Laplace transform of cos2(t-1/8π), help?

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SUMMARY

The discussion focuses on finding the Laplace transform of the function f(t) = e^(-t/2)cos2(t-1/8π) using the translation theorem. The correct interpretation of "cos2(t-1/8π)" is crucial, as it could refer to either cos^2(t-1/8π) or cos(2(t-1/8π)). The Laplace transform of cos(at) is given by L{cos(at)} = s/(s^2 + a^2), which is essential for solving the problem. The expected answer is √(2) (2s+5)/(4s²+4s+17).

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Homework Statement


I'm working on some Differential Equations homework and I'm stuck.

The question is apply the translation theorem to find the Laplace transform.

f(t) = e^(-t/2)cos2(t-1/8π)

I know how to apply the method, I just need to figure out how to transform the cosine function. (I think)

The answer should be: √(2) (2s+5)/(4s²+4s+17)


Homework Equations


L{f(t)*e^at)} = f(s-a)


The Attempt at a Solution



I'm completely stuck on how to transform cos2(t-1/8π).
 
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When you write "cos2(t-1/8π)", which of the following, if any, do you mean?

[tex]\begin{align*}<br /> &\cos^2 \left(t-\frac{1}{8\pi}\right) \\<br /> &\cos \left[2\left(t-\frac{1}{8\pi}\right)\right] \\<br /> &\cos^2 \left(t-\frac{\pi}{8}\right) \\<br /> &\cos \left[2\left(t-\frac{\pi}{8}\right)\right]<br /> \end{align*}[/tex]

For any of those cases, try using trig identities to turn it into a form where you can see how to get the Laplace transform.
 
vela said:
When you write "cos2(t-1/8π)", which of the following, if any, do you mean?

[tex]\begin{align*}<br /> &\cos^2 \left(t-\frac{1}{8\pi}\right) \\<br /> &\cos \left[2\left(t-\frac{1}{8\pi}\right)\right] \\<br /> &\cos^2 \left(t-\frac{\pi}{8}\right) \\<br /> &\cos \left[2\left(t-\frac{\pi}{8}\right)\right]<br /> \end{align*}[/tex]

For any of those cases, try using trig identities to turn it into a form where you can see how to get the Laplace transform.

The answer really depends on which one you mean. Remember that the laplace transform of cos(a*t)=s/(s^2+a^2) for a real number a and s/(s^2+a) for complex s.
 

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