Laplace transform of trig + Heaviside

Question

So I am trying to take the laplace transform of $\cos(t)u(t-\pi)$. Is it valid for me to treat it as $((\cos(t)+\pi)-\pi)u(t-\pi)$ and treat $\cos(t)-\pi$ as $f(t)$ and use the 2nd shifting property, or is this not the correct procedure? Thanks in advance.

Answer 1

Here is one approach:

$$\mathcal{L}(\cos(t)) = \dfrac{s}{s^2+1}$$

$$\mathcal{L}\{\cos(t)u(t-\pi)\} = e^{-\pi s}\mathcal{L}(\cos(t-\pi)) = e^{-\pi s}\mathcal{L}(-\cos(t)) = -e^{-\pi s} \dfrac{s}{s^2+1}$$

Recall from the sum formula:

$$\cos(t - \pi) = \cos t \cos \pi + \sin t \sin \pi = -\cos t$$

Answer 2

Note that $\cos(t) = \cos((t-\pi) + \pi) = -\cos(t-\pi)$.

Hence, $$\mathcal{L}\{\cos(t)u(t-\pi)\} = -\mathcal{L}\{\cos(t-\pi)u(t-\pi)\} = \ldots$$