Math Problem Statement

cos(2pi/n) + cos(4pi/n) + .... + cos(2pi(n-1)/n) = -1 prove by mathematical induction

Solution

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

Sn=cos(2πn)+cos(4πn)++cos(2π(n1)n).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: Sn=1,for n2.S_n = -1, \quad \text{for } n \geq 2.


Base Case (n=2n = 2):

For n=2n = 2, the terms in the summation are: S2=cos(2π2)=cos(π).S_2 = \cos\left(\frac{2\pi}{2}\right) = \cos(\pi). Using the property cos(π)=1\cos(\pi) = -1: S2=1.S_2 = -1.

Thus, the base case holds.


Inductive Hypothesis:

Assume the statement holds for n=kn = k, i.e., Sk=cos(2πk)+cos(4πk)++cos(2π(k1)k)=1.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+1n = k+1: Sk+1=cos(2πk+1)+cos(4πk+1)++cos(2πkk+1)=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.


Proof for n=k+1n = k+1:

Key Idea:

The terms cos(2πk+1),cos(4πk+1),,cos(2πkk+1)\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)th(k+1)^\text{th} roots of unity (excluding 11).

Roots of Unity Representation:

The (k+1)th(k+1)^\text{th} roots of unity are given by: e2πim/(k+1)for m=0,1,2,,k.e^{2\pi i m / (k+1)} \quad \text{for } m = 0, 1, 2, \ldots, k. The summation of all (k+1)th(k+1)^\text{th} roots of unity is: m=0ke2πim/(k+1)=0.\sum_{m=0}^k e^{2\pi i m / (k+1)} = 0. Excluding the root m=0m = 0 (which corresponds to 11), we have: m=1ke2πim/(k+1)=1.\sum_{m=1}^k e^{2\pi i m / (k+1)} = -1.

Separating Real and Imaginary Parts:

The real part of each root is: Re(e2πim/(k+1))=cos(2πmk+1).\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(2πk+1)+cos(4πk+1)++cos(2πkk+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).

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


Conclusion:

By induction, we have shown that: Sn=1for all n2.S_n = -1 \quad \text{for all } n \geq 2.


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

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Mathematical Induction
Complex Numbers
Roots of Unity

Formulas

Sum of cosines: S_n = cos(2π/n) + cos(4π/n) + ... + cos(2π(n-1)/n)
Roots of Unity: e^(2πi*k/n)
Euler's Formula: e^(iθ) = cos(θ) + i*sin(θ)

Theorems

Roots of Unity Theorem
Properties of Complex Numbers

Suitable Grade Level

Grades 11-12 and Undergraduate Mathematics