The correct option is B n
From (x−α)(x−¯¯¯¯α)=x2−2xcosπn+1
and x2n−1x2−1=x2n−2+x2n+3+...+x+1
We get
(x2−2xcosπn+1)(x2−2xcos2πn+1)...(x2−2xcosn−1nπ+1)=1+x+...+x2n−2
Divide both sides by xn−1 to obtain
(x+1x−2cosπn)(x+1x−2cos2πn)...(x+1x−2cosn−1nπ)=xn−x−nx−x−1
Now put x=cosθ+isinθ
we get
2n−1(cosθ−cosπn)(cosθ−cos2πn)...(cosθ−cosn−1nπ)=sinnθsinθ
Take limθ→0
n=2n−1(2sin2π2n)(2sin22π2n)...(2sin2n−1nπ)
=2n−1(2sinπ2nsinn−12nπ)(2sin2π2nsinn−22nπ)...(2sinn−12nπsinπn)
Now, use
2sinkπ2nsin(n−k2nπ)=sin(kπn)
we get
2n−1sinπnsin2πn...sinn−1nπ=n