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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: 7,305 Thanks: 2443 Math Focus: Mainly analysis and algebra 
Sorry, it's too early in the morning for me!
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: 7,305 Thanks: 2443 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 $\sin{(A+B)}  \sin{(AB)} = 2\cos{A}sin{B}$. 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 $\cos{(A+B)} + \cos{(AB)} = 2\cos{A}\cos{B}$ 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: 18,954 Thanks: 1601 
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. 
May 1st, 2014, 05:20 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,305 Thanks: 2443 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 $$ \sum_{n=0}^{m1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = \frac{1}{2} $$ My method for getting to the posted question also suggests the hypothesis $$ \prod_{n=0}^{m1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = (2)^{m} $$ Last edited by v8archie; May 1st, 2014 at 05:31 PM. 
May 1st, 2014, 09:12 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,954 Thanks: 1601 
$$ \prod_{n=0}^{m1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = (1)^k2^{m},\text{ where }k\text{ is the integer part of }\frac{m}{2}. $$


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