17cos(2arctg(-1/4)) ; 75cos(2 arccos 4/5) - вопрос №70167741721475245 от emir07 09.01.2022 00:52

Question

Алгебра: 17cos(2arctg(-1/4)) ; 75cos(2 arccos 4/5)

emir07 · Answer

17cos(2arctg(-1/4))=17cos2t=17*(cos^2 t -sin^2 t)=17*(16/17-1/17)=15;
t=arctg(-1/4)
tgt=-1/4; 1+tg^2 t=1/cos^2 t
1+(-1/4)^2=1/cos^2 t; 1+1/16=(16+1)/16=17/16
1/cos^2 t=17/16; cos^2 t=16/17
sin^2 t+cos^2 t=1; sin^2 t=1-cos^2 t; sin^2 t=
=1-16/17=1/17