$\displaystyle We\ have\ from\ the\ recurrence\ formula\ IV, \\[2ex] \displaystyle J_{n+1}\left(x\right)=\frac{2n}{x}J_n\left(x\right)-J_{n-1}\left(x\right) \\[2ex] \displaystyle Putting\ n=\frac{1}{2},\ we\ get\ \\[2ex] \displaystyle J_{\frac{3}{2}}\left(x\right)=\frac{1}{x}J_{\frac{1}{2}}\left(x\right)-J_{-\frac{1}{2}}\left(x\right) \\[2ex] $

$\displaystyle Now, \\[2ex] \displaystyle Jn(x)=\sum_{m=0}^{\infty{}}\frac{{\left(-1\right)}^m{\left(\frac{x}{2}\right)}^{2m+n}}{m!\vert{}\bar{n+m+1}} \\[2ex] \displaystyle J\frac{1}{2}(x)=\sum_{m=0}^{\infty{}}\frac{{\left(-1\right)}^m{\left(\frac{x}{2}\right)}^{2m+\left(\frac{1}{2}\right)}}{m!\vert{}\bar{m+\frac{3}{2}}} \\[2ex] $

$\displaystyle But \\[2ex] \displaystyle \vert{}\bar{m+\frac{3}{2}}=\left(m+\frac{1}{2}\right)\bullet{}\left(m-\frac{1}{2}\right)\bullet{}\left(m-\frac{3}{2}\right)\bullet{}\left(\frac{3}{2}\right)\bullet{}\left(\frac{1}{2}\right)\bullet{}\vert{}\bar{\frac{1}{2}} \\[5ex] \displaystyle =\frac{\left(2m+1\right)\left(2m\right)\left(2m-1\right)……3\bullet{}1}{2^{m+1}\ }\bullet{}\sqrt{\pi{}} \\[2ex] $

$\displaystyle Multiplying\ and\ dividing\ by\ 2\bullet{}4\bullet{}6 \\[2ex] \displaystyle =\frac{\left(2m+1\right)\left(2m\right)\left(2m-1\right)……3\bullet{}2\bullet{}1}{2^{m+1}\ \bullet{}2\bullet{}4\bullet{}6……….\left(2m-2\right)\bullet{}2m}\bullet{}\sqrt{\pi{}} \\[4ex] \displaystyle =\frac{\left(2m+1\right)!}{2^{m+1\ }\bullet{}\ m!}\bullet{}\sqrt{\pi{}} \\[2ex] $

$\displaystyle J\frac{1}{2}(x)=\sum_{m=0}^{\infty{}}\frac{{\left(-1\right)}^m2^{2m+1}}{m\left(2m+1\right)!}m!\bullet{}\frac{1}{\sqrt{\pi{}}}\bullet{}\frac{{\left(x\right)}^{2m+\left(\frac{1}{2}\right)}}{{\left(2\right)}^{2m+\left(\frac{1}{2}\right)}} \\[2ex] \displaystyle =\sum_{m=0}^{\infty{}}\frac{{\left(-1\right)}^mx^{2m+1}}{\left(2m+1\right)!}\bullet{}\frac{x^{-\frac{1}{2}}}{\sqrt{\pi{}}}\bullet{}2^{\frac{1}{2}} \\[2ex] \displaystyle =\sqrt{\frac{2}{\pi{}x}}\bullet{}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-…\right) \\[2ex] \displaystyle =\sqrt{\frac{2}{\pi{}x}}\sin x \\[2ex] $

$\displaystyle For \\[2ex] \displaystyle J-\frac{1}{2}(x) \\[2ex] $

$ \displaystyle From\ Recurrence\ Formula: \\[2ex] \displaystyle \frac{d}{dx}\left(x^n\bullet{}Jn(x)\right)=x^n\bullet{}Jn-1(x) \\[2ex] \displaystyle Let\ n=\frac{1}{2} \\[2ex] \displaystyle \frac{d}{dx}\left(x^{\frac{1}{2}}\bullet{}J\frac{1}{2}\left(x\right)\right)=x^{\frac{1}{2}}\bullet{}J-\frac{1}{2}\left(x\right) \\[2ex] \displaystyle \therefore{}\sqrt{\frac{2}{x}}\bullet{}\frac{d}{dx}\left(\sin x\right)=x^{\frac{1}{2}}\bullet{}J-\frac{1}{2}(x) \\[2ex] \displaystyle \therefore{}\sqrt{\frac{2}{x}}\bullet{}\cos x=x^{\frac{1}{2}}\bullet{}J-\frac{1}{2}(x) \\[2ex] \displaystyle \therefore{}\ J-\frac{1}{2}(x)=\ \sqrt{\frac{2}{x\pi{}}}\bullet{}\cos x \\[2ex] $

$\displaystyle J_{\frac{3}{2}}\left(x\right)=\frac{1}{x}\sqrt{\frac{2}{\pi{}x}}.\sin x{\ }-\sqrt{\frac{2}{\pi{}x}}.\cos x\ \\[2ex] \displaystyle J_{\frac{3}{2}}\left(x\right)=\ \sqrt{\frac{2}{\pi{}x}}.\left(\frac{\sin x}{x}-\cos x\right) \\[2ex] $