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??
Mar 28, 2007 #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??
Mar 28, 2007 #2 HallsofIvy Science Advisor Homework Helper 43,008 974 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, 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.
Mar 28, 2007 #3 tehno 367 0 Ahh that it was... almost racked my brains out 'cause "π" I read as n ( not [itex]\pi[/itex]) ...
Mar 28, 2007 #4 robert Ihnot 1,059 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: Mar 28, 2007
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]
Mar 28, 2007 #5 HallsofIvy Science Advisor Homework Helper 43,008 974 I interpreted as sum of [itex] cos(i\pi/n)[/tex] for i= 1 to n-1.