√128*cos^2*3π/8-√32=√32(2cos^2*3π/8-1)=√32(cos2 - вопрос №3992703128238323222381322383234322 от satokostanay 27.02.2020 00:30

Question

Алгебра: √128*cos^2*3π/8-√32=√32(2cos^2*3π/8-1)=√32(cos2 3π/8)= √32cos 3π/4=√32*√2/2=√16=4

satokostanay · Answer

= √32*4 * cos^2 3π8- √32= √32*√4*cos^2 3π/8-√32=√32(2* cos^2 3π8- 1)=√32cos(90+45)=√32(-sin45)=√31/1 * -√2/2= -√32*√2/1*2=√64/2= -8/2=-4...