Laplace transform

**action259** · August 27th 2006, 10:18 AM

show that

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

**CaptainBlack** · August 27th 2006, 10:27 AM

Originally Posted by action259

show that

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

I don't understand your notation.

RonL

**topsquark** · August 27th 2006, 12:11 PM

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

**action259** · August 27th 2006, 12:17 PM

sorry about the confusion it should be

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

**topsquark** · August 27th 2006, 12:24 PM

Originally Posted by action259

sorry about the confusion it should be

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

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

**galactus** · August 27th 2006, 12:38 PM

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].

**topsquark** · August 27th 2006, 12:43 PM

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

-Dan

**CaptainBlack** · August 27th 2006, 12:49 PM

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

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