(cos 23pi/12-sin 11pi/12)*(sin 17pi/12-cos19pi/12) - вопрос №91874622312111217121912 от millyby1 16.02.2022 05:05

Question

Математика: (cos 23pi/12-sin 11pi/12)*(sin 17pi/12-cos19pi/12)

millyby1 · Answer

(cos23π/12-sin11π/12)(sin17π/12-cos19π/12)
=(cos(2π-π/12)-sin(π-π/12))*
(sin(π+5π/12)-cos(2π-5π/12)=
(cosπ/12-sinπ/12)(-sin5π/12-cos5π/12)=
(cosπ/12-sinπ/12)(-sin(π/2-π/12)-
cos(π/2-π/12))==
(cosπ/12-sinπ/12)(-1)(cosπ/12+sinπ/12)=
-(cos²π/12-sin²π/12)=-cosπ/6=-√3/2