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Find value of $$\prod^{n}_{r=1}\sin \left(\frac{\left(2r-1\right)\pi}{2n}\right)$$ Where $n\in \mathbb{N}$ and $n>1$

$\bf{My\; Try::}$ Let $$P = \sin \left(\frac{\pi}{2n}\right)\cdot \sin \left(\frac{3\pi}{2n}\right)\cdot \sin \left(\frac{5\pi}{2n}\right)\cdot \cdot \cdot \cdot \cdot \sin\left(\frac{n\pi}{2n}\right)\cdot\cdot \cdot \cdot \cdot \sin\left(\frac{(2n-1)\pi}{2n}\right)$$

Now we can write $\displaystyle \sin\left(\frac{(2n-1)\pi}{2n}\right)=\sin \left(\frac{\pi}{n}\right)$ Similarly $\displaystyle \sin\left(\frac{(2n-3)\pi}{2n}\right)=\sin \left(\frac{3\pi}{2n}\right)$

So we get $$P = \left[\sin \left(\frac{\pi}{2n}\right)\cdot \sin \left(\frac{3\pi}{2n}\right)\cdot \sin \left(\frac{5\pi}{2n}\right)\cdot \cdot \cdot \cdot \cdot \sin\left(\frac{(n-2)\pi}{2n}\right)\right]^2$$

Now How can i Solve after that, Help me

Thanks

Martin Sleziak
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juantheron
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1 Answers1

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$$\prod^{n}_{r=1}\sin\frac{(2r-1)\pi}{2n}=\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{2v\pi}{2n}} =\dfrac{\prod_{u=1}^{2n-1}\sin\dfrac{u\pi}{2n}}{\prod_{v=1}^{n-1}\sin\dfrac{v\pi}n}=\dfrac{F(2n)}{F(n)}$$

where $F(m)=\prod_{v=1}^{m-1}\sin\dfrac{v\pi}m$

Now from How to prove those "curious identities"?,

$$F(m)=\dfrac m{2^{m-1}}$$

Community
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lab bhattacharjee
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