Вычислить: 1) \(\frac{2-\sin^2\left(-\frac{\pi}{6}\right)+\cos^2\left(-\frac{\pi}{3}\right)}{2\cos\left(-\frac{\pi}{3}\right)+\sin\left(-\frac{\pi}{6}\right)}\); 2) \(\sqrt3\sin\left\(-\frac{\pi}{3}\right\)-2\(\operatorname{ctg}\)\(\left\)\(-\frac{\pi}{4}\right\)+4\(\cos\)\(\left\)\(-\frac{3\pi}{2}\right\).

Question

Вопрос:

Вычислить: 1) \(\frac{2-\sin^2\left(-\frac{\pi}{6}\right)+\cos^2\left(-\frac{\pi}{3}\right)}{2\cos\left(-\frac{\pi}{3}\right)+\sin\left(-\frac{\pi}{6}\right)}\); 2) \(\sqrt3\sin\left\(-\frac{\pi}{3}\right\)-2\(\operatorname{ctg}\)\(\left\)\(-\frac{\pi}{4}\right\)+4\(\cos\)\(\left\)\(-\frac{3\pi}{2}\right\).

Accepted Answer

1)

Используем значения тригонометрических функций:

\(\sin^2\left(-\frac{\pi}{6}\right)=\frac14\), \(\cos^2\left(-\frac{\pi}{3}\right)=\frac14\), \(\cos\left(-\frac{\pi}{3}\right)=\frac12\), \(\sin\left(-\frac{\pi}{6}\right)=-\frac12\).

\[\frac{2-\frac14+\frac14}{2\cdot\frac12-\frac12}=\frac{2}{\frac12}=4.\]

2)

\[\sqrt3\cdot\left(-\frac{\sqrt3}{2}\right)-2\cdot(-1)+4\cdot0\]

\[=-\frac32+2=\frac12.\]

Ответ: 1) 4; 2) \(\frac12\).