1) (sin2a+sin4a)^2+(cos2a+cos4a)^2=4cos^2a 2)sina-sin3a/cosa-cos3a= - вопрос №4067132124224242233232 от Muffs 07.12.2022 19:21

Question

Алгебра: 1) (sin2a+sin4a)^2+(cos2a+cos4a)^2=4cos^2a 2)sina-sin3a/cosa-cos3a= -ctg2a 3)2tga/1-tg^2 (2cos^2a-1)=sin2a

Muffs · Answer

1)(2sin3acosa)²+(2cos3acosa)²=4cos²a(sin²3a+cos²3a)=4cos²a4cos²a=4cos²a2)-2sinacos2a/2sinasin3a=-cos3a/sin3a=-ctg3a-ctg3a=-ctg3a3)tg2a*cos2a=sin2a/cos2a * cos2a=sin2asin2a=sin2a...