Question

Обчислити :
a) tg (-330°)
б) cos 510°​

Answer 1

$\displaystyle\bf\\1)\\\\tg(-330^\circ)=-tg(270^\circ+60^\circ)=-\Big(-Ctg60^\circ\Big)=Ctg60^\circ=\frac{\sqrt{3} }{3} \\\\\\2)\\\\Cos510^\circ=Cos(360^\circ+150^\circ)=Cos150^\circ=Cos(180^\circ-30^\circ)=\\\\=-Cos30^\circ=-\frac{\sqrt{3} }{2}$