Prove that Summation (n, k=1) of cos (2?k/n) = Summation (n, k=1) of sin (2?k/n)

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cos(0)+cos(2p/N)+cos(4p/N)+... +cos(2p(N-1)) = 0 and sin(0)+sin(2p/N)+sin(4p/N)+... +sin(2p(N-1)) = 0, because cos(0)+cos(2p/N)+cos(4p/N)+... +cos(2p(N-1))+ i*[sin(0)+sin(2p/N)+sin(4p/N)+... +sin(2p(N-1))]= e^(i*0)+e^(i*2p/N)+e^(i*4p/N)+... +e^(i*2p(N-1)) = [Sum of the first N terms of the Geometric progression with a1=e^(i*0)=1 and r=e^(i*2p/N)] = ?[k=0 to N-1] e^(i*2kp/N)=a1*(r^n -1)/(r -1)= {[e^(i*2p/N)]^N -1}/[e^(i*2p/N)-1]= {[e^(i*2p) -1}/[e^(i*2p/N) -1]= (1-1)/[e^(i*2p/N) -1]=0 ==> ?[k=0 to N-1] cos(2kp/N)=0 and ?[k=0 to N-1] sin(2kp/N)=0 BTW, ?[k=0 to N-1] e^(i*2kp/N)=0 is the sum of all roots of the equation z^n=1. In Electrotechnics any function y=cos(?t+f) or y=sin(?t+f) from the Real domain is uniquely represented by Y=e^(if) in the Complex domain. So y=cos(?t–(2pk)/N)) or y=sin(?t–(2pk)/N)) is represented by Y=e^(-i(2pk)/N) and ?[0,N-1] cos(?t–(2pk)/N) or ?[0,N-1] sin(?t–(2pk)/N)) is represented by ?[0,N-1] e^(-i(2pk)/N) = ?[0,N-1] e^(i(2pk)/N) = 0, where N=2

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