Get Triggy With It!

First, let's isolate the sum

$\sum _{ n=1 }^{ k }{ \cos { \left( \frac { 2\pi n }{ 2k+1 } \right) } } =s$

Then, let's look at a different sum, call it $r$

$\sum _{ n=1 }^{ 2p+1 }{ \cos { \left( \frac { 2 \pi n }{ 2p+1 } \right) } } +i\sin { \left( \frac { 2 \pi n }{ 2p+1 } \right) } =r$

By Euler's Formula we know that the above summation is equal to

$\Large \sum _{ n=1 }^{ 2p+1 }{ { e }^{ \frac {2i\pi n }{ 2p+1 } } }=r$

And so we get that the summation is equal to

$\omega+\omega^{2}+\omega^{3}+\omega^{4}+\omega^{5}+\omega^{6}+...+\omega^{k-1}+1=r$

for $\large \omega= e ^{ \frac {2i\pi }{ 2p+1 }}$ .

We can see that $\omega$ is a solution to the equation $x^{2p+1}-1=0$ . Since

$\omega^{2p+1}-1=(\omega-1)(1+\omega+\omega^{2}+\omega^{3}+\omega^{4}+...+\omega^{2p})$

$\omega^{2p+1}-1=(\omega-1)(r)$

And we get that $r=0$ .

Now, let's apply some trigonometric identities to our sum $r$ . We know that $\cos(\theta)=\cos(2\pi-\theta)$ and that $\sin(\theta)=-\sin(2\pi-\theta)$ and so we can cleverly combine some of the terms together to get cancellation.

$0=\cos { \left( \frac { 2 \pi }{ 2p+1 } \right) } +i\sin { \left( \frac { 2 \pi }{ 2p+1 } \right) }+\cos { \left( \frac { 4 \pi }{ 2p+1 } \right) } +i\sin { \left( \frac { 4 \pi }{ 2p+1 } \right) }+\cos { \left( \frac { 6 \pi }{ 2p+1 } \right) } +i\sin { \left( \frac { 6 \pi }{ 2p+1 } \right) }+...+\cos { \left( 2\pi \right) } +i\sin { \left( 2\pi \right) }$

We know that $\cos(2\pi)+i\sin(2\pi)=1$ , and as for the rest of our sum, for every value $i\sin(\theta)$ there is a value $i\sin(2\pi-\theta)$ , so all of the imaginary parts cancel out. Also, for every value of $\cos(\theta)$ , there is a value of $\cos(2\pi-\theta)$ , and so we get that

$r=2\sum _{ n=1 }^{ p }{ \cos { \left( \frac { 2\pi n }{ 2p+1 } \right) } }+1$

$r=2s+1$

And so we can conclude that our original sum $s=-\frac{1}{2}$ for all values of $p$ .

This means we simply have to evaluate

$\left| \sum_{k=1}^{2015}-\frac{1}{2} \right|=\left| -1007.5 \right|=1007.5=z$

Therefore $\left\lfloor z \right\rfloor =\boxed{1007}$

My solution is analogous to Brandon's; let me write a "compressed' version.

We know that $\sum_{n=0}^{2k}\cos\left(\frac{2\pi n}{2k+1}\right)=0$ ; those are just the real parts of the $(2k+1)$ th roots of unity whose sum is zero by Viete. Omitting $\cos(0)=1$ from the sum, we have $\sum_{n=1}^{2k}\cos\left(\frac{2\pi n}{2k+1}\right)=-1$ . Because of the symmetry $\cos(2\pi-x)=\cos(x)$ , we can conclude that $\sum_{n=1}^{k}\cos\left(\frac{2\pi n}{2k+1}\right)=-1/2$ . The sum for $k=1..2015$ comes out to be $-\frac{2015}{2}$ , and the floor of the absolute value is $\boxed{1007}$ .