Sine, Cosine, and Any Function
General Formula
Given function $f(t)$ with Laplace Transform $F(s)$ and constant $b \in \mathbb{R}$
Recall the identity \(\sin(t) = \frac{e^{it} - e^{-it}}{2i}\).
\[\begin{align} \mathcal{L}\{ f(t) \sin(bt) \} &= \mathcal{L} \left \{ f(t) \cdot \frac{1}{2i} (e^{ibt} - e^{-ibt}) \right \} \\[10pt] &= \frac{1}{2i} \left ( \mathcal{L}\{ f(t) e^{ibt} \} - \mathcal{L}\{ f(t) e^{-ibt} \} \right ) \\[10pt] &= \frac{F(s-ib) - F(s+ib)}{2i} \end{align}\]Recall the identity \(\cos(t) = \frac{e^{it} + e^{-it}}{2}\).
\[\begin{align} \mathcal{L}\{ f(t) \cos(bt) \} &= \mathcal{L} \left \{ f(t) \cdot \frac{1}{2} (e^{ibt} + e^{-ibt}) \right \} \\[10pt] &= \frac{1}{2} \left ( \mathcal{L}\{ f(t) e^{ibt} \} + \mathcal{L}\{ f(t) e^{-ibt} \} \right ) \\[10pt] &= \frac{F(s-ib) + F(s+ib)}{2} \end{align}\]Much shorter than the previous two posts. Also notice that the above result agrees with the special case we proved for $f(t) = t^n$.