Prove that for all $n \in \mathbb N$ the following is true: \[2^n \prod_{k=1}^n \sin \frac{k \pi}{2n+1} = \sqrt{2n+1}\]
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Tags: trigonometry, algebra, polynomial, Trigonometric Identities, IMO Shortlist
Gib Z
22.09.2010 15:14
First we prove the following: \[ \prod_{k=1}^{n-1}\sin\frac{k\pi}{n}= n \cdot 2^{1-n} \]
Let $ \omega = e^{\frac{i\pi}{n}}$ . Since $ \sin x = \frac{ e^{ix} - e^{-ix}}{2i} $ the product is then equal to \[ \left( \frac{ \omega - \omega^{-1} }{2i}\right) \left( \frac{ \omega^2 - \omega^{-2} }{2i} \right) \left( \frac{ \omega^3 - \omega^{-3} }{2i} \right) \cdots \left( \frac{ \omega^{n-1} - \omega^{- (n-1)} }{2i} \right) \]
\[ = (1- \omega^{-2})(1-\omega^{-4})(1-\omega^{-6}) \cdots (1-\omega^{-2(n-1)}) \omega^{ n(n-1)/2} (2i)^{-(n-1)} \]
Since $ \omega^{n/2} = i $, it remains to show that \[ (1- \omega^{-2})(1-\omega^{-4})(1-\omega^{-6}) \cdots (1-\omega^{-2(n-1)}) = n \]
Consider the polynomial given by \[ P(x) = (x- \omega^{-2})(x-\omega^{-4})(x-\omega^{-6}) \cdots (x-\omega^{-2(n-1)}) \]
We know it's roots are all complex n-th roots of unity except 1, so we must have $ P (x) = \frac{x^n-1}{x-1} = 1 + x + x^2 + \cdots + x^{n-1} $. Putting $ x=1 $ completes the proof of the product formula.
Now we may proceed with the problem: \[ \prod_{k=1}^n \sin \left( \frac{k\pi}{2n+1} \right) = \frac{ \prod_{k=1}^{2n} \sin \left( \frac{k\pi}{2n+1} \right)}{ \prod_{k=n+1}^{2n} \sin \left( \frac{k\pi}{2n+1} \right)} \] Using $ \sin (\pi - x) = \sin x $ on each term in the denominator, and the product formula we derived before gives : \[ \prod_{k=1}^n \sin \left( \frac{k\pi}{2n+1} \right) = \frac{(2n+1)\cdot 2^{1-(2n+1)}}{ \prod_{k=1}^{n} \sin \left( \frac{k\pi}{2n+1} \right)} \] Solving for the product gives \[ \prod_{k=1}^n \sin \left( \frac{k\pi}{2n+1} \right) = 2^{-n} \sqrt{2n+1} \] as required.
opptoinfinity
21.10.2018 07:25
If $z$ be a complex number then $z-z^{-1}=2isin\theta$. Putting $z=cos\frac{\pi}{2n+1}+isin\frac{\pi}{2n+1}$ and using De Moivre's formula we have $A=\prod_{k=1}^{n}(z^k-z^{(-k)}=i^n\sqrt{2n+1}$ Also the roots of the polynomial $z^{2n+1}-1=0$ are in the set $z^{2k}$ now we know $\frac{x^{2n+1}-1}{x-1}=x^2n+x^{2n-1}+. . .+x+1$ So actually we have $(-1)^nz^{\frac{n(n+1)}{2}}A=\prod_{k=1}^{n}(1-z^{2k)}$ and also $z^{\frac{-n(n+1)}{2}}A=\prod_{k=1}^{n}(1-z^{-2k})$. Multiply this two equations we get $\prod_{k=1}^{n}(1-z^{2k})(1-z{-2k})=2n+1$ therefore $A=\frac{\sqrt{2n+1}{i^n}}. Proved!
opptoinfinity
21.10.2018 07:30
opptoinfinity wrote: If $z$ be a complex number then $z-z^{-1}=2isin\theta$. Putting $z=cos\frac{\pi}{2n+1}+isin\frac{\pi}{2n+1}$ and using De Moivre's formula we have $A=\prod_{k=1}^{n}(z^k-z^{-k})=i^n\sqrt{2n+1}$ Also the roots of the polynomial $z^{2n+1}-1=0$ are in the set $z^{2k}$ now we know $\frac{x^{2n+1}-1}{x-1}=x^{2n}+x^{2n-1}+. . .+x+1$ So actually we have $(-1)^nz^{\frac{n(n+1)}{2}}A=\prod_{k=1}^{n}(1-z^{2k})$ and also $z^{\frac{-n(n+1)}{2}}A=\prod_{k=1}^{n}(1-z^{-2k})$. Multiply this two equations we get $\prod_{k=1}^{n}(1-z^{2k})(1-z^{-2k})=2n+1$ therefore $A=\frac{\sqrt{2n+1}}{i^n}$. Proved! Actually edited.
SPHS1234
04.07.2021 21:43
Consider a $2n+1 $ sided regular polygon $A_1(1),A_2,(\alpha),\cdots A_{2n+1}(\alpha^{2n})$ Recognize the LHS as $A_1A_2 .A_1A_3.\cdots A_1A_{n+1}$ $=|1-\alpha||1-\alpha^2|\cdots |1-\alpha^n|$ $=|1-\alpha^{2n}||1-\alpha^{2n-1}|\cdots |1-\alpha^{n+1}|$ =$\sqrt{|1-\alpha||1-\alpha^2|\cdots |1-\alpha^n| |1-\alpha^{2n}||1-\alpha^{2n-1}|\cdots |1-\alpha^{n+1}|}$ $=\sqrt{2n+1}$ as $|x-\alpha||x-\alpha^2|\cdots |x-\alpha^n| |x-\alpha^{2n}||x-\alpha^{2n-1}|\cdots |x-\alpha^{n+1}|$ $=x^{2n}+x^{2n-1}+\cdots 1$