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Question Number 67524 by mathmax by abdo last updated on 28/Aug/19

prove that ∀z ∈C we have sinz =z Π_(n=1) ^∞ (1−(z^2 /(n^2 π^2 )))

$${prove}\:{that}\:\forall{z}\:\in{C}\:\:{we}\:{have} \\ $$$${sinz}\:={z}\:\prod_{{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right) \\ $$

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

The complement formulas give us (π/(sin(πz)))=Γ(z)Γ(1−z) Now sin(πz)=(π/(Γ(z)Γ(1−z))) =(π/(−zΓ(z)Γ(−z))) cause Γ(x+1)=xΓ(x) knowing that (1/(Γ(z)))= ze^(γz) Π_(n=1) ^∞ (1+(z/n))e^(−(z/n)) (1/(Γ(z)Γ(−z)))= ze^(γz) Π_(n=1) ^∞ (1+(z/n))e^(−(z/n)) . (−z)^ e^(γ(−z)) Π_(n=1) ^∞ (1+((−z)/n))e^(−((−z)/n)) =−z^2 Π_(n=1) ^∞ (1+(z/n))(1−(z/n))=−z^2 Π_(n=1) ^∞ (1−(z^2 /n^2 )) So sin(πz)=πzΠ_(n=1) ^∞ (1−(z^2 /n^2 )) finally with w=πz sin(w)=wΠ_(n=1) ^∞ (1−(w^2 /((nπ)^2 )))

$${The}\:{complement}\:{formulas}\:{give}\:{us}\:\:\:\frac{\pi}{{sin}\left(\pi{z}\right)}=\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right) \\ $$$${Now}\:\:{sin}\left(\pi{z}\right)=\frac{\pi}{\Gamma\left({z}\right)\Gamma\left(\mathrm{1}−{z}\right)}\:=\frac{\pi}{−{z}\Gamma\left({z}\right)\Gamma\left(−{z}\right)}\:\:\:{cause}\:\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\ $$$${knowing}\:{that}\:\:\frac{\mathrm{1}}{\Gamma\left({z}\right)}=\:{ze}^{\gamma{z}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{z}}{{n}}\right){e}^{−\frac{{z}}{{n}}} \\ $$$$\frac{\mathrm{1}}{\Gamma\left({z}\right)\Gamma\left(−{z}\right)}=\:{ze}^{\gamma{z}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{z}}{{n}}\right){e}^{−\frac{{z}}{{n}}} \:\:.\:\left(−{z}\right)^{} {e}^{\gamma\left(−{z}\right)} \underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{−{z}}{{n}}\right){e}^{−\frac{−{z}}{{n}}} \\ $$$$\:\:\:=−{z}^{\mathrm{2}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{z}}{{n}}\right)\left(\mathrm{1}−\frac{{z}}{{n}}\right)=−{z}^{\mathrm{2}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$${So}\:\:{sin}\left(\pi{z}\right)=\pi{z}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{z}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\:\: \\ $$$${finally}\:\:\:{with}\:\:{w}=\pi{z}\: \\ $$$${sin}\left({w}\right)={w}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{w}^{\mathrm{2}} }{\left({n}\pi\right)^{\mathrm{2}} }\right) \\ $$

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