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How can I do this?\[\mathcal{L}\left \{ \delta(t-\pi)\cos(t) \right \}\]
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I know that\[\mathcal{L}\left \{ \cos(t)\right \}=\frac{s}{s^2+1}\]and\[\mathcal{L}\left \{ \delta(t-\pi)\right \}=e^{-\pi s}.\]But I have no idea how to proceed.
Judging from the answer to this question (back of the book), can I assume that\[\mathcal{L}\left \{ \delta(t-a)f(t) \right \}=e^{-as}f(a)?\]
It says\[\mathcal{L}\left \{ \delta(t-\pi)\cos(t) \right \}=-e^{-\pi s}\]and\[\mathcal{L}\left \{ \delta(t-\pi)\sin(t) \right \}=0.\]
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