Laplace of \$f(t)\cos(\omega t)\$
First, note that: \$\small cos(\omega t) = \dfrac{e^{j\omega t}+e^{-j\omega t}}{2} \normalsize\$

The s-shifting property of the Laplace Transform is: \$\small\mathscr{L}[f(t).e^{at}]=F(s-a)\normalsize\$

Hence:

$$\mathscr{L} [f(t).cos(\omega t)]=\dfrac{\mathscr{L}[f(t).e^{j\omega t}]+\mathscr{L}[f(t).e^{-j\omega t}]}{2}=\dfrac{F(s-j\omega)+F(s+j\omega)}{2}$$



Example 1.

\$\small f(t)=H(t)=\$ Heaviside function = unit step function, find expression for \$\small\mathscr{L} [H(t).cos(\omega t)]\$

We know from LT tables that: \$\small\mathscr{L}cos(\omega t)= \dfrac{s}{(s^2+\omega^2)}\$. To derive this from above relationship:

\$\small F(s)=H(s)=\dfrac{1}{s}\$

\$\small\mathscr{L} [H(t).cos(\omega t)]=\dfrac{F(s-j\omega)+F(s+j\omega)}{2}=\dfrac{1}{2}\big(\dfrac{1}{s-j\omega}+\dfrac{1}{s+j\omega}\big)=\dfrac{s}{(s^2+\omega^2)}\normalsize\$



Example 2.

\$\small f(t)=e^{-t}\$, find expression for \$\small\mathscr{L} [e^{-t}.cos(t)]\$

\$\small F(s)=\dfrac{1}{s+1}\$

\$\small\mathscr{L} [e^{-t}.cos(t)]=\dfrac{F(s-j)+F(s+j)}{2}=\dfrac{1}{2}\big(\dfrac{1}{s+1-j}+\dfrac{1}{s+1+j}\big)=\dfrac{s+1}{s^2+2s+2}\normalsize\$