把[0,π]平均分成n份,每小份取右端点作为函数的值∑sin(kπ/n)*(π/n)(k从1到n)=(π/n)/sin(π/n)∑sin(kπ/n)sin(π/n)=(π/n)/sin(π/n)∑-1/2{[cos(k+1)π/n]-[cos(k-1)π/n]}=(-1/2)(π/n)/sin(π/n){cosπ+cos[...