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
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(cosπn+2−cos(2k+3n+2)π)∑k=0n2sin2(k+1n+2)π=(n+1)cosπ(n+2)−cos(n+3n+2)πsin(n+1n+2)πsinπn+2(n+1)−cosπ.sin(n+1n+2)πsinπn+2=(n+1)cosπ(n+2)+cos(n+3n+3)π(n+1)+1=(n+1)cos(πn+2)+cos(πn+2)n+2=(1)f(4)=cosπ6=32correct(2)α=tan(cos−1;f(6))=tan|cos−1;(cosπ8)|=tanπ8tanπ4=2tanπ81−tan2π8⇒1=2α1−α2⇒α2+2α−1;=0(A) correct(3)sin(7cos−1;f(5)=sin(7cos−1;(cosπ7))=sinπ=0(A) , (B), (C) are correct(4)limn→∞f(n)=cos(πn+2)=1OptionDis Incorrect