Is Z-transform of $\sin(\omega_0n)$ same as that of $\sin(\omega_0n)u[n]$
The two sequences $\sin(\omega_0n)u[n]$ and $\sin(\omega_0n)$ are very different, so their transforms can't be identical. The $\mathcal{Z}$-transform of $x[n]=\sin(\omega_0n)u[n]$ can be computed as

$$X(z)=\sum_{n=0}^{\infty}\sin(\omega_0n)z^{-n}\tag{1}$$

which converges for $|z|&gt;1$ to the expression given in the table you linked to. However, the $\mathcal{Z}$-transform of $x[n]=\sin(\omega_0n)$ would be

$$X(z)\overset{?}{=}\sum_{n=-\infty}^{\infty}\sin(\omega_0n)z^{-n}\tag{2}$$

which doesn't converge anywhere, and hence does not exist.

Consequently, the $\mathcal{Z}$-transform is no suitable tool for computing the response of a discrete-time LTI system to a sinusoidal input. However, note that the discrete-time Fourier transform (DTFT) of $x[n]=\sin(\omega_0n)$ does exist. It contains Dirac delta impulses:

$$\text{DTFT}\{\sin(\omega_0n)\}=-j\pi\left[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)\right]\tag{3}$$