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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 Tags product, trigonometric Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post srecko Linear Algebra 1 October 27th, 2016 11:25 AM panky Trigonometry 6 December 28th, 2011 10:57 AM otaniyul Linear Algebra 0 October 30th, 2009 06:40 PM ^e^ Real Analysis 6 May 6th, 2007 04:27 PM

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