Question

-cos (π/7)*cos(4π/7)*cos(2π/7) равно?

Answer 1

-cos(π/7)*cos(2π/7)*cos(4π/7)=
=-sin(π/7)*cos(π/7)*cos(2π/7)*cos(4π/7)/sin(π/7)=
=-sin(2π/7)*cos(2π/7)*cos(4π/7)/[2sin(π/7)]=
=-sin(4π/7)*cos(4π/7)/[4sin(π/7)]=-sin(8π/7)/[-8sin(π/7)]=
=-sin(π-π/7)/[8sin(π/7)]=-sin(π/7)/[8sin(π/7)]=-1/8