My Math Forum Product of sine

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 August 11th, 2010, 05:00 AM #1 Newbie   Joined: Aug 2010 Posts: 7 Thanks: 0 Product of sine Hi! I have been thinking about this problem for a rather long time, it is taken from a textbook: Prove that $\displaystyle \displaystyle \prod_{k=1}^{13} \sin \frac{k\pi}{27}=\frac{3\sqrt{3}}{2^{13}}$. Any help is appreciated! Last edited by skipjack; February 28th, 2018 at 02:38 PM.
 August 11th, 2010, 07:55 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Product of sine The solution to this is a rather neat one, so I don't want to give it away straight away. Instead, I'll give you a hint, and see how far you can get. Let $\displaystyle n\in\mathbb{N}$ be odd, and consider the polynomial equation $\displaystyle x^n-1=0.$ We know that the $\displaystyle n$ roots of this equation are $\displaystyle 1,\omega_n,\omega_n^2,\ldots,\omega_n^{n-1}$ where $\displaystyle \omega_n=e^{2i\pi/n}.$ Therefore, by the FTOA we can write $\displaystyle x^n-1=(x-1)(x-\omega_n)(x-\omega_n^2)\cdots(x-\omega_n^{n-1}),$ or, since $\displaystyle \omega_n^{n-k}=\omega_n^n\omega_n^{-k}=1\cdot\omega_n^{-k},$ \displaystyle \begin{align}x^n-1&=(x-1)(x-\omega_n)(x-\omega_n^{-1})(x-\omega_n^2)(x-\omega_n^{-2})\cdots(x-\omega_n^{(n-1)/2})(x-\omega_n^{-(n-1)/2})\\ &=(x-1)[x^2-(\omega_n+\omega_n^{-1})x+1][x^2-(\omega_n^2+\omega_n^{-2})x+1]\cdots[x^2-(\omega_n^{(n-1)/2}+\omega_n^{-(n-1)/2})x+1].\end{align} Now divide both sides by $\displaystyle (x-1),$ evaluate the equation at $\displaystyle x=1$ (either take the limit on the left or express the fraction as a sum of powers of $x$) and write the RHS in terms of trigonometric functions. You will need to remember that $\displaystyle \cos\theta=(e^{i\theta}+e^{-i\theta})/2$ and that $\displaystyle \sin\theta=\sin(\pi-\theta).$ Last edited by skipjack; February 28th, 2018 at 02:41 PM.
 August 11th, 2010, 07:59 AM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Product of sine FTOC or FTOA ?? I'm either really stupid, or...
 August 11th, 2010, 08:01 AM #4 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Product of sine No, you're right. Pesky Fundamental Theorems cropping up everywhere and confusing me.
 August 11th, 2010, 08:15 AM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Product of sine And I assume you mean "algebra" and not "arithmetic"!
 August 11th, 2010, 08:24 AM #6 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: Product of sine Oh, shut up
 August 11th, 2010, 07:42 PM #7 Newbie   Joined: Aug 2010 Posts: 7 Thanks: 0 Re: Product of sine Get it! Thanks mattpi!!

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