Have you tried using the double angle formula for cosine?

[anemone]

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.

 

[Sudharaka]

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.

 

[anemone]

Hi Sudharaka,

Wow, your suggestion works great!(Smile)

Thank you so much!:D

-anemone

 

[Sudharaka]

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. :)

 

[CellCoree]

I would suggest trying to approach this problem using the double angle formula for cosine. This formula states that $\cos(2x) = 2\cos^2(x) - 1$. By substituting this formula into the original equation, we can rewrite it as:

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

We can then expand and simplify this equation to see if it leads us to a solution. Additionally, we could also try using other trigonometric identities, such as the Pythagorean identity, to manipulate the equation and hopefully find a solution.

It is also important to note that sometimes, the most efficient way to solve a problem is not always obvious. It is important to keep an open mind and try different approaches until a solution is found. Good luck with your problem!