MHB Have you tried using the double angle formula for cosine?

AI Thread Summary
The discussion revolves around proving the equation involving the sum of cosine to the fourth power. The original poster struggles with the equation and expresses frustration over encountering sine and cosine terms that complicate the proof. A user suggests using the power reduction formula for cosine to simplify the summation. This method proves effective, leading to a positive response from the original poster, who thanks the user for the helpful suggestion. The exchange highlights the collaborative nature of problem-solving in mathematical discussions.
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.
 
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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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