Have you tried using the double angle formula for cosine?

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SUMMARY

The forum discussion centers on proving the equation $\displaystyle \sum_{k=1}^{n}\cos^4\left(\frac{k\pi}{2n+1} \right)=\frac{6n-5}{16}$. User anemone struggles with the proof, particularly when substituting $n=2$, leading to complex terms involving $\sin \frac {\pi}{10}$ and $\cos \frac {\pi}{10}$. User Sudharaka suggests employing the power reduction formula for cosine, transforming the summation into a more manageable form, which successfully aids in solving the problem.

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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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