Question

Помогите найти знак выражения

Answer 1

$sin590^\circ =sin(360^\circ +230^\circ )=sin230^\circ <0\\\\cos480^\circ =cos(360^\circ +120^\circ )=cos120^\circ <0\\\\tg130^\circ <0\\\\\\\frac{sin590^\circ \cdot cos480^\circ }{tg130^\circ }=\frac{sin230^\circ \, \cdot \, cos120^\circ }{tg130^\circ } =\frac{(-)\cdot (-)}{(-)}<0$