Calculate the Laplace transform Lleft{sin(t-k) cdot H(t-k)right}

Step 1
To find the Laplace transform of
$L\{\sin (t-k)\cdot H(t-k)\}$
Step 2
Since,
For the Heaviside function (or unit step function)
$(H_a(t)=H(t-a)=\{\begin{array}{ll}0& ta\end{array})$The second shifting theorem,
$\text{If }Lf(t)=F(s)\text{ then }LH(t-k)f(t-a)=e^{-as}F(s)$
First we need to find the laplace transform of $\sin (t)$.
Step 3
$\text{Let },f(t)=\sin (t)$
$L\{f(t)\}=L\sin (t)$
$\text{Since ,}L\{f(t)\}=\frac{a}{(s^2+a^2)}$
$L\{f(t)\}=\frac{1}{(s^2+1)}$
$\text{Thus ,}F(s)=\frac{1}{(s^2+1)}$
Step 4
By second shifting theorem,
$\text{If }Lf(t)=F(s)\text{then }LH(t-k)f(t-a)=e^{-as}F(s)$
Thus,
$L\{H(t-k)f(t-a)\}=e^{-ks}F(s)$
$=e^{-ks}\left(\frac{1}{(s^2+1)}\right)$
$=\frac{e^{-ks}}{(s^2+1)}$
Step 5
Therefore, $L\{\sin (t-k)\cdot H(t-k)\}=\frac{e^{-ks}}{(s^2+1)}$