(( cos 2п/31*cos 4п/31*cos 8п/31*cos 16/31*cos 32п/31 - вопрос №871979223143183116313231 от AAndrey7600 24.10.2021 19:11

Question

Алгебра: (( cos 2п/31*cos 4п/31*cos 8п/31*cos 16/31*cos 32п/31 =? ?

AAndrey7600 · Answer

Cos 2п/31*cos 4п/31*cos 8п/31*cos 16π/31*cos 32п/31 =
=sin2π/31*cos 2п/31*cos 4п/31*cos 8п/31*cos 16π/31*cos 32п/31/(sin2π/31 )=
=sin4π/31*cos4π/31*cos8π/31*cos16π/31*cos32π/31/(2sin2π/31)=
=sin8π/31*cos8π/31*cos16π/31*cos32π/31/(4sin2π/31)=
=sin16π/31*cos16π/31*cos32π/31/(8sin2π/31)=
=sin32π/31*cos32π/31/(16sin2π/31)=sin64π/31/(32sin2π/31)=
=sin(2π+2π/31)/(32sin2π/31)=sin2π/31/(32sin2π/31)=1/32