- #1
xuying1209
- 4
- 0
if n is an odd, cosπ/n+cos3π/n+cos5π/n+...+cos(2n-1)π/n is equal to what?
And how can I prove it??
And how can I prove it??
Odd Cosn Problem: What is cosπ/n+cos3π/n+...+cos(2n-1)π/n? Prove It!
- Context: Graduate
- Thread starter xuying1209
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SUMMARY
The discussion centers on the mathematical expression involving the sum of cosines: cos(π/n) + cos(3π/n) + ... + cos((2n-1)π/n) for odd values of n. It is established that for n > 1, this sum equals zero due to the symmetry of the 2nth roots of unity about the imaginary axis. The initial confusion arose from misinterpreting the symbol "π" as "n," which led to clarifications on the correct interpretation of the summation. The conclusion is that the sum of these cosine terms for odd n simplifies to zero.
PREREQUISITES- Understanding of trigonometric functions, specifically cosine.
- Familiarity with complex numbers and roots of unity.
- Basic knowledge of summation notation and series.
- Concept of symmetry in mathematical functions.
- Study the properties of 2nth roots of unity in complex analysis.
- Explore the implications of symmetry in trigonometric functions.
- Learn about summation techniques in trigonometry.
- Investigate the relationship between cosine functions and complex exponentials.
Mathematicians, students studying trigonometry and complex analysis, and anyone interested in the properties of roots of unity and their applications in summation problems.
- #2
HallsofIvy
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For n= 1, cos(pi/1)= -1.
For n> 1, n odd, essentially you are adding the real parts of the 2nth roots of unity. Since those roots are symmetric about the imaginary axis, the sum is 0.
For n> 1, n odd, essentially you are adding the real parts of the 2nth roots of unity. Since those roots are symmetric about the imaginary axis, the sum is 0.
- #3
tehno
- 375
- 0
Ahh that it was... almost racked my brains out 'cause "π" I read as n ( not \pi) ...
- #4
robert Ihnot
- 1,057
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I am not sure what is being asked. Is this \sum cos(n_i)/n_i, or \sum cos(pi*n_i/n_i), or what?
Last edited:
- #5
HallsofIvy
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I interpreted as sum of cos(i\pi/n)[/tex] for i= 1 to n-1.
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