Question Number 115136 by Algoritm last updated on 23/Sep/20

Answered by Olaf last updated on 23/Sep/20

sin2θ = 2sinθcosθ cosθ = (1/2).((sin2θ)/(sinθ)) θ = ((kπ)/(2n+1)), cos(((kπ)/(2n+1))) = (1/2).((sin(((2kπ)/(2n+1))))/(sin(((kπ)/(2n+1))))) Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n )Π_(k=1) ^n ((sin(((2kπ)/(2n+1))))/(sin(((kπ)/(2n+1))))) Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n ).((sin(((2π)/(2n+1))))/(sin((π/(2n+1)))))×((sin(((4π)/(2n+1))))/(sin(((2π)/(2n+1)))))...×((sin(((2πn)/(2n+1))))/(sin(((πn)/(2n+1))))) Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n ).((sin(((2πn)/(2n+1))))/(sin((π/(2n+1))))) sin(((2πn)/(2n+1))) = sin(π−(π/(2n+1))) = sin((π/(2n+1))) ⇒ Π_(k=1) ^n cos(((kπ)/(2n+1))) = (1/2^n )