Question

12tg * 600 градусов * sin 1020 градусов

Answer 1

$12~tg~600textdegree cdot sin1020textdegree=\=12~tg~(3cdot 180textdegree + 60textdegree) cdot sin(3cdot 360textdegree-60textdegree)=\=12~tg~60textdegree cdot sin(-60textdegree)=-12~tg~60textdegree cdot sin60textdegree=\\=-12sqrt3cdot dfrac{sqrt3}2=-6cdot 3=-18$

======================================

Использована периодичность и нечётность функций

tg (180°·n + α) = tg α,    n∈Z

sin (360°·n + α) = sin α,    n∈Z

sin (-α) = -sin α