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, ... }.