Question

Помогите
1) sin(2π-π/6) =?
2) sin(-2π-π/3) = ?

Answer 1

$1)Sin(2\pi-\frac{\pi }{6})=-Sin\frac{\pi }{6}=-\frac{1}{2}\\\\2) Sin(-2\pi-\frac{\pi }{3})=-Sin(2\pi+\frac{\pi }{3})=-Sin\frac{\pi }{3}=-\frac{\sqrt{3}}{2}$