问题标题:
一道数学题···麻烦各位高手··求证:cos(π/2k+1)+cos(2π/2k+1)+…+cos(2k-1)π/2k+1+cos2kπ/2k+1=0
问题描述:
一道数学题···麻烦各位高手··
求证:cos(π/2k+1)+cos(2π/2k+1)+…+cos(2k-1)π/2k+1+cos2kπ/2k+1=0
齐家月回答:
注意到cos(π-x)=-cos(x)所以左端首尾相加是0,所以2(cos(π/2k+1)+cos(2π/2k+1)+…+cos(2k-1)π/2k+1+cos2kπ/2k+1)=(cos(π/2k+1)+cos2kπ/2k+1)+(cos(2π/2k+1)+cos(2k-1)π/2k+1)...+(cos2kπ/2k+1+cos(π/2k+...
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