Is the Z-Transform of sin(Bn) Really Zero?

[jpm]

Homework Statement

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

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

Homework Equations

The Attempt at a Solution

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

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

The Z transform of the first part of x(n) was found from a table. The z transform of the 2nd was found from the fact that the z transform of $$-sin(Bn)u(-n-1)$$ equals the z transform of $$sin(Bn)u(n)$$

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

So the z transform is zero? Could this be?

 

[CEL]

Remember that $$sin x = \frac {e^{jx}-e^{-jx}}{2j}$$