Problem 24 *

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