Find the value of `sum_(r=1)^(n-1)sin^(2)""(rpi)/(n)`.

Options

A. `(n+1)/(2)`
B. `(n-1)/(2)`
C. `n/2`
D. none of these

Answer

C

Explanation

We have,
`underset(r=0)overset(n)(sum)sin^(2)""(rpi)/(n)`
`=1/2underset(r=0)overset(n)(sum){1-cos""(2rpi)/(n)}`
`=1/2underset(r=0)overset(n)(sum)1-1/2underset(r=0)overset(n)(sum)cos""(2rpi)/(n)`
`=((n+1))/(2)-1/2{1+cos""(2pi)/(n)+cos""(4pi)/(n)+...+cos""(2npi)/(n)}`
`=((n+1))/(2)-1/2{cos""(4pi)/(n)+cos""(6pi)/(n)+...+cos""(2(n-1)pi)/(n)}`
`=((n-1)/(2))=1/2xx(cos pisin(pi-(pi)/(n))=(n-1)/(2)+1/2=n/2`