Consider the recurrence relation in the title and its solution by induction.
$$ \begin{align} &g(n)=\sin(n-1)g(n-2),\quad g(0), g(1) \text{ given}\\ &g(2)=\sin(1)g(0)\\ &g(3)=\sin(2)g(1)\\ &g(4)=\sin(3)g(2)=\sin(1)\sin(3)g(0)\\ &g(5)=\sin(4)g(3)=\sin(2)\sin(4)g(1)\\ &...\\ &g(n_{even})=g(0)\prod_{k_{odd}}^{n-1}\sin(k)\\ &g(n_{odd})=g(1)\prod_{k_{even}}^{n-1}\sin(k) \end{align} $$
Not the most desirable form of solution, being in two parts, but it is correct as verified numerically by comparison with the recurrence for random values of $g(0,1)$.
Next, I tried to compare this with a closed form solution obtained from WolfranAlpha. That solution is given as
$$ g_n=(1+i)^ne^{(in^2)/4}\big(c_2(-1)^n+c_1\big)(e^{-2i};e^{-4i})_{\frac{n}{2}} $$
where $c_{1,2}$ are arbitrary parameters (to be determined from $g_{0,1}$) and $(a;q)_n$ is the q-Pochhammer symbol, which is defined as
$$ (a;q)_n=\prod_{k=0}^{n-1}(1-aq^k)=(1-a)(1-aq)(1-aq^2)...(1-aq^{n-1}) $$
which leaves me flummoxed by the $\frac{n}{2}$ in the WA solution. Any thoughts?