(( cos 2п/31*cos 4п/31*cos 8п/31*cos 16/31*cos - вопрос №231431831163132318719115 от DahenkaSi05 05.06.2022 16:55

Question

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

DahenkaSi05 · 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