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 May 1st, 2014, 04:58 AM #1 Member   Joined: Jan 2014 From: Tashkent, Uzbekistan Posts: 52 Thanks: 2 Multiplication of 3 cosines Hello everyone! Could you help to calculate: $\displaystyle \cos\frac{\pi}{7}\cos\frac{3\pi}{7}\cos\frac{5\pi} {7}$ or give a tip how to solve such kind of problems.
 May 1st, 2014, 06:07 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,614 Thanks: 2603 Math Focus: Mainly analysis and algebra Sorry, it's too early in the morning for me! Thanks from Olinguito Last edited by v8archie; May 1st, 2014 at 06:10 AM.
 May 1st, 2014, 11:16 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,614 Thanks: 2603 Math Focus: Mainly analysis and algebra Right, I have a solution. First, we will need to know the sum of the three terms. You can work this out by multiplying $\cos{\frac{\pi}{7}}+\cos{\frac{3\pi}{7}}+\cos{\fr ac{5\pi}{7}}$ by $\sin{\frac{\pi}{7}}$ and using $\sin{(A+B)} - \sin{(A-B)} = 2\cos{A}sin{B}$. Then for the product, first you square it and multiply by -8. Then use $\cos{A} = -\cos{(\pi - A)}$ to get a product of $\cos{\frac{n\pi}{7}}$ for $n = 1\cdots 6$. Next, pair up the terms and use $\cos{(A+B)} + \cos{(A-B)} = 2\cos{A}\cos{B}$ to get a product of three terms. When you expand these terms, and manipulate then using one of the identities mentioned already, you will get and expresssion in the product sought, the sum solved at the start and constants. I make the answer $-\frac{1}{8}$
 May 1st, 2014, 04:47 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 8sin(π/7)cos(π/7)cos(3π/7)cos(5π/7) = 4sin(2π/7)cos(3π/7)cos(5π/7) = 4sin(5π/7)cos(5π/7)cos(3π/7) = 2sin(10π/7)cos(3π/7) = -sin(6π/7) = -sin(π/7) Hence cos(π/7)cos(3π/7)cos(5π/7) = -1/8. Thanks from v8archie
 May 1st, 2014, 05:20 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,614 Thanks: 2603 Math Focus: Mainly analysis and algebra Oh, that's a bit smoother. Although, I'd have said the penultimate line is better as $\sin{(\frac{13\pi}{7})}$. Doing some working in my head earlier, I came to the hypothesis that $$\sum_{n=0}^{m-1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = \frac{1}{2}$$ My method for getting to the posted question also suggests the hypothesis $$\prod_{n=0}^{m-1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = (-2)^{-m}$$ Last edited by v8archie; May 1st, 2014 at 05:31 PM.
 May 1st, 2014, 09:12 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,370 Thanks: 2007 $$\prod_{n=0}^{m-1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = (-1)^k2^{-m},\text{ where }k\text{ is the integer part of }\frac{m}{2}.$$

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