Question

cos(n/7)cos(2n/7)cos(3n/7)

Answer 1

[sinπ/7*cosπ/7*cos2π/7*cos3π/7]/sinπ/7=[sin2π/7*cos2π/7*cos3π/7]/2sinπ/7=
=[sin4π/7*cos3π/7]/4sinπ/7=[sin(π-3π/7)*cos3π/7)/4sinπ/7=
=[sin3π/7*cos3π/7]/4sinπ/7=(sin6π/7)/8sinπ/7=[sin(π-π/7)]/8sinπ/7=
=(sinπ/7)/8sinπ/7=1/8