求﹙sinπ/7﹚·﹙sin2π/7﹚·﹙sin3π/7﹚的值

有一个结论sin(π/n)*sin(2π/n)*sin(3π/n)*…*sin[(n-1)π/n] = n/ 2^(n-1)于是 sin(π/7)*sin(2π/7)*sin(3π/7)*sin(4π/7)*sin(5π/7)*sin(6π/7) = 7/64sin(π/7)*sin(2π/7)*sin(3π/7)*sin(3π/7)*sin(2π/7...参考这个zhidao.baidu.com/question/513837105.html或者下面的（都是复数思想）w=cos(2π/n)+isin(2π/n)w'=cos(2π/n)-isin(2π/n)z^n=1(z-1)(z^(n-1)+z^(n-2)＋……＋z+1)=0z^(n-1)+z^(n-2)＋……＋z+1=(z-w)(z-w^2)(z-w^3)……（z-w^(n-1)）令z=1n＝(1-w)(1-w^2)(1-w^3)…(1-w^(n-1)）1-w^k=2sinkπ/n(sinkπ/n+icoskπ/n)|1-w^k|=|2sinkπ/n(sinkπ/n+icoskπ/n)|=|2sinkπ/n||(sinkπ/n+icoskπ/n)|=|2sinkπ/n|=2sin(kπ/n)取模|n|＝|(1-w)(1-w^2)(1-w^3)…(1-w^(n-1))||n|＝|(1-w)||(1-w^2)||(1-w^3)|…|(1-w^(n-1))|n=2^(n-1)sin(π/n)sin(2π/n)……sin[(n-1)π/n]说实话我也不是特别明白，所以你要是没看懂的话，可以去问问老师