Question

laplace transform of cos^3a*t

Answer 1

best proof i could find for manipulating cos^3(at), which is the only tricky part here:



We can write,
cos(3x)=cos(2x+x)
=cos(2x)cos(x) - sin(2x)sin(x)
={2cos^2(x) - 1}cos(x) - 2sin(x)cos(x)sin(x)
=2cos^3(x) - cos(x) - 2cos(x).{1-cos^2(x)}
=2cos^3(x) - cos(x) - 2cos(x) + 2cos^3(x)
=4cos^3(x) - 3cos(x)

Now,
4cos^3(x) = cos(3x)+3cos(x)
so,
cos^3(x)= 1/4[cos(3x)+3cos(x)]


from yahoo answers: http://au.answers.yahoo.com/question/index?qid=20110404031547AABJeNW

laplace transform of that should be trivial :)

Answer 2

or you could just derive it
\[F(s)= \int\limits_{0}^{\infty}e^{-st}f(x)dx\]

Answer 3

implying you have all day :)

Answer 4

*Little correction \[F(s)= \int\limits_{0}^{\infty}e^{-st}f(t)dt\]