Differential Equations

$\mathcal{L}^{-1}\left\{ \frac{s}{s^2 + \omega^2} \right\} = \cos(\omega t)$ Here $\omega = 4$: $f(t) = \cos(4t)$ At $t = \frac{\pi}{8}$: $f\left(\frac{\pi}{8}\right) = \cos\left(4 \times \frac{\pi}{8}\right) = \cos\left(\frac{\pi}{2}\right) = 0.0$