Tg23*tg293+sin52*sin128-sin322*sin142 - вопрос №28324252329352128322142 от Katriysha 14.11.2020 12:15

Question

Математика: Tg23*tg293+sin52*sin128-sin322*sin142

Katriysha · Answer

Tg23*tg(270+23)+sin52*sin(180-52)-sin(360-38)*sin(180-38)=-1+1=0
tg23*tg293=sin293sin23/cos293cos23=
=1/2*(cos270-cos316))/[1/2*(cos270+cos316)]=-cos316/cos316=-1
sin52*sin52-(-sin38)*sin38=sin²52+sin²38=sin²(90-38)+sin²38=cos²38+sin²38=1