# Derive the Laplace transform of the following function sin(h(at))

Derive the Laplace transform of the following function
$\mathrm{sin}\left(h\left(at\right)\right)$
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Step 1 We have to derive the Laplace transform of function:
$\mathrm{sin}\left(h\left(at\right)\right)$
We know from definition of hyperbolic function,
$\mathrm{sin}hx=\frac{{e}^{x}-{e}^{-x}}{2}$
If x=at then,
$\mathrm{sin}hx=\frac{{e}^{x}-{e}^{-x}}{2}$
$\mathrm{sin}hat=\frac{{e}^{at}-{e}^{-at}}{2}$
$=\frac{1}{2}\left({e}^{at}-{e}^{-at}\right)$
Step 2
Now finding Laplace transform,
$L\left(\mathrm{sin}hat\right)=L\left(\frac{1}{2}\left({e}^{at}-{e}^{-at}\right)$
$=\frac{1}{2}\left(L{e}^{at}-L\left({e}^{-at}\right)\right)$
We know the Laplace transform of exponential function,
$L\left({e}^{ax}\right)=\frac{1}{\left(s-a\right)}$
if a=-a then $L\left({e}^{-ax}\right)=\frac{1}{\left(s+a\right)}$
Putting Laplace transform of exponential function, we get
$\frac{1}{2}\left(L{e}^{at}-L\left({e}^{-at}\right)\right)=\frac{1}{2}\left(\left(\frac{1}{\left(s-a\right)}\right)-\left(\frac{1}{\left(s+a\right)}\right)$
$=\frac{1}{2}\frac{\left(s+a-\left(s-a\right)\right)}{\left(s-a\right)\left(s+a\right)}$

$=\frac{1}{2}\frac{2a}{\left({s}^{2}-{a}^{2}\right)}$
$=\frac{a}{{s}^{2}-{a}^{2}}$
Hence, Laplace transform of $\mathrm{sin}hat$ is $\frac{a}{\left({s}^{2}-{a}^{2}\right)}$