1.cos(3п/8)sin(5п/24) - cos(5п/24)sin(3п/8) 2.tg13+tg32/1-tg13tg32 - вопрос №51729181385245243821332113323 от Lol666kek 23.10.2020 20:23

Question

Алгебра: 1.cos(3п/8)sin(5п/24) - cos(5п/24)sin(3п/8) 2.tg13+tg32/1-tg13tg32 3.cos(13п/12) - cos (5п/12)

Lol666kek · Answer

Cos(3П/8)sin(5П/24) - cos(5П/24)sin(3П/8)=sin(3π/8 -5π/24)=sin(4π/24)=sin(π/6)=1/2
(tg13+tg32)/1-tg13tg32=tg(13°+32°)=1
cos(13П/12) - cos (5П/12)=-2 ×sin((13π-5π)/12) ×sin((13π+5π)/12) =
=-2 sin(8π/12)×sin(18π/12)=-2sin(2π/3)×sin(3π/2)=-2sin(π-π/3)×(-1)=2×√3/2=√3