(2sin35°sin55°)/cos20°= tg35°×tg55°= - вопрос №538329323555203555 от feroprosto 05.01.2021 16:47

Question

Алгебра: (2sin35°sin55°)/cos20°= tg35°×tg55°=

feroprosto · Answer

(2sin35.sin55)/cos20=2.sin35.cos35/cos20=
=sin70/cos20=sin70/sin70=1

(sin2x=2.sinx.cosx, sin(90-x)=cosx, cos(90-x)=sinx)

tg35.tg55=sin35/cos35 . sin55/cos55=(sin35.sin55)/(cos35.cos55)=
=(sin35.sin55)/(sin55.sin35)=1

(tgx = sinx/cosx, cos35=sin(90-35)=sin55,cos55=sin(90-55)=sin35)