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What is the value of
$\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?

The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (180-\theta)$, This reduces the problem as the products till $89^\circ$. But that doesn't help proceed.

Thanks in advanced.

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    $\begingroup$ You probably mean $180°$ ? $\endgroup$ – Yves Daoust May 4 '18 at 16:00
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    $\begingroup$ Are those degree signs or exponents? $\endgroup$ – Andrew Li May 4 '18 at 16:00
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    $\begingroup$ $90^\circ$ can be written with 90^\circ $\endgroup$ – Henry May 4 '18 at 16:04
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    $\begingroup$ @RobbieVanDerzee Something just tells me that that'll work! $\endgroup$ – SmarthBansal May 4 '18 at 16:10
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    $\begingroup$ The answer seems to be $2^{-89}$. See WA. $\endgroup$ – lhf May 4 '18 at 16:11
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Let us use $\sin(1^\circ) = \sin(179^\circ) = \cos(89^\circ)$, etc., to rewrite the product as $$\prod_{i=1}^{45} \cos^2 \left( \frac{\pi}{180} (2i - 1) \right).$$

Now, $\pm \cos \left( \frac{\pi}{180} (2i-1) \right)$ for $i = 1, \ldots, 45$ are roots of the polynomial $P_{180}(x) + 1$, where $P_n$ is the Chebyshev polynomial such that $P_n(\cos \theta) = \cos (n\theta)$. In fact, since $-1$ is the minimum possible value of $P_n(x)$ for $-1 \le x \le 1$, they are all double roots; and this accounts for all 180 roots of the polynomial. On the other hand, $P_{180}(x)$ has the form $2^{179} x^{180} + \cdots + 1$, so $P_{180}(x) + 1$ has the form $2^{179} x^{180} + \cdots + 2$. Therefore, the square of the product above is equal to the product of roots of this polynomial, which is $\frac{2}{2^{179}} = 2^{-178}$; and the original desired product is $2^{-89}$.

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  • $\begingroup$ +1 Thanks for answering. I will take a look. $\endgroup$ – SmarthBansal May 8 '18 at 18:40
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As mentioned in a comment by Hans Lundmark in this question, we have $$ \sin nx=2^{n-1}\prod_{k=0}^{n-1} \sin\left( x + \frac{k\pi}{n} \right) $$ The product we want is $$ \prod_{k=0}^{89} \sin\left(\frac{(2k+1)\pi}{180} \right) = \prod_{k=0}^{90-1} \sin\left(\frac{\pi}{180} + \frac{k\pi}{90} \right) = \frac{\sin\left(90\frac{\pi}{180}\right)}{2^{90-1}} = \frac{\sin\left(\frac{\pi}{2}\right)}{2^{89}} = \frac{1}{2^{89}} $$

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  • $\begingroup$ +1 Thanks for the answer. I would say using that first equation does give me the goosebumps. I'm not used to using that equation. :P $\endgroup$ – SmarthBansal May 4 '18 at 17:30
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    $\begingroup$ @SmarthBansal, we all learn something every day. I learned that equation today researching your question. That's the fun of MSE. $\endgroup$ – lhf May 4 '18 at 17:45
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enter image description here

enter image description here

The product of the chords in this figure is

$(2\cos \frac {2\pi}{5})(2\cos \frac {\pi}{5})(2\cos 0)(2 \cos -\frac {\pi}{5})(2\cos -\frac {2\pi}{5}) = 2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5}$

If we map this figure to the complex plane the product of those lengths = $|(1+e^{\frac {\pi i}{5}})(1+e^{\frac {3\pi i}{5}})(1+e^{\frac {5\pi i}{5}})(1+e^{\frac {7\pi i}{5}})(1+e^{\frac {7\pi i}{5}})|$

Note: $(z+e^{\frac {\pi i}{5}})(z+e^{\frac {3\pi i}{5}})(z+e^{\frac {5\pi i}{5}})(z+e^{\frac {7\pi i}{5}})(z+e^{\frac {7\pi i}{5}}) = z^5 + 1$

Evaluated at $z= 1$

$2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5} = 2\\ \prod_\limits {n=-2}^2 \cos \frac {n\pi}{5} = 2^{-4}$

And this generalizes:

$\prod_\limits {n=1}^k \cos \frac {(2n-1)\pi}{2k} = 2^{-(k-1)}$

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  • $\begingroup$ +1 Thanks for the answer! I don't understand how you map those up in the complex plane? (I mean how do you know for instance that the second chord is $1+ e^{3\pi i /5}$?) (Sry new to this type) $\endgroup$ – SmarthBansal May 8 '18 at 18:51
  • $\begingroup$ The "roots of unity." en.wikipedia.org/wiki/Root_of_unity It falls out of DeMoivre's theorem. if $z = \rho (\cos \theta + i\sin \theta) =\rho e^{i\theta}$ then $z^n = \rho^n (\cos n\theta + i\sin n\theta) =\rho^n e^{i n\theta}$ working backwards if $z^n = -1$ we can back into the possible values of $\theta$ $\endgroup$ – Doug M May 8 '18 at 19:15
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We have to calculate $$\prod^{45}_{k=1}\sin^2((2k-1)^\circ)$$

Because $\sin(180-\theta)=\sin(\theta)$

Now Let $$P=\prod^{45}_{k=1}\sin((2k-1)^\circ)$$

Then $$\prod^{45}_{k=1}\sin(2k^\circ)\cdot P=\prod^{45}_{k=1}\sin((2k-1)^\circ)\cdot \prod^{45}_{k=1}\sin(2k^\circ)$$

So $$\prod^{45}_{k=1}\sin(2k^\circ)\cdot P=\frac{1}{2^{44}}\cdot \frac{1}{\sqrt{2}}\cdot \prod^{45}_{k=1}\sin(2k^\circ)$$

So we get $$P=\frac{1}{2^{\frac{89}{2}}}$$

So $$\prod^{45}_{k=1}\sin^2((2k-1)^\circ)=\frac{1}{2^{89}}$$

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  • $\begingroup$ +1 Thanks for the answer! How'd you get the third-last step from the fourth-last one? $\endgroup$ – SmarthBansal May 4 '18 at 17:20
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    $\begingroup$ -1 This answer contains absolutely nothing... In one of the steps you seem to assume we already know $\prod \sin((2k-1)^\circ)$ in order to compute $P = \prod \sin((2k-1)^\circ)$ ? You assume you know the answer to get the answer $\endgroup$ – Winther May 4 '18 at 19:00

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