Evaluating a product of sines [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 21 '18 at 17:49

• They are the same. $\prod_{k=1}^{n-1}(2a_k)=2^{n-1}\prod_{k=1}^{n-1}a_k$ – CY Aries Mar 21 '18 at 17:40
You should think about what the product sign actually means: $$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}=2\sin\frac{1\pi}{n}\cdot 2\sin\frac{2\pi}{n}\cdot \cdots \cdot 2\sin\frac{(n-2)\pi}{n}\cdot2\sin\frac{(n-1)\pi}{n}$$
$$\prod_{k=1}^{n-1}2\sin\frac{k\pi}{n}=2^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}=(2\cdot2\cdot\cdots\cdot2)\sin\frac{1\pi}{n}\cdot \sin\frac{2\pi}{n}\cdot \cdots \cdot\sin\frac{(n-1)\pi}{n}$$
• Yes. It is clear now, thus taken off the question. How about the evaluation of $(1)$? – John Glenn Mar 21 '18 at 17:44