Laplace Transform of squared trig functions help?

[chief10]

now say we have cos^2(3t), how would you go about computing it with the 3t?

i can manage cos^2(t) but I'm not sure how to take it that one step further

in the link below is what I've managed so far.. SOLVEDI worked it out.

If anyone's interested in the future, Just start it off as cos^2(t). Solve it all the way through using the identity (1/2)(cos2t+1) <------- this identity can be split up and solved using your standard Laplace identities.

Then identify the 3t from cos^2(3t). Realize that it's 3*ANGLE so multiply your integers in your final laplace transform of cos^2(t) by 3^2=9.

Your two integers should end up as 18 in the numerator and 36 in your denominator.

Hope this helps.

-chief10

 

[Mute]

You could always use the identity

$$\cos x = \frac{e^{ix}+e^{-ix}}{2},$$
for x = 3t, and expand the square to get four purely exponential terms (of complex argument) which you can Laplace transform.

 

[chief10]

You could always use the identity

$$\cos x = \frac{e^{ix}+e^{-ix}}{2},$$
for x = 3t, and expand the square to get four purely exponential terms (of complex argument) which you can Laplace transform.

so set the power of the exp as 3it?

how would it look if you had four exp terms though? I'm having trouble conceptualizing that.