Expand $\sin z=z-z^3/3!+z^5/5!-\cdots$. Dividing by $z^2$ gives $f(z)=1/z-z/6+z^3/120-\cdots$.
The residue is the coefficient of $z^{-1}$ in the Laurent series, hence **1**. The apparent squared denominator leaves only a simple pole because sin z has a zero at the origin.
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