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I saw this product from a question, but got deleted. $$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}$$ Naturally, I was curious, and evaluated this in mathematica, which suprisingly turns out to be: $$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}=n\tag{1}$$ From this question, we have: $$\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=\frac n{2^{n-1}}$$ My question is how is $(1)$ evaluated?


marked as duplicate by Jack D'Aurizio trigonometry Mar 21 '18 at 17:49

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  • 3
    $\begingroup$ They are the same. $\prod_{k=1}^{n-1}(2a_k)=2^{n-1}\prod_{k=1}^{n-1}a_k$ $\endgroup$ – CY Aries Mar 21 '18 at 17:40

You should think about what the product sign actually means: $$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}=2\sin\frac{1\pi}{n}\cdot 2\sin\frac{2\pi}{n}\cdot \cdots \cdot 2\sin\frac{(n-2)\pi}{n}\cdot2\sin\frac{(n-1)\pi}{n}$$

so if we wish to take out the two to the front, we'd get

$$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}=2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=(2\cdot2\cdot\cdots\cdot2)\sin\frac{1\pi}{n}\cdot \sin\frac{2\pi}{n}\cdot \cdots \cdot\sin\frac{(n-1)\pi}{n}$$

  • $\begingroup$ Yes. It is clear now, thus taken off the question. How about the evaluation of $(1)$? $\endgroup$ – John Glenn Mar 21 '18 at 17:44
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    $\begingroup$ Simply use the linked question, and rewrite using what I wrote in my answer $\endgroup$ – vrugtehagel Mar 21 '18 at 17:45

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