OMG, the first time for years to hold the Abramowitz/Stegun book in my hands again
In[193]:=
Sin[z Sin[x]] == 2 Sum[BesselJ[2 k + 1, z] Sin[(2 k + 1) x], {k, 0, Infinity}]
Out[193]= True
well, so go ahead with this, the even coefficients are as expected equal to zero
In[223]:= Table[
FourierSinCoefficient[Sin[z Sin[x]], x, k], {k, 0, 18, 2}]
Out[223]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
and the odd coefficients fit into the relation as expected, too
In[225]:= Table[
FullSimplify[FourierSinCoefficient[Sin[z Sin[x]], x, k] - 2 BesselJ[k, z]], {k, 1, 19, 2}]
Out[225]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
from a computational point of view you have it if you are willing to presume, that the relation valid for {1,3,5,...,17,19} is also valid for {21, 23, .... ,2k +1, ... }.
