- #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!
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- Thread starter xuying1209
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Discussion Overview
The discussion revolves around the sum of cosines of the form cos(π/n), cos(3π/n), ..., cos((2n-1)π/n) for odd values of n. Participants are exploring how to prove the value of this sum and clarifying the interpretation of the expression.
Discussion Character
- Exploratory, Technical explanation, Debate/contested
Main Points Raised
- One participant poses the initial question about the sum and asks for a proof.
- Another participant suggests that for n=1, the sum equals -1, and for n>1 (where n is odd), the sum is 0 due to the symmetry of the 2nth roots of unity.
- A third participant expresses confusion over the interpretation of the original question, questioning whether it refers to a specific summation format.
- Another participant interprets the sum as the cosine of multiples of π/n for i ranging from 1 to n-1, indicating a potential misunderstanding of the limits of the summation.
Areas of Agreement / Disagreement
There is no consensus on the interpretation of the sum or its value. Multiple competing views exist regarding the correct formulation and the resulting sum.
Contextual Notes
Participants have not fully clarified the limits of the summation or the specific form of the expression being evaluated, leading to uncertainty in the discussion.
Who May Find This Useful
Individuals interested in trigonometric sums, roots of unity, or those studying properties of cosine functions in mathematical contexts may find this discussion relevant.
- #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 [itex]\pi[/itex]) ...
- #4
robert Ihnot
- 1,057
- 1
I am not sure what is being asked. Is this [tex]\sum cos(n_i)/n_i, or \sum cos(pi*n_i/n_i), or what?[/tex]
Last edited:
- #5
HallsofIvy
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I interpreted as sum of [itex]cos(i\pi/n)[/tex] for i= 1 to n-1.[/itex]
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