Frobenius & Special Functions · Hard problem

Given $J_{1/2}(x) = \sqrt{\dfrac{2}{\pi x}}\sin x$, $J_{-1/2}(x) = \sqrt{\dfrac{2}{\pi x}}\cos x$ and the Bessel recurrence $J_{\nu-1}(x) + J_{\nu+1}(x) = \dfrac{2\nu}{x}J_\nu(x)$, compute $J_{3/2}(\pi)$.