To prove the given equation by mathematical induction, let's denote the summation as:

$S_n = \cos\left(\frac{2\pi}{n}\right) + \cos\left(\frac{4\pi}{n}\right) + \cdots + \cos\left(\frac{2\pi(n-1)}{n}\right).$

We aim to show that: $S_n = -1, \quad \text{for } n \geq 2.$

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Base Case ($n = 2$):

For $n = 2$, the terms in the summation are: $S_2 = \cos\left(\frac{2\pi}{2}\right) = \cos(\pi).$ Using the property $\cos(\pi) = -1$: $S_2 = -1.$

Thus, the base case holds.

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Inductive Hypothesis:

Assume the statement holds for $n = k$, i.e., $S_k = \cos\left(\frac{2\pi}{k}\right) + \cos\left(\frac{4\pi}{k}\right) + \cdots + \cos\left(\frac{2\pi(k-1)}{k}\right) = -1.$

We need to prove that the statement also holds for $n = k+1$: $S_{k+1} = \cos\left(\frac{2\pi}{k+1}\right) + \cos\left(\frac{4\pi}{k+1}\right) + \cdots + \cos\left(\frac{2\pi k}{k+1}\right) = -1.$

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Proof for $n = k+1$:

Key Idea:

The terms $\cos\left(\frac{2\pi}{k+1}\right), \cos\left(\frac{4\pi}{k+1}\right), \ldots, \cos\left(\frac{2\pi k}{k+1}\right)$ form the real parts of the $(k+1)^\text{th}$ roots of unity (excluding $1$).

Roots of Unity Representation:

The $(k+1)^\text{th}$ roots of unity are given by: $e^{2\pi i m / (k+1)} \quad \text{for } m = 0, 1, 2, \ldots, k.$ The summation of all $(k+1)^\text{th}$ roots of unity is: $\sum_{m=0}^k e^{2\pi i m / (k+1)} = 0.$ Excluding the root $m = 0$ (which corresponds to $1$), we have: $\sum_{m=1}^k e^{2\pi i m / (k+1)} = -1.$

Separating Real and Imaginary Parts:

The real part of each root is: $\operatorname{Re}\left(e^{2\pi i m / (k+1)}\right) = \cos\left(\frac{2\pi m}{k+1}\right).$ Thus, the summation of the real parts is: $\cos\left(\frac{2\pi}{k+1}\right) + \cos\left(\frac{4\pi}{k+1}\right) + \cdots + \cos\left(\frac{2\pi k}{k+1}\right).$

By the roots of unity property, the sum of all the complex roots (real and imaginary parts together) is $-1$. Hence, the real part of the sum is also: $S_{k+1} = -1.$

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Conclusion:

By induction, we have shown that: $S_n = -1 \quad \text{for all } n \geq 2.$

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Questions for Further Exploration:

1. What are the roots of unity, and how are they used in this proof?
2. How does the property of symmetry simplify summations involving cosines?
3. Could this result be extended to sums involving sine terms?
4. What is the geometric interpretation of summing roots of unity?
5. Can this proof be adapted for cases where the summation starts with different terms?

Tip:

Understanding the link between trigonometric sums and complex numbers (roots of unity) is key to solving many cyclic summation problems effectively