= Solution
<Laurent's theorem> states that a <holomorphic function> on $R<|z-a|<S$ has a unique expansion $\sum_{n\in\mathbb Z}a_n(z-a)^n$, converging absolutely there and uniformly on <compact> subannuli, with
$$
a_n=\frac1{2\pi i}\int_{|z-a|=r}\frac{f(z)}{(z-a)^{n+1}}dz\quad(R<r<S).
$$
Here $\sin z=z(1-z^2/6+\cdots)$ has a <simple zero> at zero and no other zero in $|z|<\pi$. Hence $1/\sin z$ has only a <simple pole> there, giving the stated series with no powers below $z^{-1}$. Multiplying by $\sin z$ and comparing the constant, linear and quadratic coefficients yields
$$
\boxed{c_{-1}=1,\qquad c_0=0,\qquad c_1=1/6.}
$$
Use the coefficient integrals on the two circles. The annular region between them contains exactly the <simple poles> $\pi,-\pi$, and the <residue> of $1/\sin z$ at each is $1/\cos(\pm\pi)=-1$. Apply the <residue theorem> to $1/(z^{n+1}\sin z)$ between the circles:
$$
\boxed{d_n=c_n-\pi^{-n-1}-(-\pi)^{-n-1}\quad(n\ge-1).}
$$
For lower indices, set $c_n=0$ because the inner <Laurent series> has no such terms; the same contour argument then still applies. Since $f$ is odd, its even-index coefficients vanish. A complete explicit list for the requested range is
$$
\boxed{\begin{aligned}
d_1&=\tfrac16-\frac2{\pi^2},&d_0&=0,&d_{-1}&=-1,\\
d_{-2k}&=0\ (k\ge1),&&d_{-(2k+1)}&=-2\pi^{2k}\ (k\ge1).
\end{aligned}}
$$
The subtraction accounts for the two <poles> crossed; using the inner coefficients unchanged outside their <annulus> would miss precisely these terms.
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