Have you tried using the double angle formula for cosine?

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Discussion Overview

The discussion revolves around the equation $\displaystyle \sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1} \right)=\frac{6n-5}{16}$, with participants exploring methods to prove its validity. The scope includes mathematical reasoning and problem-solving techniques related to trigonometric identities.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in proving the equation, particularly when substituting $n=2$, leading to terms involving $\sin \frac {\pi}{10}$ and $\cos \frac {\pi}{10}$, which they find unhelpful.
  • Another participant suggests using the power reduction formula for cosine within the summation, breaking it down into simpler components and applying Lagrange's trigonometric identity for further simplification.
  • A later reply indicates that the suggested method was effective for the initial poster, expressing gratitude for the assistance.
  • There is no indication that the method proposed has been tested by the suggestor, leaving some uncertainty about its general applicability.

Areas of Agreement / Disagreement

Participants generally agree on the approach suggested by one member, but there is no consensus on the overall validity of the original equation or the effectiveness of the proposed method, as it has not been independently verified by all participants.

Contextual Notes

The discussion does not clarify the assumptions or limitations of the proposed methods, nor does it address potential dependencies on specific definitions or mathematical steps that may be unresolved.

anemone
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Please consider the following equation:

$\displaystyle \sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1} \right)=\frac{6n-5}{16}$

For this particular equation, which I am trying to prove is true, I have found no way to crack it, even if I let $n=2$ and begin to try to combine the terms together, I end up with the annoying terms $\displaystyle \sin \frac {\pi}{10}$ and $\displaystyle \cos \frac {\pi}{10}$ and I am quite certain that this is not the way to go.

I have referred back to Opalg's great posts at this site to search for ideas, but also to no avail...

Any suggestions are welcome to help me to work this problem.

Thanks in advance.
 
Last edited:
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anemone said:
Please consider the following equation:

$\displaystyle \sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1} \right)=\frac{6n-5}{16}$

For this particular equation, which I am trying to prove is true, I have found no way to crack it, even if I let $n=2$ and begin to try to combine the terms together, I end up with the annoying terms $\displaystyle \sin \frac {\pi}{10}$ and $\displaystyle \cos \frac {\pi}{10}$ and I am quite certain that this is not the way to go.

I have referred back to Opalg's great posts at this site to search for ideas, but also to no avail...

Any suggestions are welcome to help me to work this problem.

Thanks in advance.


Hi anemone, :)

Here's a method that I thought of. This may not be the most elegant method however. :)

Use the power reduction formula for the cosine inside the summation.

\[\sum_{k=1}^{n}\cos^4\theta = \sum_{k=1}^{n}\left(\frac{3 + 4 \cos 2\theta + \cos 4\theta}{8}\right)=\frac{3}{8}\sum_{k=1}^{n}1+ \frac{1}{2}\sum_{k=1}^{n}\cos{2\theta}+\frac{1}{8}\sum_{k=1}^{n}\cos{4\theta}\]

where \(\displaystyle\theta=\frac{k\pi}{2n+1}.\)

Then use Lagrange's trigonometric identity for each summation.

Kind Regards,
Sudharaka.
 
Hi Sudharaka,

Wow, your suggestion works great!(Smile)

Thank you so much!:D

-anemone
 
anemone said:
Hi Sudharaka,

Wow, your suggestion works great!(Smile)

Thank you so much!:D

-anemone

You are welcome! Nice to hear that it works; I never actually tried it. :)
 

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