InterviewSolution
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InterviewSolution
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For non-negative inger n, let `f(n)=sum_(k=)^(n) sin((k+1)/(n+1)pisin((k+2)/(n+1)pi))/(sum_(k=0)^(n)sin^(2)((k+1)/(n+1)pi))` Assuming `cos^(-1)x` takes values in `[0,pi]` which of the following options is/are correct?A. if `alpha=tan(cos^(-1)f(6))`, then `alpha^(2)+2alpha-1=0`B. `underset(ntoinfty)(lim)f(x)=(1)/(2)`C. `f(4)=(sqrt(3))/(2)`D. `sin(7cos^(-1)f(5))=0` |
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Answer» Correct Answer - A::C::D `f(n)=(underset(k=0)overset(n)sum sin((k+1)/(n+1)pi)sin((k+2)/(n+2)pi))/(underset(k=0)overset(n)sum2sin^(2)((k+1)/(n+2))pi)=(underset(k=0)overset(n)sum(cos((pi)/(n+2))-cos((2k+3)/(n+2))pi))/(underset(k=0)overset(n)sum2sin^(2)((k+1)/(n+2))pi)` `=((n+1)cos((pi)/(n+2))-(cos((n+3)/(n+2))pisin((n+1)/(n+2))pi)/(sin((pi)/(n+2))))/((n+1)-(cospisin((n+1)/(n+2))pi)/(sin((pi)/(n+2))))` `=((n+1)cos((pi)/((n+2)))+cos((n+3)/(n+2))pi)/((n+1)+1)=((n+1)cos((pi)/((n+2)))+cos((pi)/(n+2)))/(n+2)=cos((pi)/(n+2))` (A) `alpha=tan(cos^(-1)f(6))=tan|cos^(-1)(cos((pi)/(8)))|=tan((pi)/(8))` `tan((pi)/(4))=(2tan((pi)/(8)))/(1-tan^(2)((pi)/(8)))implies1=(2alpha)/(1-alpha^(2))impliesalpha^(2)+2alpha-1=0` (A) correct (C). `f(4)=cos((pi)/(6))=(sqrt(3))/(2)` correct (D). `sin(7cos^(-1)f(5))=sin(7cos^(-1)(cos((pi)/(7))))=sinpi=0` correct (A), (C), (D) correct |
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