Вариант 1 1. 3 - cos^2 (-π/4) + sin^2 (-π/6) --------------------------------- 3 sin (-π/6) + cos (-π/4);

Вариант 1 1. $3 - \cos^2\left(-\frac{\pi}{4}\right) + \sin^2\left(-\frac{\pi}{6}\right) = 3 - \left(\frac{\sqrt{2}}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 = 3 - \frac{2}{4} + \frac{1}{4} = 3 - \frac{1}{4} = 2,75$ $3 \sin\left(-\frac{\pi}{6}\right) + \cos\left(-\frac{\pi}{4}\right) = 3\left(-\frac{1}{2}\right) + \frac{\sqrt{2}}{2} = -1,5 + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}-3}{2}$ Результат: $\frac{2,75}{\frac{\sqrt{2}-3}{2}} = \frac{11}{4} \cdot \frac{2}{\sqrt{2}-3} = \frac{5,5}{\sqrt{2}-3} = \frac{5,5(\sqrt{2}+3)}{2-9} = \frac{5,5(\sqrt{2}+3)}{-7} = -\frac{11(\sqrt{2}+3)}{14}$ 2. $2\sqrt{2} \cos\left(-\frac{\pi}{4}\right) + 3 \operatorname{tg}\left(-\frac{\pi}{3}\right) - 5 \sin\left(-\frac{\pi}{2}\right) = 2\sqrt{2} \cdot \frac{\sqrt{2}}{2} + 3 \cdot (-\sqrt{3}) - 5(-1) = 2 - 3\sqrt{3} + 5 = 7 - 3\sqrt{3}$ Вариант 2 1. $4 - \sin^2\left(-\frac{\pi}{3}\right) + \cos^2\left(-\frac{\pi}{6}\right) = 4 - \left(-\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = 4 - \frac{3}{4} + \frac{3}{4} = 4$ $2 \cos\left(-\frac{\pi}{6}\right) - \sin\left(-\frac{\pi}{3}\right) = 2\left(\frac{\sqrt{3}}{2}\right) - \left(-\frac{\sqrt{3}}{2}\right) = \sqrt{3} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$ Результат: $\frac{4}{\frac{3\sqrt{3}}{2}} = 4 \cdot \frac{2}{3\sqrt{3}} = \frac{8}{3\sqrt{3}} = \frac{8\sqrt{3}}{9}$ 2. $\sqrt{3} \operatorname{tg}\left(-\frac{\pi}{6}\right) - 4 \cos\left(-\frac{\pi}{3}\right) + 6 \sin\left(-\frac{3\pi}{2}\right) = \sqrt{3} \left(-\frac{\sqrt{3}}{3}\right) - 4 \cdot \frac{1}{2} + 6(1) = -1 - 2 + 6 = 3$