Question

evaluate \(\large \mathcal L \lbrace \sin kt \cos kt \rbrace\) with the aid of a trigonometric identity.

Answer 1

\[\sin a\cos b=\frac12(\sin(a+b)+\sin(a-b))\]is probably the identity to use
then it becomes like our earlier problem

Answer 2

let me think this thorigh...

Answer 3

through(

Answer 4

through*

Answer 5

thu*

Answer 6

so im going to treat kt as a?

Answer 7

k would be like a if you want to use de moivre theorem like @eliassaab did, but I would just split this sucker in two and integrate by parts like I did in the other

Answer 8

so it's \[L \lbrace \frac{\sin (2kt)}{2} \rbrace\]
??

Answer 9

well im not really familiar with that de moivre thingy

Answer 10

me neither, that's why I'm trying to avoid it
probably help here though
I have not done a problem quite like this in deriving laplaces, so I will ponder it over my dinner :)

Answer 11

While I'm eating I hope to watch it get solved, otherwise I'll try when I'm done :)

Answer 12

hmm okay

Answer 13

review example 4 for a basic idea of how to deal with the sine Laplace transform
http://tutorial.math.lamar.edu/Classes/DE/LaplaceDefinition.aspx

Answer 14

well i need to ask a few questions for checking so i guess i'll close this question for now :(

Answer 15

Now as you did your other problem for Cos, you can do
\[
\mathcal{L}_t[\sin (k t)](s)=\frac{k}{k^2+s^2}\\
Hence\\
\mathcal{L}_t[\sin (2 k t)](s)=\frac{2 k}{4
k^2+s^2}\\
\mathcal{L}_t\left[\frac{1}{2} \sin (2 k
t)\right](s)=\frac{k}{4 k^2+s^2}\\

\]

We are done since
\[
\sin( k t )\cos(kt)=\frac{1}{2} \sin (2 k
t)
\]

Answer 16

wow cool *_* so i was a bit on the track..

Answer 17

so does this mean \[\mathcal L \lbrace \sin (nkt) \rbrace = \frac{nk}{s^2 + (nk)^2}\]
@eliassaab ?

because if yes i think i may be able to use it in the future

Answer 18

Yes

Answer 19

wow cool thanks