July 13th, 2017, 07:13 PM  #1 
Senior Member Joined: Jul 2011 Posts: 395 Thanks: 15  Trigonometric Product
If $\displaystyle \theta = \frac{2\pi}{2009},$ Then $\displaystyle \cos \theta \cdot \cos 2\theta\cdot \cos 3 \theta \cdot\cdots \cos 1004 \theta$ is

July 16th, 2017, 11:51 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1384 
1/2^1004

July 16th, 2017, 12:49 PM  #3 
Senior Member Joined: Oct 2009 Posts: 141 Thanks: 59 
This can be easily solved using complex numbers, let me start by giving a hint: $$z^n  1 = \prod_{\nu = 1}^n (z  e^{2\pi\nu/n})$$ Try to rewrite the right hand side to the product of cosines. 
July 17th, 2017, 02:10 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,919 Thanks: 1384 
If $\theta = 2\pi/9$, using cos(A) ≡ sin(2A)/(2sin(A)) for nonzero A, $\cos(\theta)\cos(2\theta) \cos(3\theta)\cos(4\theta) = \dfrac{\sin(4\pi/9)\sin(8\pi/9)\sin(12\pi/9)\sin(16\pi/9)}{16\sin(2\pi/9)\sin(4\pi/9)\sin(6\pi/9)\sin(8\pi/9)} = 1/16$. The same method applies to the original problem. 

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product, trigonometric 
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