Question

Найдите tg150° * sin150°.​

Answer 1

$tg150^\circ \cdot sin150^\circ =tg(180^\circ -30^\circ )\cdot sin(180^\circ -30^\circ )=\\\\=-tg30^\circ \cdot sin30^\circ =-\frac{\sqrt3}{3}\cdot \frac{1}{2}=-\frac{\sqrt3}{6}$