Stuck on Laplace Transform of Odd Trig Function

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Discussion Overview

The discussion revolves around the Laplace transform of the function [cos(t)]^3, particularly focusing on the challenges associated with transforming odd trigonometric functions. Participants explore various mathematical identities and techniques to simplify the expression for the Laplace transform.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant expresses difficulty in applying the Laplace transform to [cos(t)]^3 and mentions using the half-angle formula without success.
  • Another participant suggests using the cosine addition formula to derive an identity involving [cos(t)]^3.
  • A third participant provides a detailed derivation showing how to express [cos(t)]^3 in terms of cos(3t) and cos(t), leading to a potential simplification for the Laplace transform.
  • A later reply indicates that the original poster successfully solved the problem using a half-angle formula on [cos(t)]^2 and a trigonometric product formula, confirming agreement with another participant's solution.

Areas of Agreement / Disagreement

Participants generally agree on the methods to approach the problem, but there is no consensus on the initial steps to take, as different techniques are proposed and explored.

Contextual Notes

The discussion includes various mathematical identities and transformations that may depend on specific definitions or assumptions about trigonometric functions, which are not fully resolved.

Underhill
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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?
 
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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!
 

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