Why is: \sum_{k=1}^{[\frac{n}{2}]}\sin^2((2k-1)\frac{\pi}{n})=\frac{n}{4}

Scoopedepalj 2022-02-28 Answered
Why is:
k=1[n2]sin2((2k1)πn)=n4
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Expert Answer

chezpepina87j
Answered 2022-03-01 Author has 6 answers
Let
T=k=1[n2]cos((2k1)2πn)
S=k=1[n2]sin2((2k1)πn)
We can get the relation T+2S=[n2]
If n is even then using summation of cosines whose angles are in AP we get,
T=k=1[n2]cos((2k1)2πn)=sin[n2]2πnsin2πncos[2πn+([n2]1)2πn]=0
S=n4
If n is odd then let n=2m+1. We get,
T=k=1mcos((2k1)2π2m+1)
=sin2πm2m+1sin2π2m+1cos[2π2m+1+(m1)2π2m+1]
=sin2πm2m+1cos2πm2m+1sin2π2m+1
=sin2π2m+12sin2π2m+1=12
S=2m+12=n4
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