Summing Products Of Cosines

Relevant wiki: Roots of Unity

First off, we evaluate $\large \displaystyle \sum_{i=1}^{2016} (\cos ia)$ .

Consider the polynomial $x^{2016}+x^{2015}+...+1$ . Then, this polynomial has roots (  cos (ia)+ \sin ia  for i=1,2,...,2016 . So, the sum of the roots of this equation is \frac{-b}{a} = \frac{-1}{1} = -1 by Vieta's formulas. But, since \cos (2\pi - \theta) = -\sin(\theta) , the sum of the roots becomes

$\large \displaystyle \sum_{i=1}^{2016} (\cos ia) + \large \displaystyle \sum_{k=1}^{2016} (i\sin ka)$

$= \large \displaystyle \sum_{i=1}^{2016} (\cos ia) + \large \displaystyle \sum_{k=1}^{1008} (i\sin ka) + \large \displaystyle \sum_{k=1}^{1008} (i\sin (2\pi-ka))$

$= \large \displaystyle \sum_{i=1}^{2016} (\cos ia) + \large \displaystyle \sum_{k=1}^{1008} (i\sin ka) - \large \displaystyle \sum_{k=1}^{1008} (i\sin ka)$

$= \large \displaystyle \sum_{i=1}^{2016} (\cos ia) = -1$ (Since the sum of the roots is $-1$ ).

Next, we evaluate $\large \displaystyle \sum_{i=1}^{2016} (\cos^2 ia)$

By the identity $\large \displaystyle \sum_{i=1}^{n-1} (\cos^2 \frac{2i\pi}{n}) = \frac{n-2}{2}$ ,

$\large \displaystyle \sum_{i=1}^{2016} (\cos^2 ia) = \frac{2017-2}{2} = 1007.5$

Also, by a little algebraic manipulation,

$\large ( \displaystyle \sum_{i=1}^{2016} (\cos ia))^2 =\displaystyle \sum_{i=1}^{2016} (\cos^2 ia) + 2 (\displaystyle \sum_{1\leq i < j \leq 2016} (\cos ia)(\cos ja))$

Now, plugging in the values in the above equation, i.e. $\large \displaystyle \sum_{i=1}^{2016} (\cos ia) = -1$ and $\large \displaystyle \sum_{i=1}^{2016} (\cos^2 ia) =1007.5,$

$\large (-1)^2 = 1007.5 + 2 (\displaystyle \sum_{1\leq i < j \leq 2016} (\cos ia)(\cos ja))$

$\large 2 (\displaystyle \sum_{1\leq i < j \leq 2016} (\cos ia)(\cos ja)) = -1006.5$

$\large \displaystyle \sum_{1\leq i < j \leq 2016} (\cos ia)(\cos ja) = \frac {-1006.5}{2} = -503.25$

So, our answer is $\boxed {-503.25}$