show that

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

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- Aug 27th 2006, 10:18 AMaction259Laplace transform
show that

cost*(1/a*sinat) = sint*cosat - Aug 27th 2006, 10:27 AMCaptainBlackQuote:

Originally Posted by**action259**

RonL - Aug 27th 2006, 12:11 PMtopsquarkQuote:

Originally Posted by**action259**

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

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 PMaction259
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 PMtopsquarkQuote:

Originally Posted by**action259**

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

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 PMgalactus
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 PMtopsquark
Hmmm...I appear to be missing something in the notation. Ah well. :)

-Dan - Aug 27th 2006, 12:49 PMCaptainBlackQuote:

Originally Posted by**topsquark**

RonL