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:
We aim to show that:
Base Case ():
For , the terms in the summation are: Using the property :
Thus, the base case holds.
Inductive Hypothesis:
Assume the statement holds for , i.e.,
We need to prove that the statement also holds for :
Proof for :
Key Idea:
The terms form the real parts of the roots of unity (excluding ).
Roots of Unity Representation:
The roots of unity are given by: The summation of all roots of unity is: Excluding the root (which corresponds to ), we have:
Separating Real and Imaginary Parts:
The real part of each root is: Thus, the summation of the real parts is:
By the roots of unity property, the sum of all the complex roots (real and imaginary parts together) is . Hence, the real part of the sum is also:
Conclusion:
By induction, we have shown that:
Questions for Further Exploration:
- What are the roots of unity, and how are they used in this proof?
- How does the property of symmetry simplify summations involving cosines?
- Could this result be extended to sums involving sine terms?
- What is the geometric interpretation of summing roots of unity?
- 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
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