Let $\theta = \frac{\pi}{2n+1}$. Then the desired sum equals $$S=\cos{2\theta} + \cos{4\theta} +\cdots +\cos{2n\theta}$$ Multiplying $S$ by $2\sin\theta$ and applying the sum to product transformation, \begin{align} 2\sin\theta\cos{2\theta} &= \sin{3\theta}-\sin\theta\\ 2\sin\theta\cos{4\theta} &= \sin{5\theta}-\sin{3\theta}\\ \cdots\\ 2\sin\theta\cos{2n\theta} &= \sin{(2n+1)\theta}-\sin{(2n-1)\theta} \end{align} Adding these equations yields and also noting $\sin(2n+1)\theta=0$: $$2\sin\theta\cdot S = \sin{(2n+1)\theta}-\sin\theta=-\sin\theta\implies S=\boxed{-\frac{1}{2}}$$