$\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)$
$\mathcal{L} (\cos t) = \dfrac{s}{s^2 + 1}$
Thus,
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{\dfrac{s}{4}}{\left( \dfrac{s}{4} \right)^2 + 1} \right]$
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{\dfrac{s}{4}}{\dfrac{s^2}{16} + 1} \right]$
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{s}{4 \left( \dfrac{s^2 + 16}{16} \right)} \right]$
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{s}{\dfrac{s^2 + 16}{4}} \right]$
$\mathcal{L} (\cos 4t) = \dfrac{1}{4} \left[ \dfrac{4s}{s^2 + 16} \right]$
$\mathcal{L} (\cos 4t) = \dfrac{s}{s^2 + 16}$ answer
