show that
cost*(1/a*sinat) = sint*cosat
Is thisOriginally 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
When I meddle with this expression a bit I get thatOriginally 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
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].