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高一数学证明题cosπ/(2n+1)*cos2π/(2n+1)*cos3π/(2n+1)*...*cosnπ/(2n+1)=1/(2^n),n是正整数（要有详细过程哈!~谢谢!~）
题目详情
高一数学证明题
cosπ/(2n+1)*cos2π/(2n+1)*cos3π/(2n+1)*...*cosnπ/(2n+1)=1/(2^n),n是正整数
（要有详细过程哈!~谢谢!~）
cosπ/(2n+1)*cos2π/(2n+1)*cos3π/(2n+1)*...*cosnπ/(2n+1)=1/(2^n),n是正整数
（要有详细过程哈!~谢谢!~）
▼优质解答
答案和解析
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
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
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