1. $6\sqrt{6} \cdot \text{tg}(\frac{\pi}{6}) \cdot \cos(\frac{\pi}{4}) = 6\sqrt{6} \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{2}}{2} = 6 \cdot \frac{\sqrt{12}}{2\sqrt{3}} = 3 \cdot \frac{2\sqrt{3}}{\sqrt{3}} = 6$
2. $\frac{-7\sqrt{3}}{\sin(\frac{5\pi}{3})\cos(-\frac{2\pi}{3})} = \frac{-7\sqrt{3}}{(-\frac{\sqrt{3}}{2}) \cdot (-\frac{1}{2})} = \frac{-7\sqrt{3}}{\frac{\sqrt{3}}{4}} = -28$
3. $6\sin(240^\circ)\text{tg}(-840^\circ) = 6(-\frac{\sqrt{3}}{2})\text{tg}(-120^\circ) = -3\sqrt{3} \cdot \sqrt{3} = -9$
4. $-6\sqrt{3}\text{tg}(\frac{11\pi}{3})\cos(-\frac{14\pi}{3}) = -6\sqrt{3} \cdot \text{tg}(-\frac{\pi}{3}) \cdot \cos(-\frac{2\pi}{3}) = -6\sqrt{3} \cdot (-\sqrt{3}) \cdot (-\frac{1}{2}) = 18 \cdot (-\frac{1}{2}) = -9$
5. $\cos \alpha = -\frac{2\sqrt{5}}{5}$, $\alpha \in (\frac{\pi}{2}; \pi)$ (II четверть, синус положительный). $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \frac{20}{25}} = \sqrt{\frac{5}{25}} = \frac{\sqrt{5}}{5}$. $\text{ctg} \alpha = \frac{\cos \alpha}{\sin \alpha} = \frac{-2\sqrt{5}/5}{\sqrt{5}/5} = -2$. $3\text{ctg} \alpha = 3(-2) = -6$
6. $\cos \alpha = -\frac{4}{5}$, $\alpha \in (\frac{\pi}{2}; \pi)$ (II четверть). $\sin \alpha = \sqrt{1 - (-0.8)^2} = 0.6$. $45 \sin \alpha = 45 \cdot 0.6 = 27$
7. $\text{ctg} \alpha = 0.5 = \frac{1}{2}$. $\sin \alpha = \frac{1}{\sqrt{1 + \text{ctg}^2 \alpha}} = \frac{1}{\sqrt{1 + 0.25}} = \frac{1}{\sqrt{1.25}} = \frac{1}{\sqrt{5}/2} = \frac{2}{\sqrt{5}}$. $\cos \alpha = \text{ctg} \alpha \cdot \sin \alpha = \frac{1}{2} \cdot \frac{2}{\sqrt{5}} = \frac{1}{\sqrt{5}}$. $40\sqrt{5} \cos \alpha = 40\sqrt{5} \cdot \frac{1}{\sqrt{5}} = 40$
8. $60\sqrt{10}\sin(9\pi - \alpha) = 60\sqrt{10}\sin(\pi - \alpha) = 60\sqrt{10}\sin \alpha$. $\cos \alpha = \frac{3\sqrt{10}}{10}$, $\alpha \in (0; \frac{\pi}{2})$. $\sin \alpha = \sqrt{1 - \frac{90}{100}} = \sqrt{0.1} = \frac{1}{\sqrt{10}}$. Ответ: $60\sqrt{10} \cdot \frac{1}{\sqrt{10}} = 60$
9. $16\text{tg}(-9\pi - \alpha) = 16\text{tg}(-\pi - \alpha) = 16\text{tg}(-\pi - \alpha) = 16\text{tg}(-(\pi + \alpha)) = -16\text{tg} \alpha$. $\cos \alpha = -\frac{4}{5}$, $\alpha \in (\frac{\pi}{2}; \pi)$. $\sin \alpha = \sqrt{1 - 0.64} = 0.6$. $\text{tg} \alpha = \frac{0.6}{-0.8} = -0.75$. Ответ: $-16(-0.75) = 12$
10. $21\text{ctg}(\frac{\pi}{2} - \alpha) = 21\text{tg} \alpha$. $\cos \alpha = \frac{3\sqrt{10}}{10}$, $\sin \alpha = \sqrt{1 - 0.9} = \frac{1}{\sqrt{10}}$. $\text{tg} \alpha = \frac{1/\sqrt{10}}{3\sqrt{10}/10} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$. Ответ: $21 \cdot \frac{1}{3} = 7$
11. $40\sqrt{5}\cos(2\pi - \alpha) = 40\sqrt{5}\cos \alpha$. $\sin \alpha = \frac{\sqrt{5}}{5}$, $\alpha \in (0; \frac{\pi}{2})$. $\cos \alpha = \sqrt{1 - 0.2} = \sqrt{0.8} = \frac{\sqrt{4}}{\sqrt{5}} = \frac{2}{\sqrt{5}}$. Ответ: $40\sqrt{5} \cdot \frac{2}{\sqrt{5}} = 80$
12. $38\sqrt{2}\sin(\frac{3\pi}{8})\cos(\frac{3\pi}{8}) = 19\sqrt{2} \cdot 2\sin(\frac{3\pi}{8})\cos(\frac{3\pi}{8}) = 19\sqrt{2}\sin(\frac{3\pi}{4}) = 19\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 19$
13. $69\sqrt{2}(\cos^2 \frac{\pi}{8} - \sin^2 \frac{\pi}{8}) = 69\sqrt{2} \cdot \cos(2 \cdot \frac{\pi}{8}) = 69\sqrt{2}\cos(\frac{\pi}{4}) = 69\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 69$
14. $-35\cos 2\alpha = -35(2\cos^2 \alpha - 1) = -35(2(0.1)^2 - 1) = -35(0.02 - 1) = -35(-0.98) = 34.3$
15. $18\sqrt{2}\cos^2(\frac{5\pi}{8}) - \sqrt{162} = 18\sqrt{2} \cdot \frac{1 + \cos(\frac{5\pi}{4})}{2} - 9\sqrt{2} = 9\sqrt{2}(1 - \frac{\sqrt{2}}{2}) - 9\sqrt{2} = 9\sqrt{2} - 9 - 9\sqrt{2} = -9$
16. $\sqrt{147} - 14\sqrt{3}\sin^2(\frac{19\pi}{12}) = 7\sqrt{3} - 14\sqrt{3}(\frac{1 - \cos(\frac{19\pi}{6})}{2}) = 7\sqrt{3} - 7\sqrt{3}(1 - \cos(3\pi + \frac{\pi}{6})) = 7\sqrt{3} - 7\sqrt{3}(1 - (-\cos \frac{\pi}{6})) = 7\sqrt{3} - 7\sqrt{3}(1 + \frac{\sqrt{3}}{2}) = 7\sqrt{3} - 7\sqrt{3} - 7\sqrt{3} \cdot \frac{\sqrt{3}}{2} = -\frac{21}{2} = -10.5$
