1. **State the problem:** Prove that $$\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2}$$. 2. **Recall the formula for sum of cosines of equally spaced angles:** For integer $n$, the sum $$\sum_{k=1}^{n-1} \cos\frac{2\pi k}{n} = -1$$. 3. For $n=7$, the sum of cosines of $\frac{2\pi k}{7}$ for $k=1$ to $6$ is: $$\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} + \cos\frac{8\pi}{7} + \cos\frac{10\pi}{7} + \cos\frac{12\pi}{7} = -1$$ 4. Use the periodicity and symmetry of cosine: $$\cos\frac{8\pi}{7} = \cos\left(2\pi - \frac{6\pi}{7}\right) = \cos\frac{6\pi}{7}$$ $$\cos\frac{10\pi}{7} = \cos\left(2\pi - \frac{4\pi}{7}\right) = \cos\frac{4\pi}{7}$$ $$\cos\frac{12\pi}{7} = \cos\left(2\pi - \frac{2\pi}{7}\right) = \cos\frac{2\pi}{7}$$ 5. Substitute these back: $$\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} + \cos\frac{6\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{2\pi}{7} = -1$$ 6. Group like terms: $$2\left(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}\right) = -1$$ 7. Divide both sides by 2: $$\cancel{2}\left(\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7}\right) = \frac{-1}{\cancel{2}}$$ 8. Simplify: $$\cos\frac{2\pi}{7} + \cos\frac{4\pi}{7} + \cos\frac{6\pi}{7} = -\frac{1}{2}$$ **Final answer:** $$\boxed{-\frac{1}{2}}$$