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...
求证:cos(π/2k+1)+cos(2π/2k+1)+…+cos(2k-1)π/2k+1+cos2kπ/2k+1=0
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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+1))=0
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+1))=0
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