Tg ( п+arcsin(-1/2)) cos (п-arcsin (-1)) tg (п/2+arctg - вопрос №12123321006390181 от danchik1106 29.04.2021 03:00

Question

Алгебра: Tg ( п+arcsin(-1/2)) cos (п-arcsin (-1)) tg (п/2+arctg √3) sin (3п/2-arccos (-1) ))0)0

danchik1106 · Answer

Tg(π+asin(-1/2))= tg(arcsin(-1/2))= tg(-arcsin(1/2))= -tg(arcsin1/2)=
= -sin(arcsin(1/2))/(√1-sin²(1/2)) =
= (-1/2)/(√1-1/4) = (-1/2)/(√3/2) = -1/√3 = -√3/3

cos(π-arcsin(-1)) = -cos(arcsin(-1)) = -cos(-arcsin1) = -cos(arcsin1) =
= -√(1-sin²(arcsin1)) = 0

tg(π/2+arctg√3) = -ctg(arctg√3) = -1/tg(arctg√3) = -1/√3 = -√3/3

sin(3π/2 - arccos(-1)) = -cos(arccos(-1))= -cos(π-arccos1))=
=cos(arccos1) = 1