My Math Forum  

Go Back   My Math Forum > High School Math Forum > Trigonometry

Trigonometry Trigonometry Math Forum


Thanks Tree2Thanks
  • 1 Post By v8archie
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
May 1st, 2014, 05:58 AM   #1
sf7
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.
sf7 is offline  
 
May 1st, 2014, 07:07 AM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,512
Thanks: 2514

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 07:10 AM.
v8archie is offline  
May 1st, 2014, 12:16 PM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,512
Thanks: 2514

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 $-\frac{1}{8}$
v8archie is offline  
May 1st, 2014, 05:47 PM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 19,983
Thanks: 1853

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
skipjack is offline  
May 1st, 2014, 06:20 PM   #5
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,512
Thanks: 2514

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 06:31 PM.
v8archie is offline  
May 1st, 2014, 10:12 PM   #6
Global Moderator
 
Joined: Dec 2006

Posts: 19,983
Thanks: 1853

$$ \prod_{n=0}^{m-1}{\cos{\frac{(2n+1)\pi}{2m+1}}} = (-1)^k2^{-m},\text{ where }k\text{ is the integer part of }\frac{m}{2}. $$
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Trigonometry

Tags
cosine, cosines, multiplication



Search tags for this page
Click on a term to search for related topics.
Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Law of sines/cosines jkh1919 Algebra 1 November 26th, 2011 02:27 PM
Law of Cosines PWW molokach Algebra 0 March 18th, 2011 10:30 AM
Hyperbolic law of cosines Turloughmack Algebra 4 November 24th, 2010 02:56 AM
Hyperbolic law of cosines Turloughmack Applied Math 1 November 23rd, 2010 02:26 AM
law of cosines MG85 Number Theory 3 November 24th, 2009 10:15 PM





Copyright © 2018 My Math Forum. All rights reserved.