2cos75*sin75= 6-sin^2α-cos^2α= (-1+sin^2α)*tg^2α 1-tg67*tg7/tg67-tg7= - вопрос №2440411275756221221677677 от lolabrikoslol 31.12.2022 23:54

Question

Алгебра: 2cos75*sin75= 6-sin^2α-cos^2α= (-1+sin^2α)*tg^2α 1-tg67*tg7/tg67-tg7=

lolabrikoslol · Answer

1)sin150=sin(180-30)=sin30=1/22)6-(sin²a+cos²a)=6-1=53)-cos²a*sin²a/cos²a=-sin²a4)cos74/cos67*cos7:sin60/cos67*cos7=cos74/cos67*cos7*cos67*cos7/sin60=cos74:√3/2=2cos74/√3=(2√3cos74)/3По формулам:1-tgatgb=cos(a+b)/cosacosbtga-tgb=sin(a-b)/cosacosb...