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 $\displaystyle x^n1=0.$ We know that the $\displaystyle n$ roots of this equation are $\displaystyle 1,\omega_n,\omega_n^2,\ldots,\omega_n^{n1}$ where $\displaystyle \omega_n=e^{2i\pi/n}.$ Therefore, by the FTOA we can write $\displaystyle x^n1=(x1)(x\omega_n)(x\omega_n^2)\cdots(x\omega_n^{n1}),$ or, since $\displaystyle \omega_n^{nk}=\omega_n^n\omega_n^{k}=1\cdot\omega_n^{k},$ $\displaystyle \begin{align}x^n1&=(x1)(x\omega_n)(x\omega_n^{1})(x\omega_n^2)(x\omega_n^{2})\cdots(x\omega_n^{(n1)/2})(x\omega_n^{(n1)/2})\\ &=(x1)[x^2(\omega_n+\omega_n^{1})x+1][x^2(\omega_n^2+\omega_n^{2})x+1]\cdots[x^2(\omega_n^{(n1)/2}+\omega_n^{(n1)/2})x+1].\end{align}$ Now divide both sides by $\displaystyle (x1),$ 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: 2,767 Thanks: 5  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: 2,767 Thanks: 5  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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