数学
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!~)

2019-05-23

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!~)
优质解答
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
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
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