I was asked to prove, with the power serie of $J_{\pm \nu}(x)$ that $J_{\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}\sin(x)$ and $J_{-\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}\cos(x)$.
We know that $J_{\frac{1}{2}}=\Sigma_{r=0}^{\infty}\frac{(-1)^r}{r!\Gamma(r+\frac{3}{2})}(\frac{x}{2})^{2r+\frac{1}{2}}$, multypling and diving by $\frac{2}{\sqrt{x}}$ one can obtain $\frac{\sqrt{2}}{\sqrt{x}}\Sigma_{r=0}^{\infty}\frac{(-1)^r}{r!\Gamma(r+\frac{3}{2})}(\frac{x}{2})^{2r+1}$. However, i don't know how to deal with $\Gamma$ (gamma) function, in order to "remove" it, and get a factorial (in order to get then Taylor's polynomial of sine).
With $J_{\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}\sin(x)$ i'll figure out how the other solution will be, there's no need of two explanations.
You need the duplication formula for the $\Gamma$-function, $$ \Gamma(z)\Gamma(z+1/2) = 2^{1-2z}\sqrt{\pi} \Gamma(2z), $$ so $$ r!\Gamma(r+3/2) = \Gamma(r+1)\Gamma((r+1)+1/2) = 2^{-1-2r} \sqrt{\pi} \Gamma(2r+2) = 2^{-1-2r} \sqrt{\pi}(2r+1)!, $$ and the result follows. The other one can be done similarly.