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January 25, 2015

January 25, 2015

Posted by **chuck** on Thursday, December 3, 2009 at 2:38pm.

- math -
**MathMate**, Thursday, December 3, 2009 at 4:20pmIf f(t)=g(t)=sin(t)

The convolution would be

∫ sin(u)sin(t-u) du from 0 to t

Use the identity:

sin(x)sin(y)=(1/2)(cos(x-y)-cos(x+y))

where x=u, y=t-u

∫ sin(u)sin(t-u) du from 0 to t

=∫ (1/2)(cos(u-t+u)-cos(u+t-u))du

=∫ (1/2)(cos(2u-t)-cos(t))du

=(1/2)[(1/2)sin(2u-t)-ucos(t)] (from 0 to t)

=(1/2)[(1/2)sin(t-(1/2)sin(-t)-tcos(t)]

=(1/2)(sin(t)- t*cos(t))

- math -
**chuck**, Thursday, December 3, 2009 at 5:10pmI figured it out before I got your answer, and I used the trig identity sin(t-u)=sin(t)cos(u)-cos(t)sin(u). I ended up having to use four or five more trig substitutions before finally getting to the answer you have there. Your substitution is much easier to compute. Too bad I didn't check back here before going through all of that! Thank you!

- math -
**MathMate**, Thursday, December 3, 2009 at 5:23pmYou're most welcome!

Glad that it helped, and thank you for your feedback.

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