# TrigonometryTrigonometric Summation

#### anemone

##### MHB POTW Director
Staff member

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

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.

Last edited:

#### Sudharaka

##### Well-known member
MHB Math Helper

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

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.

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

##### MHB POTW Director
Staff member
Hi Sudharaka,

Thank you so much!

-anemone

MHB Math Helper
Hi Sudharaka,