Trigonometric formulas doesn't work here

For any positive integer $n$ , the roots of $X^n + 1 = 0$ are $-1$ and $e^{\pm\frac{(2k+1)\pi i}{2n+1}}$ for $0 \le k \le n-1$ . Thus the sum of these roots is $0$ , and hence $0 \; = \; -1 + \sum_{k=0}^{n-1} \left(e^{\frac{(2k+1)\pi i}{2n+1}} + e^{-\frac{(2k+1)\pi i}{2n+1}}\right) \; = \; -1 + 2\sum_{k=0}^{n-1} \cos\left(\frac{(2k+1)\pi}{2n+1}\right)$ so that $\sum_{k=0}^{n-1} \cos\left(\frac{(2k+1)\pi}{2n+1}\right) \; = \; \boxed{\frac12}$

In general, $\displaystyle \sum_{k=0}^{n-1} \cos \left(\frac {2k+1}{2n+1} \right) = \dfrac 12$ (see proof here ).