Question

laplace transform:

L{sin^3 (at)}

so for I have this:
L{ sin(at) * [1/2 ( 1 - cos(2at)]}
1/2 L{sin(at)} - 1/2 L{cos(2at)*sin(at)}
1/2 * (a^2)/(s^2 + a^2) - 1/2 L{cos(2at)*sin(at)}

How do i get rid of the cos(2at)*sin(at) ?

Answer 1

http://www.wolframalpha.com/input/?i=laplace+transform&a=*C.laplace+transform-_*Calculator.dflt-&f2=Sin%28a*t%29^3+&x=0&y=0&f=LaplaceTransformCalculator.transformfunction_Sin%28a*t%29^3+&f3=t&f=LaplaceTransformCalculator.variable1_t&f4=s&f=LaplaceTransformCalculator.variable2_s


Here i have the result, but I need the step :/

Answer 2

try converting the trig functions to their complex equivalents (e^(ix))

Answer 3

using Eulers formula does the job, posting the work in a sec.

Answer 4

im trying to get a trignometric relation with sin^3(at) instead of doing
sin(at)*sin^2(at) = sin(at)*(1/2 - 1/2 * cos(2at)

Answer 5

wtf did you just make happen there? Huge thanks!
I was trying to use this explanation
http://answers.yahoo.com/question/index?qid=20080723124442AAcZCTX

but i didnt get that either. Anyway, thanks for the relation

Answer 6

Here are variations of Euler's Formulas that may help with problems like these:

\[e^{i\theta}=\cos\theta+i \sin\theta\]
\[\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}\]
\[\sin\theta=\frac{e^{i\theta}-e^{-i\theta}}{2i}\]

Answer 7

oh now i get it. Thanks sir

Answer 8

help pls....Laplace transform: L{cos^3 (at)}

Answer 9

http://www.wolframalpha.com/input/?i=laplace+transform&a=*C.laplace+transform-_*Calculator.dflt-&f2=cos(a*t)%5E3&f=LaplaceTransformCalculator.transformfunction_cos(a*t)%5E3&f3=t&f=LaplaceTransformCalculator.variable1%5Cu005ft&f4=s&f=LaplaceTransformCalculator.variable2%5Cu005fs i get the result here too,. but i need the step : (credits to Mr. Diogo for this site^^)