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underset(r=0)overset(n)(sum)sin^(2)""(rp...

`underset(r=0)overset(n)(sum)sin^(2)""(rpi)/(n)` is equal to

Text Solution

Verified by Experts

The correct Answer is:
`n//2`
`sum_(r=1)^(n-1)sin^(2)""(rpi)/(n)=sum_(r=1)^(n-1)(1-cos(2rpi//n))/(2)`
`=(1)/(2)(n-1-(sin(n-1)(pi)/(n))/(sin((pi)/(n)))cos(((2pi)/(n)+(n-1)(2pi)/(n))/(2)))`
`=(1)/(2)(n-1-(sin((pi)/(n)))/(sin((pi)/(n)))(-1))=(n)/(2)`
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