Question

Find the Laplace Transform using trig identities for:
t sin(2t) sin(5t)

this is what I've got so far:
(1/4i)[(s-7i)^-2 + (s+7i)^-2 - (s-3i)^-2 - (s+3i)^-2]

Answer 1

\[2sinAsinB = \cos(A-B) - \cos(A+B)\]

\[tsin(2t)\sin5(t) = \frac{ 1 }{ 2 }t(\cos(-3t) - \cos(7t))\]

then take laplace of that. is this what you did?

not sure what your step is though and why you have i

Answer 2

this table is very good btw, in case you don't have a good one already: http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf

you'll want to refer to number 10 or number 30 [the general case]

Answer 3

thank you i'll try to solve it using what you've taught me

Answer 4

I used the formula that says:
\[\sin(a)=(e^{ai}-e^{-ai})/2i\]

Answer 5

Now by using \[(1/2)t [\cos(-3t) - \cos(7t)]\]
I got \[-20s/[(s^2+9)(s^2+49)]\]
is it the right answer ??

Answer 6

i believe that is sinh(a) not sin(a)
\[\frac{ 1 }{ 2 }t[\cos(-3t) - \cos(7t)]\ using\ equation\ 10\]

\[\frac{ 1 }{ 2 }\left( \frac{ s^2 - 9 }{ (s^2 +9)^2 }-\frac{ s^2 - 49 }{ (s^2 + 49)^2 } \right)\]

Answer 7

thank you now i can understand what you were doing to solve it
but the only problem is i didn't get how you compute the function from the above trig identity

\[2sinAsinB=\cos(A−B)−\cos(A+B)\]
into
\[tsin(2t)\sin(5t)=(1/2)t[\cos(−3t)−\cos(7t)]\]

Answer 8

A = 2t
B = 5t

cos(A-B) = cos(2t - 5t) = cos(-3t)
cos(A+B) = cos(2t + 5t) = cos(7t)

Answer 9

thank you so much, that was very helpful =)