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}