Properties of gamma function What is the simplest way to prove the two following properties of $
\mathit{\Gamma}{\mathrm{(}}{z}{\mathrm{)}}
$
a) 
$
\mathit{\Gamma}{\mathrm{(}}\frac{3}{2}{\mathrm{)}}\mathrm{{=}}\frac{1}{2}\mathit{\Gamma}{\mathrm{(}}\frac{1}{2}{\mathrm{)}}\mathrm{{=}}\frac{\sqrt{\mathit{\pi}}}{2}
$
b) $
\mathit{\Gamma}{\mathrm{(}}{z}{\mathrm{)}}\mathit{\Gamma}{\mathrm{(}}{1}\mathrm{{-}}{z}{\mathrm{)}}\mathrm{{=}}\frac{\mathit{\pi}}{\sin{\mathrm{(}}\mathit{\pi}{z}{\mathrm{)}}}
$
Thanks very much in advance.

Both are classical results. $\Gamma\left(\frac{3}{2}\right)$ is related with $\Gamma\left(\frac{1}{2}\right)$ through the function relation $\Gamma(z+1)=z\,\Gamma(z)$ and the value of $\Gamma\left(\frac{1}{2}\right)$ is
$$ \int_{0}^{+\infty}\frac{dz}{\sqrt{z}}e^{-z}\,dz = \int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}$$
by Fubini's theorem:
$$ \left(\int_{-\infty}^{+\infty}e^{-x^2}\,dx\right)^2 = \iint_{\mathbb{R}^2}e^{-(x^2+y^2)}\,dx\,dy = 2\pi \int_{0}^{+\infty}\rho e^{-\rho^2}\,d\rho.$$
The reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ is a consequence of the Weierstrass product for the sine function and Euler's product
$$\Gamma(z+1)=\prod_{n\geq 1}\frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}.$$
