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

Question
Laplace transform
Derive the Laplace transform of the following function
$$\sin(h(at))$$

2021-02-06
Step 1 We have to derive the Laplace transform of function:
$$\sin(h(at))$$
We know from definition of hyperbolic function,
$$\sin hx=\frac{e^x-e^{-x}}{2}$$
If x=at then,
$$\sin h x =\frac{e^x-e^{-x}}{2}$$
$$\sin h at =\frac{e^{at}-e^{-at}}{2}$$
$$=\frac{1}{2}(e^{at}-e^{-at})$$
Step 2
Now finding Laplace transform,
$$L(\sin h at)=L(\frac{1}{2}(e^{at}-e^{-at})$$
$$=\frac{1}{2}(L{e^{at}}-L(e^{-at}))$$
We know the Laplace transform of exponential function,
$$L(e^{ax})=\frac{1}{(s-a)}$$
if a=-a then $$L(e^{-ax})=\frac{1}{(s+a)}$$
Putting Laplace transform of exponential function, we get
$$\frac{1}{2}(L{e^{at}}-L(e^{-at}))=\frac{1}{2}((\frac{1}{(s-a)})-(\frac{1}{(s+a)})$$
$$=\frac{1}{2}\frac{(s+a-(s-a))}{(s-a)(s+a)}$$
$$=\frac{1}{2}\frac{(s+a-s+a)}{(s^2-a^2)} \text{ since } (\sin ce (a+b)(a-b)=a^2-b^2)$$
$$=\frac{1}{2}\frac{2a}{(s^2-a^2)}$$
$$=\frac{a}{s^2-a^2}$$
Hence, Laplace transform of $$\sin h at$$ is $$\frac{a}{(s^2-a^2)}$$

### Relevant Questions

Find laplace transform of each following
a) $$\displaystyle{t}^{n}$$
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d) $$\displaystyle \cos{{h}}{\left({c}{t}\right)}$$
Use integration by parts to find the Laplace transform of the given function
$$f(t)=4t\cos h(at)$$
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$$\ddot{{y}}+{4}\dot{{y}}+{4}{y}=-{x}+{2}\dot{{x}}$$
Note: Assume that all initial conditions are zero
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$$y'''(0)=1$$
The inverse Laplace transform for
$$\displaystyle{F}{\left({s}\right)}=\frac{8}{{{s}+{9}}}-\frac{6}{{{s}^{2}-\sqrt{{3}}}}$$ is
a) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \sin{{h}}{{\left({3}{t}\right)}}$$
b) $$\displaystyle{8}{e}^{{-{9}{t}}}-{6} \cos{{h}}{\left({3}{t}\right)}$$
c) $$\displaystyle{8}{e}^{{{9}{t}}}-{6} \sin{{h}}{\left({3}{t}\right)}$$
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$$L\left\{\sin(t-k) \cdot H(t-k)\right\}$$
$$\text{Find the Laplace transform }\ F(s)=L\left\{f(t)\right\}\ \text{of the function }\ f(t)=6+\sin(3t) \ \text{defined on the interval }\ t\geq0$$
$$te^{-4t}\sin 3t$$
$$a) t^2 \sin kt$$
$$b) t\sin kt$$