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Question Number 116626 by frc2crc last updated on 05/Oct/20

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(\frac{a^2{n}^2}{{\left(an\right)}^2-1)}\right)=???$$$$$$

Answered by mnjuly1970 last updated on 05/Oct/20

$$solution$$$$\frac{sin\left(\pi x\right)}{\pi x}=\underset{n=}{\prod ^{\mathrm{\infty }}}(1-\frac{x^2}{n^2})$$$$\frac{\pi x}{sin\left(\pi x\right)}=\underset{n=1}{\prod ^{\mathrm{\infty }}}\frac{n^2}{n^2-x^2}$$$$x=\frac{1}{a}\Rightarrow \frac{\pi }{asin\left(\frac{\pi }{a}\right)}=\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(\frac{a^2{n}^2}{a^2{n}^2-1}\right)$$$$m.n.1970$$$$$$

Answered by Dwaipayan Shikari last updated on 05/Oct/20

$$\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{x^2}{n^2})=\frac{sin\pi x}{\pi x}$$$$\prod ^{\mathrm{\infty }}\left(\frac{a^2{n}^2}{a^2{n}^2-1}\right)=\frac{1}{\underset{n=1}{\prod ^{\mathrm{\infty }}}\left(\frac{a^2{n}^2-1}{a^2{n}^2}\right)}$$$$\frac{1}{\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{1}{a^2{n}^2})}=\frac{1}{\frac{sin\frac{\pi }{a}}{\frac{\pi }{a}}}=\frac{\pi }{a}cosec\frac{\pi }{a}$$

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