text: 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