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

$$provethat\mathrm{\forall }z\in Cwehave$$$$sinz=z\prod _{n=1}^{\mathrm{\infty }}(1-\frac{z^2}{n^2{\pi }^2})$$

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

$$Thecomplementformulasgiveus\frac{\pi }{sin\left(\pi z\right)}=\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(1-z)$$$$Nowsin\left(\pi z\right)=\frac{\pi }{\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(1-z)}=\frac{\pi }{-z\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(-z)}cause\mathrm{\Gamma }(x+1)=x\mathrm{\Gamma }\left(x\right)$$$$knowingthat\frac{1}{\mathrm{\Gamma }\left(z\right)}={ze}^{\gamma z}\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{z}{n})e^{-\frac{z}{n}}$$$$\frac{1}{\mathrm{\Gamma }\left(z\right)\mathrm{\Gamma }(-z)}={ze}^{\gamma z}\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{z}{n})e^{-\frac{z}{n}}.{(-z)}^{}{e}^{\gamma (-z)}\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{-z}{n})e^{-\frac{-z}{n}}$$$$=-z^2\underset{n=1}{\prod ^{\mathrm{\infty }}}(1+\frac{z}{n})(1-\frac{z}{n})=-z^2\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{z^2}{n^2})$$$$Sosin\left(\pi z\right)=\pi z\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{z^2}{n^2})$$$$finallywithw=\pi z$$$$sin\left(w\right)=w\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{w^2}{{\left(n\pi \right)}^2})$$

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