# I am trying to take the laplace transform of \cos(t)u(t-\pi).

I am trying to take the laplace transform of $\mathrm{cos}\left(t\right)u\left(t-\pi \right)$. Is it valid for me to treat it as $\left(\left(\mathrm{cos}\left(t\right)+\pi \right)-\pi \right)u\left(t-\pi \right)$ and treat $\mathrm{cos}\left(t\right)-\pi$ as f(t) and use the 2nd shifting property, or is this not the correct procedure?
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utloverej
Here is one approach:
$\mathcal{L}\left(\mathrm{cos}\left(t\right)\right)=\frac{s}{{s}^{2}+1}$
$\mathcal{L}\mathrm{cos}\left(t\right)u\left(t-\pi \right)={e}^{-\pi s}\mathcal{L}\left(\mathrm{cos}\left(t-\pi \right)\right)={e}^{-\pi s}\mathcal{L}\left(-\mathrm{cos}\left(t\right)\right)=-{e}^{-\pi s}\frac{s}{{s}^{2}+1}$
Recall from the sum formula:
$\mathrm{cos}\left(t-\pi \right)=\mathrm{cos}t\mathrm{cos}\pi +\mathrm{sin}t\mathrm{sin}\pi =-\mathrm{cos}t$
###### Not exactly what you’re looking for?
Note that $\mathrm{cos}\left(t\right)=\mathrm{cos}\left(\left(t-\pi \right)+\pi \right)=-\mathrm{cos}\left(t-\pi \right)$
Because
Hence,
$\mathcal{L}\left\{\mathrm{cos}\left(t\right)u\left(t-\pi \right)\right\}=-\mathcal{L}\left\{\mathrm{cos}\left(t-\pi \right)u\left(t-\pi \right)\right\}=\dots$