当n趋向于无穷时，为什么lim（1-cos（1/n)=1/2n^2_百度知道求证:cos1/2cos1/4cos1/5...cos1/n&2/3_百度知道2(sinx/2)*∑sinkx=(cosx/2)－cos(2n＋1)*x/2这怎么证明_百度知道高一数学证明题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!~）
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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扫描下载二维码cos2π/2n+1+cos4π/2n+1+......+cos2nπ/2n+1=_百度知道}