Using either$L2$or$L3$and$L4$verify$L9$and$L10$.

$\sinh at=\frac{e^{at}-e^{-at}}{2}$and$\cosh at=\frac{e^{at}+e^{-at}}{2}$

Laplace transform is a powerful integral transform used to switch a function from the time domain to the s-domain.

The expression is given by,

$\sinh at=\frac{e^{at}-e^{-at}}{2}$

Take Laplace transform on both sides as,

$$\begin{gathered} L\left(\sinh at\right)=L\left(\frac{e^{at}-e^{-at}}{2}\right) \\ =L\left(\frac{e^{at}}{2}\right)-L\left(\frac{e^{-at}}{2}\right) \\ =\frac{1}{2}L\left(e^{at}\right)-\frac{1}{2}L\left(e^{-at}\right) \\ L\left(e^{-at}\right)=\frac{1}{(p+a)}\left(\text{property}\right(L2\left)\right) \end{gathered}$$

Solve further

$$\begin{gathered} L\left(\sinh at\right)=\frac{1}{2}L\left(e^{at}\right)-\frac{1}{2}L\left(e^{-at}\right) \\ =\frac{1}{2}\frac{1}{(p-a)}-\frac{1}{2}\frac{1}{(p+a)} \\ =\frac{1}{2}\left(\frac{1}{(p-a)}-\frac{1}{(p+a)}\right) \\ =\frac{1}{2}\left(\frac{p+a-p+a}{(p-a)(p+a)}\right) \end{gathered}$$

Solve further

role="math" localid="1664284886527" $$\begin{gathered} =\frac{1}{2}\left(\frac{2a}{p^2-a^2}\right) \\ L\left(\sinh at\right)=\left(\frac{a}{p^2-a^2}\right) \end{gathered}$$

Hence, the expression $L\left(\sinh at\right)=\left(\frac{a}{p^2-a^2}\right)$is verified.

For $\cosh at$

$\cosh at=\frac{e^{at}+e^{-at}}{2}$

Take Laplace transform on both sides as,

$$\begin{gathered} L\left(\cosh at\right)=L\left(\frac{e^{at}+e^{-at}}{2}\right) \\ =L\left(\frac{e^{at}}{2}\right)+L\left(\frac{e^{-at}}{2}\right) \\ =\frac{1}{2}L\left(e^{at}\right)+\frac{1}{2}L\left(e^{-at}\right) \end{gathered}$$

Using

localid="1664285200314" $$\begin{gathered} L\left(e^{-at}\right)=\frac{1}{(p+a)}\left(\text{property}\right(L2\left)\right) \\ L\left(\cosh at\right)=\frac{1}{2}L\left(e^{at}\right)+\frac{1}{2}L\left(e^{-at}\right) \\ =\frac{1}{2}\frac{1}{(p-a)}+\frac{1}{2}\frac{1}{(p+a)} \\ =\frac{1}{2}\left(\frac{1}{(p-a)}+\frac{1}{(p+a)}\right) \end{gathered}$$

Solve further

$$\begin{gathered} =\frac{1}{2}\left(\frac{p+a+p-a}{(p-a)(p+a)}\right) \\ =\frac{1}{2}\left(\frac{2p}{p^2-a^2}\right) \end{gathered}$$