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This question already has an answer here:

Evaluate: $$\prod_{i=1}^{89} \sin (i)$$

My attempt:We know that

$\sin \alpha \cdot \sin(90- \alpha)=\sin \alpha \cos \alpha=\frac{1}{2}\sin 2\alpha$

Then we have:

$$ (\sin1\cdot\sin89)(\sin2\cdot\sin88)\cdots (\sin44\cdot\sin46)\sin45 =\left(\frac{1}{2}\sin2\right)\cdots\left(\frac{1}{2}\sin88\right)\frac{\sqrt{2}}{2} $$

Factorize from $\frac{1}{2}$ then do the same thing again You will get:

$$ \frac{\sqrt{2}}{2^{67}}(\sin4\cdot\sin8\cdots\sin88) $$

Now what to do?Any hints?

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marked as duplicate by Batominovski, Dragonemperor42, Henrik, Daniel W. Farlow, user223391 Aug 7 '16 at 1:02

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ I assume you mean $\sin 1^\circ$ etc. instead of $\sin 1$ etc? $\endgroup$ – Hagen von Eitzen Aug 6 '16 at 14:09
  • $\begingroup$ Here is an answer (but no proof): $$\prod_{n=1}^{89} \sin (n) = \frac{3 \sqrt{10}}{2^{89}}$$ $\endgroup$ – Tolaso Aug 6 '16 at 14:10
  • $\begingroup$ @Taha Akbari Why $2^{67}$ ? $\endgroup$ – Behrouz Maleki Aug 6 '16 at 14:13
  • $\begingroup$ I think it comes down to $\prod_{k=1}^{89}\sin {\left (\frac{k\pi}{180}\right)}=\frac{\sqrt 2}{2^{45}}(\sin {2^{\circ}}\times\sin {4^{\circ}}\times\sin {6^{\circ}}\times\sin {88^{\circ}})$ $\endgroup$ – StubbornAtom Aug 6 '16 at 14:59
  • $\begingroup$ @Behrouz Maleki Because 44 in first then 22 and one of them because $\frac{\sqrt{2}}{2}$. $\endgroup$ – Taha Akbari Aug 6 '16 at 15:00
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Not exactly a closed form, but maybe it is interesting to see the following way. Note that $$\prod_{k=1}^{n}\sin\left(k\right)=\frac{1}{2^{n}i^{n}}\prod_{k=1}^{n}\left(e^{ik}-e^{-ik}\right)=\frac{1}{2^{n}i^{n}}\prod_{k=1}^{n}e^{ik}\prod_{k=1}^{n}\left(1-e^{-2ik}\right) $$ now obviously we have $$\prod_{k=1}^{n}e^{ik}=\exp\left(i\sum_{k=1}^{n}k\right)=\exp\left(i\frac{n\left(n+1\right)}{2}\right) $$ and $$\prod_{k=1}^{n}1-e^{-2ik}=\left(e^{-2i}\right)_{n} $$ where $\left(a,q\right)_{n} $ is the $q$-Pochhammer symbol. So $$\prod_{k=1}^{n}\sin\left(k\right)=\frac{\exp\left(i\frac{n\left(n+1\right)}{2}\right)\left(e^{-2i},e^{-2i}\right)_{n}}{2^{n}i^{n}} $$ now taking $n=89$ we get $$\prod_{k=1}^{89}\sin\left(k\right)=\color{red}{\frac{\exp\left(i4005\right)\left(e^{-2i},e^{-2i}\right)_{89}}{2^{89}i}} $$ now it is interesting to note that $\frac{\exp\left(i4005\right)\left(e^{-2i},e^{-2i}\right)_{89}}{i}$ is real, but at this moment I don't know how to prove it.

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By the simmetry $\sin(180^{\circ}-k^{\circ})=\sin(k^{\circ})$, we have that $$\prod_{k=1}^{179}\sin (k^{\circ})=\sin(90^{\circ})\left(\prod_{k=1}^{89}\sin (k^{\circ})\right)\left(\prod_{k=1}^{89}\sin (180-k^{\circ})\right)=\left(\prod_{k=1}^{89}\sin (k^{\circ})\right)^2$$ On the other hand by Evaluation of a product of sines (noted by H. H. Rugh and Batominovski). $$\prod_{k=1}^{179}\sin (k^{\circ})= \frac{180}{2^{179}}$$ Hence $$\prod_{k=1}^{89}\sin (k^{\circ})=\left(\frac{180}{2^{179}}\right)^{1/2}=\frac{\sqrt{90}}{2^{89}}=\frac{3 \sqrt{10}}{2^{89}}$$ which confirms the comment by Tolaso.

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