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does sin(t-2) = e^(2-t) ?
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or \[\sin t = e^{-t}\]
I don't think so... \[e^{i t} = \cos(t) + i \sin(t)\]Also,\[\sin(t) = {1 \over {2i}} (e^{i t} - e^{-i t})\]Not sure if this was what you were going for...
well I guess there is a type in textbook's answers... I had to apply a laplace transform to \[1/(s+1)\] and answer is clearly sin t but textbooks says its e^(-t) which is weird because textbook is pretty correct
the laplace transform of sin t would be:\[\mathcal{L}(\sin t)=\frac{1}{s^2+1}\]
see my problem, thanks
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