1. ## Laplace transform

show that

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

2. Originally Posted by action259
show that

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

RonL

3. Originally Posted by action259
show that

cost*(1/a*sinat) = sint*cosat
Is this
$cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)$?

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

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

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

-Dan

4. sorry about the confusion it should be

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

5. Originally Posted by action259
sorry about the confusion it should be

$
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
$tan(at) = a \cdot tan(t)$
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

6. Using a table of LaPlace transforms.

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

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

Multiply:

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

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

Now, $sin(t)=\frac{1}{p^{2}+1}$ and $cos(at)=\frac{p}{p^{2}+a^{2}}$ and the result follows from [1].

7. Hmmm...I appear to be missing something in the notation. Ah well.

-Dan

8. 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 $\mathcal{L}$s missing here.

RonL