- #1
xuying1209
- 4
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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!
- Thread starter xuying1209
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The discussion centers on the sum of cosines for odd integers n, specifically the expression cos(π/n) + cos(3π/n) + ... + cos((2n-1)π/n). For n=1, the result is -1, while for n greater than 1, the sum equals zero due to the symmetry of the 2nth roots of unity. There is some confusion about the interpretation of the summation, with participants clarifying whether it refers to cos(iπ/n) for i from 1 to n-1. Ultimately, the key takeaway is that the sum simplifies to zero for odd n greater than one. The conversation highlights the importance of correctly interpreting mathematical expressions in such 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
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Ahh that it was... almost racked my brains out 'cause "π" I read as n ( not \pi) ...
- #4
robert Ihnot
- 1,058
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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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