分别计算：cosπ/3,cosπ/5cos2π/5,cosπ/7cos2π/7cos3π/7
问题描述：
分别计算：cosπ/3,cosπ/5cos2π/5,cosπ/7cos2π/7cos3π/7
根据上述结果猜想一个一般的结论,并给予证明
答
1.cosπ/3=1/22.cosπ/5cos2π/5=2sin(π/5)*cos(π/5)*cos(2π/5) / 2sin(π/5)=sin(2π/5)*cos(2π/5) / 2sin(π/5)=sin(4π/5) / 4sin(π/5)=sin(π/5) / 4sin(π/5)=1/43.cos(π/7)cos(2π/7)cos(3π/7)=2sin(π...cos[2nπ/(2n-1)] =sin[π/(2n-1)] 这个为什么哦，这里有个笔误。那段应该改成下面这个：=……=2sin[nπ/(2n-1)]cos[nπ/(2n-1)] / (2^n)sin[π/(2n-1)]=sin[2nπ/(2n-1)] / (2^n)sin[π/(2n-1)]