Please check my work. Did I calculate the following Laplace Transform correctly?
$$f(t)=sin(t)u(t-\frac{\pi}{2})$$
My solution:
Use the following corollary from the second shifting theorem (t-shift)...
$$\mathcal{L}\{g(t)u(t-a)\}=e^{-as}\mathcal{L}\{g(t+a)\}$$
where...
$$f(t)=g(t+a)$$
with $g(t)=sin(t)$ and $a=\pi/2$
$$g(t+a)=g(t+\frac{\pi}{2})=sin(t+\frac{\pi}{2})$$
reduce the sine term to a simpler form...
$$sin(t+\frac{\pi}{2})=cos(\frac{\pi}{2})sin(t)+sin(\frac{\pi}{2})cos(t)$$
$$=0\cdot\sin(t)+1\cdot cos(t)=cos(t)$$
hence...
$$\mathcal{L}\{sin(t)u(t-\frac{\pi}{2})\}=e^{-\frac{\pi}{2}s}\mathcal{L}\{cos(t)\}$$
$$F(s)=\frac{e^{-\frac{\pi}{2}s}}{s^2+1}$$