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Question:

For non-negative integer n, let f(n) = Σ_(k=0)^n sin((k+1)/(n+2)π)sin((k+2)/(n+2)π) / Σ_(k=0)^n sin((k+1)/(n+2)π). Assuming cos⁻¹x takes values in [0, π], which of the following option is/are correct?

f(4) = √(3)/2

sin(7cos⁻¹f(5)) = 0

If α = tan(cos⁻¹f(6)) then α² + 2α - 1 = 0

lim_(n→∞)f(n) = 1/2

Solution:

Correct option is C.sin⁡(7cos−1;⁡f(5))=0f(n)=∑K=0nsin⁡(k+1n+2π)sin⁡(k+2n+2π)∑k=0n2sin2⁡(k+1n+2)π=∑k=0n(cos⁡πn+2−cos⁡(2k+3n+2)π)∑k=0n2sin2⁡(k+1n+2)π=(n+1)cos⁡π(n+2)−cos⁡(n+3n+2)πsin⁡(n+1n+2)πsin⁡πn+2(n+1)−cos⁡π.sin⁡(n+1n+2)πsin⁡πn+2=(n+1)cos⁡π(n+2)+cos⁡(n+3n+3)π(n+1)+1=(n+1)cos⁡(πn+2)+cos⁡(πn+2)n+2=(1)f(4)=cos⁡π6=32correct(2)α=tan⁡(cos−1;⁡f(6))=tan⁡|cos−1;⁡(cos⁡π8)|=tan⁡π8tan⁡π4=2tan⁡π81−tan2⁡π8⇒1=2α1−α2⇒α2+2α−1;=0(A) correct(3)sin⁡(7cos−1;⁡f(5)=sin⁡(7cos−1;⁡(cos⁡π7))=sin⁡π=0(A) , (B), (C) are correct(4)limn→∞f(n)=cos⁡(πn+2)=1OptionDis Incorrect