Laplace transforms on cos^3 t

In summary, the conversation is about doing a Laplace transform on cos^3 t. The speaker understands Laplace transforms but is struggling with the trigonometry involved. They mention using the double angle formula to write cos^3 t as (cos t)*(cos 2t)/2 + (cos t)/2. The book's solution uses the formula for cos(a+b) and cos(a-b) to simplify the expression further. The speaker asks for an explanation of this process.
  • #1
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I need to do a laplace transform on cos^3 t. I understand laplace but the trig is tripping me up.

cos^3 t = Cos^2 t * Cos t = cos t * (cos 2t + 1)/2 (double angle formula)

so i have (cos t)*(cos 2t)/2 + (cos t)/2.

my book's solution says (cos t)*(cos 2t)/2 = (1/2)(cos (2t+1) + cost (2t-1))... how? I can't think of any formulas that give the above result. can someone please explain?

thanks.
 
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  • #2
Do you know the formulas for cos(a+b) and cos (a-b) in terms of cos and sin of a and b?

Write them down, then eliminate the sin terms.
 
  • #3
The usual way is to use the triple angle identity
Cos(3t)=4cos3(t)-3cos(t)
It could also be done with any number of identities or the product formula for Laplace transforms.
 
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1. What is a Laplace transform?

A Laplace transform is a mathematical tool used to convert a function of time into a function of complex frequency. It is particularly useful in solving differential equations and analyzing systems in engineering and physics.

2. How is a Laplace transform applied to cos^3 t?

To apply a Laplace transform to cos^3 t, we use the basic property of Laplace transforms that states: L{cos^nt} = (s^2 + n^2)^-1. In this case, n=3, so the Laplace transform of cos^3 t is given by L{cos^3t} = (s^2 + 9)^-1.

3. What is the inverse Laplace transform of (s^2 + 9)^-1?

The inverse Laplace transform of (s^2 + 9)^-1 is equal to cos 3t, which can be derived using the inverse Laplace transform property: L^-1{(s^2 + a^2)^-1} = cos at.

4. How is a Laplace transform used to solve differential equations?

A Laplace transform can be used to solve differential equations by converting the equation into an algebraic equation that can be easily solved. This is done by taking the Laplace transform of both sides of the equation, solving for the transformed function, and then using the inverse Laplace transform to find the solution in the original domain.

5. What are some applications of Laplace transforms in real-world problems?

Laplace transforms have various applications in engineering, physics, and other fields. Some examples include analyzing electrical circuits, solving heat transfer problems, and modeling mechanical systems. They are also used in signal processing to analyze signals in the frequency domain.

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