Question

Вычислите с помощью формул приведения a)sin750град-ов*ctg510градусов+tg(-120градусов) b) tg16pi /3-cos(-55pi /6)

Answer 1

$a)sin750*ctg510+tg(-120)=sin(360*2+30)*ctg(360+150)-\\-tg(90+30)=sin30*ctg(90+60)-ctg30=-sin30*tg60-ctg30=\\=-\frac{1}{2}*\sqrt3-\sqrt3=-\frac{3\sqrt3}{2}\\ b) tg\frac{16\pi}{3}-cos(-\frac{55\pi}{6})=tg(5\pi+\frac{\pi}{3})-cos(9\pi+\frac{\pi}{3})=\\=tg(\pi+\frac{\pi}{3})-cos(\pi+\frac{\pi}{3})=tg\frac{\pi}{3}+cos\frac{\pi}{3}=\sqrt3+\frac{1}{2}$