Evaluate \displaystyle \prod_{n = 1}^{1006} \sin \left(\frac {n \pi} {2013}\right).
[originally posted by Lord of the Flies on TSR]
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Note that from \sin (\pi - x) = \sin x we have:
\displaystyle \prod_{n = 1}^{1006} \sin \left(\frac {n \pi} {2013}\right) = \sqrt {\prod_{n = 1}^{2012} \sin \left(\frac {n \pi} {2013}\right)}
(see ish101’s post for a more detailed justification of this)
Note that we can rewrite:
\begin{align*}\prod_{n = 1}^{2012} \sin \left(\frac {n \pi} {2013}\right) & = \left(\frac 1 {2i}\right)^{2012} \times \prod_{n = 1}^{2012} \left(e^{n \pi i/2013} - e^{-n \pi i/2013}\right) \\ & = 2^{-2012} i^{-2012} \prod_{i = 1}^{2012} \left(-e^{-n \pi i/2013} \left(1 - e^{2 n \pi i/2013}\right)\right) \\ & = 2^{-2012} (-1)^{-1006} \prod_{i = 1}^{2012} \left(-e^{-n \pi i/2013}\right) \prod_{i = 1}^{2012} \left(1 - e^{2 n \pi i/2013}\right) \\ & = 2^{-2012} (-1)^{2012} \times \prod_{n = 1}^{2012} e^{-n \pi i/2013} \times \prod_{n = 1}^{2012} \left(1 - e^{2 n \pi i/2013}\right) \\ & = 2^{-2012} \exp \left(-\frac {2013 \times 2012} 2 \times \frac {\pi i} {2013}\right) \prod_{n = 1}^{2012} \left(1 - e^{2 n \pi i/2013}\right) \\ & = 2^{-2012} e^{-1006 \pi i} \prod_{n = 1}^{2012} \left(1 - e^{2 n \pi i/2013}\right) \\ & = 2^{-2012} \prod_{n = 1}^{2012} \left(1 - e^{2 n \pi i/2013}\right)\end{align*}
noting that i^{2012} = (-1)^{1006} = 1 and e^{1006 \pi i} = (-1)^{1006} = 1.
Note that:
\displaystyle x^n - 1 = \prod_{k = 0}^{n - 1} \left(x - e^{2\pi i k/n}\right) = (x - 1) \prod_{k = 1}^{n - 1} \left(x - e^{2\pi i k/n}\right)
So:
\displaystyle \sum_{k = 0}^{n - 1} x^k = \prod_{k = 1}^{n - 1} \left(x - e^{2\pi i k/n}\right)
Setting n = 2013, x = 1 gives:
\displaystyle \prod_{n = 1}^{2012} \sin \left(\frac {n \pi} {2013}\right) = 2^{-2012} \prod_{n = 1}^{2012} \left(1 - e^{2 n \pi i/2013}\right) = 2^{-2012} \sum_{k = 0}^{2012} 1 = \frac {2013} {2^{2012}}
So that:
\displaystyle \prod_{n = 1}^{1006} \sin \left(\frac {n \pi} {2013}\right) = \frac {\sqrt{2013}} {2^{1006}}
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i was typing up a solution too lol, it was quite similar so i’ll post the first bit
FOR FIRST BIT
we know that \displaystyle\sin\left(\frac{n\pi}{2013}\right)=\sin\left(\pi-\frac{n\pi}{2013}\right)=\sin\left(\frac{(2013-n)\pi}{2013}\right)
consider
\begin{align}
\prod_{n=1}^{2012}\sin\left(\frac{n\pi}{2013}\right)
&= \left(\prod_{n=1}^{1006}\sin\left(\frac{n\pi}{2013}\right)\right)\left(\prod_{n=1007}^{2012}\sin\left(\frac{n\pi}{2013}\right)\right)
\\&= \left(\prod_{n=1}^{1006}\sin\left(\frac{n\pi}{2013}\right)\right)\left(\prod_{n=1007}^{2012}\sin\left(\frac{(2013-n)\pi}{2013}\right)\right)
\\&= \left(\prod_{n=1}^{1006}\sin\left(\frac{n\pi}{2013}\right)\right)\left(\prod_{n=1}^{1006}\sin\left(\frac{n\pi}{2013}\right)\right)
\\&= \left(\prod_{n=1}^{1006}\sin\left(\frac{n\pi}{2013}\right)\right)^2
\\ \implies \prod_{n=1}^{1006}\sin\left(\frac{n\pi}{2013}\right)&=\sqrt{\prod_{n=1}^{2012}\sin\left(\frac{n\pi}{2013}\right)}
\end{align}
now it will suffice to evaluate \displaystyle\prod_{n=1}^{2012}\sin\left(\frac{n\pi}{2013}\right) as has been done above.
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the second line should have (-1)^{1006} right? from i^{2012}
The (-1)^{2012} was taken out of the product
which product do you mean? (also i think you mistyped it as (-1)^{1016})
corrected that now
i’ve added in some extra steps, hope it’s clearer
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This is unreal