Can you find the relation between 1007 and 2015?

Suppose, $x^{2m+1}-a^{2m+1}=0$ ... ... ... (i)

$\implies (x-a)(x^{2m}+ax^{2m-1}+a^2 x^{2m-2}+ ... +a^{2m})=0$

$\implies x=a$ , or , $x^{2m}+ax^{2m-1}+a^2 x^{2m-2}+ ... +a^{2m}=0$ ... ... ...(ii)

Again, from (i),

$x^{2m+1}=a^{2m+1}$

$\implies x^{2m+1}=a^{2m+1} \exp(i \cdot 2\pi r)$ [ $r$ is an integer]

$\implies x=a \exp(i \cdot \frac{ 2\pi r}{2m+1})$

$\implies x=a[ \cos (\frac{ 2\pi r}{2m+1})+i \sin (\frac{ 2\pi r}{2m+1})]$

$\implies x-a \cos (\frac{2\pi r}{2m+1})= i a \sin(\frac{ 2\pi r}{2m+1})$

$\implies x^2-2ax \cos (\frac{2\pi r}{2m+1}) +a^2 \cos ^2 (\frac{2\pi r}{2m+1}) = -a^2 \sin ^2 (\frac{2\pi r}{2m+1})$ [we can square the previous equation without having irrelevant root. The reason is, the reverse process gives us two equations, and both of them are valid for two different values of $r$ ]

$\implies x^2-2ax \cos (\frac{2\pi r}{2m+1}) +a^2=0$

If $r=0$ , then $x=a$ .

We have $m$ numbers of quadratic equations by putting $r=1, 2, 3 ... m$ , and the $2m$ numbers of roots of these equations are also the roots of the following $2m$ degree polynomial equation ( from (ii) ), $x^{2m}+ax^{2m-1}+a^2 x^{2m-2}+ ... +a^{2m}=0$

So, $x^{2m}+ax^{2m-1}+a^2 x^{2m-2}+ ... +a^{2m}= \prod ^m _{r=1} [x^2 - 2ax \cos (\frac{2\pi r}{2m+1})+a^2]$

Equating the the coefficient of $x^{2m-1}$ on both sides,

$-2 \sum _{r=1} ^{m} \cos (\frac{2\pi r}{2m+1}) = 1$

$\implies \sum _{r=1} ^{m} \cos (\frac{2\pi r}{2m+1}) = -\frac{1}{2}$

$In\quad general,\quad \sum _{ r=1 }^{ n }{ \cos { \frac { 2\pi r }{ 2n+1 } } } =\quad \frac { -1 }{ 2 } \\ \left\{ \because \quad 2\cos { \frac { 2\pi r }{ 2n+1 } =\quad { \omega }^{ r } } +\quad { \omega }^{ -r }\\ \Rightarrow \sum _{ r=1 }^{ n }{ \cos { \frac { 2\pi r }{ 2n+1 } } } =\quad \frac { \sum _{ r=1 }^{ 2014 }{ { \omega }^{ r } } }{ 2 } \quad =-\frac { 1 }{ 2 } \right\}$