由2sina*cosa=sin2a

　　有sina*cosa=sin2a/2

　　是故

　　cosπ/(2n+1)*cos2π/(2n+1)*cos3π/(2n+1)*...*cosnπ/(2n+1)*sinπ/(2n+1)*sin2π/(2n+1)*sin3π/(2n+1)*...*sinnπ/(2n+1)

　　=1/(2^n)*sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin2nπ/(2n+1)

　　下面证明

　　sinπ/(2n+1)*sin2π/(2n+1)*sin3π/(2n+1)*...*sinnπ/(2n+1)

　　=sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin2nπ/(2n+1)

　　事实上：

　　n为奇数时：

　　sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin2nπ/(2n+1)

　　=sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin（n-1）π/(2n+1)

　　*sin（n+1）π/(2n+1)*...*sin2nπ/(2n+1)

　　考虑到sina=sin（π-a）

　　所以sin（n+1）π/(2n+1)*...*sin2nπ/(2n+1)=sin（π-（n+1）π/(2n+1)）*sin（π-（n+3）π/(2n+1)）*...*sin（π-2nπ/(2n+1）)

　　=sinnπ/(2n+1)*sin(n-2)π/(2n+1)*...*sin（π/(2n+1）)

　　因此sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin（n-1）π/(2n+1)

　　*sin（n+1）π/(2n+1)*...*sin2nπ/(2n+1)

　　=sinπ/(2n+1)*sin2π/(2n+1)*sin3π/(2n+1)*...*sinnπ/(2n+1)

　　n为偶数时：

　　sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin2nπ/(2n+1)

　　=sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sinnπ/(2n+1)

　　*sin（n+2）π/(2n+1)*...*sin2nπ/(2n+1)

　　考虑到sina=sin（π-a）

　　所以sin（n+2）π/(2n+1)*...*sin2nπ/(2n+1)=sin（π-（n+2）π/(2n+1)）*sin（π-（n+4）π/(2n+1)）*...*sin（π-2nπ/(2n+1）)

　　=sin（n-1）π/(2n+1)*sin(n-3)π/(2n+1)*...*sin（π/(2n+1）)

　　因此sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sinnπ/(2n+1)

　　*sin（n+3）π/(2n+1)*...*sin2nπ/(2n+1)

　　=sinπ/(2n+1)*sin2π/(2n+1)*sin3π/(2n+1)*...*sinnπ/(2n+1)

　　于是

　　sinπ/(2n+1)*sin2π/(2n+1)*sin3π/(2n+1)*...*sinnπ/(2n+1)

　　=sin2π/(2n+1)*sin4π/(2n+1)*sin6π/(2n+1)*...*sin2nπ/(2n+1)

　　因此

　　cosπ/(2n+1)*cos2π/(2n+1)*cos3π/(2n+1)*...*cosnπ/(2n+1)=1/(2^n),n是正整数