Step 1

Consider the given identity,

\(\sin kt=\frac{1}{(2i)(e^{i kt}-e^{-i kt}}\)

Apply the Laplace transform on both sides,

\(L\left\{\sin kt\right\}=L\left\{\frac{1}{(2i)(e^{i kt}-e^{-i kt})}\right\}\)

Step 2

Apply the linearity of Laplace transform,

\(L\left\{\sin kt\right\}=\frac{1}{(2i)\left[L\left\{e^{ikt}\right\}-L\left\{e^{-ikt}\right\}\right]}\)

\(=\frac{1}{(2i)[\frac{1}{s-ik}-\frac{1}{s-(-ik)}]} \dots \left(L\left\{e^{at}\right\}=\frac{1}{(s-a)}\right)\)

\(=\frac{1}{(2i)\left[\frac{1}{(s-ik)}-\frac{1}{(s+ik)}\right]}\)

\(=\frac{1}{(2i)\left[\frac{(s+ik)-(s-ik)}{(s-ik)(s+ik)}\right]}\)

\(=\frac{1}{(2i)\left[\frac{(2ik)}{(s^2-i^2k^2)}\right]}\)

\(=\frac{k}{(s^2+k^2)}\)

Hence, \(L\left\{\sin kt\right\}=\frac{k}{(s^2+k^2)}\)

Consider the given identity,

\(\sin kt=\frac{1}{(2i)(e^{i kt}-e^{-i kt}}\)

Apply the Laplace transform on both sides,

\(L\left\{\sin kt\right\}=L\left\{\frac{1}{(2i)(e^{i kt}-e^{-i kt})}\right\}\)

Step 2

Apply the linearity of Laplace transform,

\(L\left\{\sin kt\right\}=\frac{1}{(2i)\left[L\left\{e^{ikt}\right\}-L\left\{e^{-ikt}\right\}\right]}\)

\(=\frac{1}{(2i)[\frac{1}{s-ik}-\frac{1}{s-(-ik)}]} \dots \left(L\left\{e^{at}\right\}=\frac{1}{(s-a)}\right)\)

\(=\frac{1}{(2i)\left[\frac{1}{(s-ik)}-\frac{1}{(s+ik)}\right]}\)

\(=\frac{1}{(2i)\left[\frac{(s+ik)-(s-ik)}{(s-ik)(s+ik)}\right]}\)

\(=\frac{1}{(2i)\left[\frac{(2ik)}{(s^2-i^2k^2)}\right]}\)

\(=\frac{k}{(s^2+k^2)}\)

Hence, \(L\left\{\sin kt\right\}=\frac{k}{(s^2+k^2)}\)