Question

cos300° + 2√3 tg (-120)°

Answer 1

Объяснение:

$cos300^0+2\sqrt{3}*tg(-120^0) =cos(360^0-60^0)-2\sqrt{3} * tg(120^0)=\\=cos60^0-2\sqrt{3} *tg(180^0-60^0)=0,5-2\sqrt{3}*(-tg60^0)= \\=0,5+2\sqrt{3*} *tg60^0=0,5+2\sqrt{3}*\sqrt{3}=0,5+2*3=0,5+6=6,5.$