I know this can be solved easily be expanding as follows, $$ \mathcal{L}(\sin{(\omega t + a)}) = \mathcal{L}(\sin{\omega t} \cos{a} + \sin{a} \cos{\omega t}) $$ which is very simple after this.
But I want to know if it can be solved using properties of Laplace transform, i.e. time shift and scaling. For e.g. $$ \mathcal{L}(e^{-at}\sin{bt}) = \frac{b^2}{(s+a)^2 + b^2}$$
I think this work $$\sin(\omega t+a) = \dfrac{1}{2i}\left(e^{i(\omega t+a)}-e^{-i(\omega t+a)}\right) = \dfrac{e^{ia}}{2i}e^{i\omega t}-\dfrac{e^{-ia}}{2i}e^{-i\omega t}$$