Answer
$\dfrac{s-3}{(s-3)^2+16}$
Work Step by Step
The Laplace transform of function $\cos (4t)$ is given as:
$L(\cos 4t)=\dfrac{s}{s^2+16}$
The first shifting Theorem for $a=3$ can be expressed as:
$L[e^{3t} \cos 4t]=\dfrac{s-3}{(s-3)^2+16}$
