Step 1: Write the Laplace transform as a direct integral.
Instead of quoting the shift theorem, start from the definition. Since \( u(t-3) \) switches the function on only for \( t \geq 3 \):
\[
\mathcal{L}\{\sin(t-3)u(t-3)\} = \int_{3}^{\infty} \sin(t-3)\, e^{-st}\, dt
\]
Step 2: Substitute \( \tau = t - 3 \).
Let \( \tau = t - 3 \), so \( t = \tau + 3 \) and \( dt = d\tau \). When \( t = 3 \), \( \tau = 0 \), and as \( t \to \infty \), \( \tau \to \infty \). The integral becomes
\[
\int_{0}^{\infty} \sin(\tau)\, e^{-s(\tau+3)}\, d\tau = e^{-3s} \int_{0}^{\infty} \sin(\tau)\, e^{-s\tau}\, d\tau
\]
Step 3: Evaluate the remaining integral.
The leftover integral is just the standard Laplace transform of \( \sin(\tau) \), which is \( \frac{1}{s^2+1} \). So
\[
\mathcal{L}\{\sin(t-3)u(t-3)\} = e^{-3s}\cdot\frac{1}{s^2+1} = \boxed{\frac{e^{-3s}}{s^2+1}}
\]
which is option (C).