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

Question

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?

I assume you mean $\sin 1^\circ$ etc. instead of $\sin 1$ etc? — Hagen von Eitzen, Aug 06 '16 at 14:09
Here is an answer (but no proof):
$$\prod_{n=1}^{89} \sin (n) = \frac{3 \sqrt{10}}{2^{89}}$$ — Tolaso, Aug 06 '16 at 14:10
I think it comes down to $\prod_{k=1}^{89}\sin {\left (\frac{k\pi}{180}\right)}=\frac{\sqrt 2}{2^{45}}(\sin {2^{\circ}}\times\sin {4^{\circ}}\times\sin {6^{\circ}}\times\sin {88^{\circ}})$ — StubbornAtom, Aug 06 '16 at 14:59
@Behrouz Maleki Because 44 in first then 22 and one of them because $\frac{\sqrt{2}}{2}$. — Taha Akbari, Aug 06 '16 at 15:00

score 6 · Answer 1 · edited Feb 11 '20 at 16:12

By the symmetry $\sin(180^{\circ}-k^{\circ})=\sin(k^{\circ})$, 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.

score 1 · Accepted Answer · answered Aug 06 '16 at 14:49

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.

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

2 Answers2

Linked

Related