Prove that `sum_(k=1)^(n-1)(n-k)cos(2kpi)/n=-n/2`, where`ngeq3` is an integer

Explanation

Let `S = sum_(k=1)^(n-1) (n-k) cos((2kpi)/n)`

`=>S = (n-1)cos((2pi)/n)+(n-2)cos(2*(2pi)/n)+...+1*cos((n-1)*(2pi)/n)->(1)`

Now, we know, `costheta = cos(2pi-theta)`

So, replacing `(2pi)/n` with `2pi-(2pi)/n` in the above equation, we get,

`=>S= (n-1)cos(n-1)((2pi)/n)+(n-2)cos(n-1)((2pi)/n)+...+cos((2pi)/n)->(2)`

Now, adding (1) and (2),

`2S = n[cos((2pi)/n) + cos(2*(2pi)/n)+...+cos(n-1)*(2pi)/n]`

`=>2S = nsum_(k=0)^(n-2) cos(((2pi)/n) +k(2pi)/n) )`

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