InterviewSolution
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InterviewSolution
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`sum_(r=0)^(n) sin^(2)""(rpi)/(n)` is equal toA. `(n+1)/(2)`B. `(n-1)/(2)`C. `n/2`D. none of these |
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Answer» Correct Answer - C 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` |
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