Question

Решите уравнение 4sin120×tg300

Answer 1

4sin(120°)tg(300°)=4sin(90°+30°)tg(120°+180°)=4cos(30°)tg(120°)=

=4cos(30°)tg(90°+30°)=-4cos(30°)ctg(30°)=-4·√3/2·√3=-6

Answer 2

Это выражение

4sin(90+30) * tg(270+30)=

4cos30*(-ctg30)=-4*√3/2 * √3=-6.