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Prove that sum(k=1)^(n-1)(n-k)cos(2kpi)...

Prove that `sum_(k=1)^(n-1)(n-k)cos(2kpi)/n=-n/2,w h e r engeq3i sa nin t ege r`

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`S=sum_(k=1)^(n-1)(n-k)cos""(2kpi)/(n)`
`=(n-1)cos""(2pi)/(n)+(n-2)cos 2 (2pi)/(n)+......+1cos(n-1)(2pi)/(n)`
We know that `cos theta=cos(2pi-theta)`. Replacing each angle `theta` by `2pi-theta` in Eq. i, we get
`S=(n-1)cos(n-1)(2pi)/(n)+(n-2)cos(n-2)(2pi)/(n)+.........+1 cos ""(2pi)/(n)` [using Eq. i]
Adding terms having the same angle and taking n common, we have
`2S=n[cos((2pi)/(n))+cos((4pi)/(n))+cos((6pi)/(n))+......+cos(n-1)(2pi)/(n)]`
`=n[(sin(n-1)((pi)/(n)))/(sin((pi)/(n)))cos((2pi)/(n))+(n-1)(2pi)/(n))/(2)`
`=n[(sin(pi-(pi)/(n)))/(sin((pi)/(n)))cospi]=-n`
`therefore S=-n//2`
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