证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1))

Question

证明sin(pi/n)*sin(2pi/n)*sin(3pi/n)*…sin((n-1)pi/n)=n/(2^(n-1))

Answer 1

用复数 w=cos(2π/n)+isin(2π/n) w'=cos(2π/n)-isin(2π/n) z^n=1 (z-1)(z^(n-1)+z^(n-2)＋……＋z+1)=0 z^(n-1)+z^(n-2)＋……＋z+1=(z-w)(z-w^2)(z-w^3)……（z-w^(n-1)） 令z=1 n＝(1-w)(1-w^2)(1-w^3)…(1-w^(n-1)） 1-w^k=2sinkπ/n(sinkπ/n+icoskπ/n) |1-w^k|=|2sinkπ/n(sinkπ/n+icoskπ/n)|=|2sinkπ/n||(sinkπ/n+icoskπ/n)|=|2sinkπ/n|=2sin(kπ/n) 取模 |n|＝|(1-w)(1-w^2)(1-w^3)…(1-w^(n-1))| |n|＝|(1-w)||(1-w^2)||(1-w^3)|…|(1-w^(n-1))| n=2^(n-1)sin(π/n)sin(2π/n)……sin[(n-1)π/n] ∴sin(π/n)sin(2π/n)……sin[(n-1)π/n]=n/2^(n-1)