# Derive the transform of f(t)=sin kt by using the identity sin kt =frac{1}{2}u (e^{i kt}-e^{-i kt})

Derive the transform of $f\left(t\right)=\mathrm{sin}kt$ by using the identity $\mathrm{sin}kt=\frac{1}{2}u\left({e}^{ikt}-{e}^{-ikt}\right)$
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Step 1
Consider the given identity,
$\mathrm{sin}kt=\frac{1}{\left(2i\right)\left({e}^{ikt}-{e}^{-ikt}}$
Apply the Laplace transform on both sides,
$L\left\{\mathrm{sin}kt\right\}=L\left\{\frac{1}{\left(2i\right)\left({e}^{ikt}-{e}^{-ikt}\right)}\right\}$
Step 2
Apply the linearity of Laplace transform,
$L\left\{\mathrm{sin}kt\right\}=\frac{1}{\left(2i\right)\left[L\left\{{e}^{ikt}\right\}-L\left\{{e}^{-ikt}\right\}\right]}$
$=\frac{1}{\left(2i\right)\left[\frac{1}{s-ik}-\frac{1}{s-\left(-ik\right)}\right]}\dots \left(L\left\{{e}^{at}\right\}=\frac{1}{\left(s-a\right)}\right)$
$=\frac{1}{\left(2i\right)\left[\frac{1}{\left(s-ik\right)}-\frac{1}{\left(s+ik\right)}\right]}$
$=\frac{1}{\left(2i\right)\left[\frac{\left(s+ik\right)-\left(s-ik\right)}{\left(s-ik\right)\left(s+ik\right)}\right]}$
$=\frac{1}{\left(2i\right)\left[\frac{\left(2ik\right)}{\left({s}^{2}-{i}^{2}{k}^{2}\right)}\right]}$
$=\frac{k}{\left({s}^{2}+{k}^{2}\right)}$
Hence, $L\left\{\mathrm{sin}kt\right\}=\frac{k}{\left({s}^{2}+{k}^{2}\right)}$