# Laplace transform

• Aug 27th 2006, 10:18 AM
action259
Laplace transform
show that

cost*(1/a*sinat) = sint*cosat
• Aug 27th 2006, 10:27 AM
CaptainBlack
Quote:

Originally Posted by action259
show that

cost*(1/a*sinat) = sint*cosat

RonL
• Aug 27th 2006, 12:11 PM
topsquark
Quote:

Originally Posted by action259
show that

cost*(1/a*sinat) = sint*cosat

Is this
$\displaystyle cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)$?

or
$\displaystyle cos \left ( t \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)$?

or
$\displaystyle cos \left ( t \frac{1}{a}sin(at) \right ) = sin \left ( t \cdot cos(at) \right )$?

In any of the above cases, I have doubts about the statement's truth.

-Dan
• Aug 27th 2006, 12:17 PM
action259
sorry about the confusion it should be

$\displaystyle cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)$
• Aug 27th 2006, 12:24 PM
topsquark
Quote:

Originally Posted by action259
sorry about the confusion it should be

$\displaystyle cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)$

When I meddle with this expression a bit I get that
$\displaystyle tan(at) = a \cdot tan(t)$
which simply isn't true.

Is there a typo here, or is there possibly some restriction on the problem? (Or are we possibly solving for a?)

-Dan
• Aug 27th 2006, 12:38 PM
galactus
Using a table of LaPlace transforms.

$\displaystyle cos(t)=\frac{p}{p^{2}+1}$

$\displaystyle \frac{sin(at)}{a}=\frac{a}{ap^{2}+a^{3}}$

Multiply:

$\displaystyle \left(\frac{p}{p^{2}+1}\right)\left(\frac{a}{ap^{2 }+a^{3}}\right)$

=$\displaystyle \frac{p}{(a^{2}+p^{2})(p^{2}+1)}$.....[1]

Now, $\displaystyle sin(t)=\frac{1}{p^{2}+1}$ and $\displaystyle cos(at)=\frac{p}{p^{2}+a^{2}}$ and the result follows from [1].
• Aug 27th 2006, 12:43 PM
topsquark
Hmmm...I appear to be missing something in the notation. Ah well. :)

-Dan
• Aug 27th 2006, 12:49 PM
CaptainBlack
Quote:

Originally Posted by topsquark
Hmmm...I appear to be missing something in the notation. Ah well. :)

-Dan

Your not the only one. Looks as though there may be a forest of $\displaystyle \mathcal{L}$s missing here.

RonL