How to find the minimum value of cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(20x)

Question

I want to find the minimum value of the series cos(2x)+cos(4x)+cos(6x)+cos(8x)+...+cos(2nx). x could be 2pit. Anyone can share a method of how to determine the minimum value of the series?

Answer 1

You can use $\displaystyle \sum_{k=1} \cos(kx)=R\left(\sum_{k=1} e^{2ikx}\right)$

Answer 2

Welcome to MSE. As a hint, you might want to look into Lagrange's trignometric identities. (They can be derived with complex numbers like Luis is suggesting.

Answer 3

Comment:

Let $t=2x$ , you get:

$s=\cos (t)+\cos(2t)+\cos(3t)+\cdot\cdot\cdot +\cos(10t)$

Now use this formula:

$$s=\cos (t)+\cos(2t)+\cos(3t)+\cdot\cdot\cdot +\cos(nt)=\frac{\sin \frac n2 t\cos\frac {n+1}2 t}{\sin (\frac t 2)}$$

So you have to find minimum of this function:

$$y=\frac{\sin(10x)\cos(11x)}{\sin(x)}$$

By plotting in Wolfram minimum of y is about -2.8.