) = 2sin(x)cos(x) to rewrite f(t) in a more manageable way.
  • One participant mentions that f(t) can be expressed as (1/4)sin^2(2t) and notes the derivative relationship with sin(2t)cos(2t).
  • Another participant states that they initially thought sin^2(t) had a known Laplace transform but later realized they were mistaken.
  • A suggestion is made to consult a transform table to find a corresponding answer and work backwards.
  • One participant claims that another's solution is correct without providing further details.

Areas of Agreement / Disagreement

There is no clear consensus on the best approach to take or the correctness of the proposed solutions, as participants present differing strategies and interpretations.

Contextual Notes

Some participants rely on trigonometric identities and transformations, while others suggest looking up known transforms, indicating a variety of methods being considered without resolution on which is optimal.

RafiS
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iv got a problem i can't seem to understand. if anyone could help me out it would be great

f(t)= sin2tcos2t

im just not sure what to do when i have the product of 2 trig functions.

the correct answer is (2/(s^2 + 16))

thanks for any help
 
Last edited:
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Is there something you can do with f(t) to make it look different?
 
yeah, like exk suggested, try to use this trig identity somehow:

[tex]sin2x=2isnxcosx[/tex] can you figure it out how to transform your f(t) into a similar form?
 
Use f(x) = 1/4sin^2(2t)
f'(x) = sin(2t)cos(2t)
 
Vid said:
Use f(x) = 1/4sin^2(2t)
f'(x) = sin(2t)cos(2t)

I don't see how would this help!
If i have gotten the op right, he just needst to take the laplace transform of

[tex]f(t)=sin2tcos2t=\frac{1}{2}sin4t[/tex]
so

[tex]L{f(t)}=L{\frac{1}{2}sin4t}=\int_{0}^{\infty}\frac{1}{2}sin(4t)e^{-st}dt[/tex]
 
I was thinking that sin^2(t) was a common laplace transform, but I was mistaken.
 
another strategy for this would be to look up in a transform table what your answer corresponds to and work backwards.

sutupidmath's solution is correct.
 

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