RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/27 15:05:32
RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11RT,求
RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
原式
=(cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11*2sinπ/11)/2sinπ/11
=(sin2π/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/2*sin4π/11*cos3π/11*cos4π/11*cos5π/11)/2sinπ/11
=(1/4*sin8π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/4*sin3π/11*cos3π/11*cos5π/11)/2sinπ/11
=(1/8*sin6π/11*cos5π/11)/2sinπ/11
=(1/8*sin5π/11*cos5π/11)/2sinπ/11
=(1/16*sin10π/11)/2sinπ/11
=1/32
RT,求值cosπ/11*cos2π/11*cos3π/11*cos4π/11*cos5π/11
求值:cosπ/7*cos2π/7*cos3π/7
cosπ/5 - cos2π/5 怎样变成 cosπ/5 + cos2π/5rt
化简cosπ/5cos2π/5RT.
求值：cosπ/17cos2π/17cos4π/17cos8π/17
求值cosπ/5cos2π/5cos4π/5cos8π/5
求值:cosπ/7·cos2π/7·cos4π/7
求值cos2π/5*tan5π
求值cos[arctan2+arcsin(-1/3)]rt
求值 cosπ/9cos2π/9cos4π/9还有一题：2·（根号下1+sin4）-（根号下2-2cos4）
求值:cosπ/7cos2π/7cos3π/7cos4π/7cos5π/7cos6π/7
cosπ/11×cos2π/11×cos3π/11×cos4π/11×cos5π/11=?求值cosπ/11×cos2π/11×cos3π/11×cos4π/11×cos5π/11=?求值 最后几步由sin6π/11×cos5π/11到sin10π/11怎样变得
根号下1减sin(π+2)cos(π+2)化简是多少啊A sin2-cos2 B cos2-sin2 C +-(sin2-cos2) D sin2+cos2
求COSπ/5*COS2π/5
cosπ/5×cos2π/5=
COSπ/5乘以COS2π/5
cosπ/5×cos2π/5
cosπ/5cos2π/5