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 July 13th, 2017, 07:13 PM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 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: 20,931 Thanks: 2205 1/2^1004
 July 16th, 2017, 12:49 PM #3 Senior Member   Joined: Oct 2009 Posts: 850 Thanks: 327 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. Thanks from greg1313
 July 17th, 2017, 02:10 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 If $\theta = 2\pi/9$, using cos(A) ≡ sin(2A)/(2sin(A)) for non-zero 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. Thanks from greg1313

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