477. Вычислить: 1) (2 − sin²(−π/6) + cos²(−π/3)) / (2 cos(−π/3) + sin(−π/6)); 2) √3 sin(−π/3) − 2 ctg(−π/4) + 4 cos(−3π/2).

1. Подставим значения: \(\sin^2(-\pi/6)=\frac14\), \(\cos^2(-\pi/3)=\frac14\), \(\cos(-\pi/3)=\frac12\), \(\sin(-\pi/6)=-\frac12\).

   \(\frac{2-\frac14+\frac14}{2\cdot\frac12-\frac12}=\frac{2}{\frac12}=4.\)

2. \(\sin(-\pi/3)=-\frac{\sqrt3}{2}\), \(\ctg(-\pi/4)=-1\), \(\cos(-3\pi/2)=0\). Тогда

   \(\sqrt3\cdot\left(-\frac{\sqrt3}{2}\right)-2(-1)+4\cdot0=-\frac32+2=\frac12.\)

Ответ: 1) \(4\); 2) \(\frac12\).