xuying1209 Messages 4 Reaction score 0 Thread starter Mar 28, 2007 #1 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??
HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983 Mar 28, 2007 #2 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.
tehno Messages 375 Reaction score 0 Mar 28, 2007 #3 Ahh that it was... almost racked my brains out 'cause "π" I read as n ( not [itex]\pi[/itex]) ...
robert Ihnot Messages 1,057 Reaction score 1 Mar 28, 2007 #4 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]
HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 983 Mar 28, 2007 #5 I interpreted as sum of [itex]cos(i\pi/n)[/tex] for i= 1 to n-1.[/itex]