三角函數的證明題

Question

試證:sin(π/n)*sin(2π/n)*...*sin((n-1)π/n)=n/2n-1

Answer 1

n為自然數 設 z = cos(2π/n) + isin(2π/n) 則 xn-1 = 0 的根為 1, z, ... , zn-1 xn-1+...+x+1 = (x-z)(x-z2)...[x-zn-1]x=1代入，兩邊取絕對值n =｜1-z｜*｜1-z2｜*...*｜1-zn-1｜ ----------------------------------------k = 1,2,...,(n-1)| 1-zk | = | 1-[cos(2kπ/n) + sin(2kπ/n)] | = | 1-cos(2kπ/n) - isin(2kπ/n)| = √{[1-cos(2kπ/n)]2 + [sin(2kπ/n)]2}= √[2-2cos(2kπ/n)]  = 2sin(kπ/n) --------------------------------------n =｜1-z｜*｜1-z2｜*...*｜1-zn-1｜ n = 2n-1 [sin(π/n)*sin(2π/n)*...*sin((n-1)π/n)]sin(π/n)*sin(2π/n)*...*sin((n-1)π/n) = n/2n-1