I'm trying to take the (two-sided, aka defined for all n) Z-transform of

$\displaystyle x(n)=sin(Bn)$

What I tried to do was split x(n) into

$\displaystyle x(n)=sin(Bn)u(n)+sin(Bn)u(-n-1)$

$\displaystyle X(z)=\frac{z sin(B)}{z^2-2zcos(B)+1}-\frac{z sin(B)}{z^2-2zcos(B)+1}$

But a problem arises now because the z-transform of the first equals the negative of z-transform of the second, which makes the z-transform = 0. Waaah? Could this be?