Question Number 67524 by mathmax by abdo last updated on 28/Aug/19
$${prove}\:{that}\:\forall{z}\:\in{C}\:\:{we}\:{have} \\ $$$${sinz}\:={z}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$
Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19
$${The}\:{complement}\:{formulas}\:{give}\:{us}\:\:\:\frac{\pi}{{sin}\left(\pi{z}\right)}=\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right) \\ $$$${Now}\:\:{sin}\left(\pi{z}\right)=\frac{\pi}{\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right)}\:=\frac{\pi}{−{z}\Gamma\left({z}\right)\Gamma\left(−{z}\right)}\:\:\:{cause}\:\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\ $$$${knowing}\:{that}\:\:\frac{\mathrm{1}}{\Gamma\left({z}\right)}=\:{ze}^{\gamma{z}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{z}}{{n}}\right){e}^{−\frac{{z}}{{n}}} \\ $$$$\frac{\mathrm{1}}{\Gamma\left({z}\right)\Gamma\left(−{z}\right)}=\:{ze}^{\gamma{z}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{z}}{{n}}\right){e}^{−\frac{{z}}{{n}}} \:\:.\:\left(−{z}\right)^{} {e}^{\gamma\left(−{z}\right)} \underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{−{z}}{{n}}\right){e}^{−\frac{−{z}}{{n}}} \\ $$$$\:\:\:=−{z}^{\mathrm{2}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{z}}{{n}}\right)\left(\mathrm{1}−\frac{{z}}{{n}}\right)=−{z}^{\mathrm{2}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$${So}\:\:{sin}\left(\pi{z}\right)=\pi{z}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\: \\ $$$${finally}\:\:\:{with}\:\:{w}=\pi{z}\: \\ $$$${sin}\left({w}\right)={w}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{w}^{\mathrm{2}} }{\left({n}\pi\right)^{\mathrm{2}} }\right) \\ $$