What is the laplace transform of $\delta(t-\pi /6)\sin (t)$

Question

What is the laplace transform of $\delta(t-\pi /6)\sin (t)$

I know that $L\{\delta(t-\pi/6) \}=e^{-s\pi/6}$ I also know that $L\{\sin (t) \}=1/(s^2+1)$

I also know that $L\{(u(t-\pi/6)f(t-\pi/6)\}=e^{-s\pi/6}F(s)$ where $F(s)=L\{f(t)\}$

Answer 1

The Laplace transform is by definition $$ \int_0^{\infty}\exp(-st)\delta(t-\pi /6)\sin (t)dt=\sin(\frac{\pi}{6})\exp(-s\frac{\pi}{6})=(1/2)\exp(-s\frac{\pi}{6})$$

Answer 2

In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

Using the "sifting property" of the Dirac Delta, for any suitable test function $\phi(t)$, we have

$$\int_{-\infty}^\infty \delta(t-t')\,\phi(t)\,dt=\phi(t') \tag 1$$

Now, let $\phi(t)=e^{-st}\sin(t)u(t)$ and $t'=\pi/6$. Then from $(1)$ we find

$$\begin{align} \int_0^\infty \delta(t-\pi/6)\,e^{-st}\sin(t)\,dt&=\int_{-\infty}^\infty \delta(t-\pi/6)\,e^{-st}\sin(t)u(t)\,dt\\\\ &=e^{-s\pi/6}\sin(\pi/6)u(\pi/6)\\\\ &=e^{-s\pi/6}\sin(\pi/6)\\\\ &=\frac12 e^{-\pi s/6} \end{align}$$

as was to be shown!