在线问 无穷级数的一道题 很简单 只是我看不懂 呵呵
来源:百度知道 编辑:UC知道 时间:2024/09/28 11:20:25
∵cos(2k-1)π/12-cos(2k+1)π/12=coskπ/6cosπ/12+sinkπ/6sinπ12
-[coskπ/6cosπ/12-sinkπ/6sinπ12]=2sinkπ/6sinπ12
∴2sinkπ/6sinπ12=cos(2k-1)π/12-cos(2k+1)π/12
∴sinkπ/6=[cos(2k-1)π/12-cos(2k+1)π/12]/sinπ/12
∴sinπ/6+sin2π/6+sin3π/6+…sinnπ/6
=[(cosπ/12-cos3π/12)+(cos3π/12-cos5π/12)+(cos5π/12-cos7π/12)+…+(cos(2n-1)π/12-cos(2n+1)π/12)]/sinπ/12
=[cosπ/12-cos(2n+1)π/12)]/sinπ/12
n=12k,12k+1,12k+2,2K+3,…12k+11。时将有不同的结果,所以无穷级数发散。
2sinαsinβ=cos(α-β)-cos(α+β)
这是和差化积算出来的.