### Задание 3
1) Дано: $\cos \alpha = -\frac{\sqrt{3}}{2}$, $\alpha \in (\frac{\pi}{2}; \pi)$ (II четверть, $\sin \alpha > 0$).
$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - (-\frac{\sqrt{3}}{2})^2 = 1 - \frac{3}{4} = \frac{1}{4}$.
$\sin \alpha = \sqrt{\frac{1}{4}} = \frac{1}{2}$ (т.к. II четверть).
2) Дано: $\cos \alpha = \frac{\sqrt{21}}{5}$, $\alpha \in (\frac{3\pi}{2}; 2\pi)$ (IV четверть, $\sin \alpha < 0$).
$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - (\frac{21}{25}) = \frac{4}{25}$.
$\sin \alpha = -\sqrt{\frac{4}{25}} = -\frac{2}{5}$.
### Задание 4
$4\cos 2\alpha = 4(1 - 2\sin^2 \alpha) = 4(1 - 2 \cdot (-0,8)^2) = 4(1 - 2 \cdot 0,64) = 4(1 - 1,28) = 4 \cdot (-0,28) = -1,12$.
### Задание 5
1) $\sin \alpha = \frac{7\sqrt{53}}{53}$, $\alpha \in (\frac{\pi}{2}; \pi)$ (II четверть, $\cos \alpha < 0$).
$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{49 \cdot 53}{53^2} = 1 - \frac{49}{53} = \frac{4}{53}$.
$\cos \alpha = -\frac{2}{\sqrt{53}}$.
$\text{tg } \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{7\sqrt{53}}{53} : (-\frac{2}{\sqrt{53}}) = -\frac{7\sqrt{53}}{53} \cdot \frac{\sqrt{53}}{2} = -\frac{7 \cdot 53}{53 \cdot 2} = -3,5$.
2) $\cos \alpha = -\frac{\sqrt{37}}{37}$, $\alpha \in (\pi; \frac{3\pi}{2})$ (III четверть, $\sin \alpha < 0$).
$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \frac{37}{37^2} = 1 - \frac{1}{37} = \frac{36}{37}$.
$\sin \alpha = -\frac{6}{\sqrt{37}}$.
$\text{tg } \alpha = \frac{-6}{\sqrt{37}} : (-\frac{\sqrt{37}}{37}) = \frac{6}{\sqrt{37}} \cdot \frac{37}{\sqrt{37}} = \frac{6 \cdot 37}{37} = 6$.
### Задание 6
1) $20\sqrt{2} \cos \frac{5\pi}{4} \cos \frac{2\pi}{3} = 20\sqrt{2} \cdot (-\frac{\sqrt{2}}{2}) \cdot (-\frac{1}{2}) = 20 \cdot \frac{2}{4} = 10$.
2) $14\sqrt{2} \text{tg} \frac{3\pi}{4} \sin \frac{5\pi}{4} = 14\sqrt{2} \cdot (-1) \cdot (-\frac{\sqrt{2}}{2}) = 14 \cdot \frac{2}{2} = 14$.
3) $9\sqrt{2} \sin \frac{\pi}{8} \cos \frac{\pi}{8} = 9\sqrt{2} \cdot \frac{1}{2} \sin \frac{\pi}{4} = \frac{9\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{9 \cdot 2}{4} = 4,5$.
4) $7\sqrt{2}(\cos^2 \frac{5\pi}{8} - \sin^2 \frac{5\pi}{8}) = 7\sqrt{2} \cos(2 \cdot \frac{5\pi}{8}) = 7\sqrt{2} \cos \frac{5\pi}{4} = 7\sqrt{2} \cdot (-\frac{\sqrt{2}}{2}) = -7$.
5) $\sqrt{48}\cos^2 \frac{17\pi}{12} - \sqrt{12} = 4\sqrt{3} \cdot \frac{1 + \cos \frac{17\pi}{6}}{2} - 2\sqrt{3} = 2\sqrt{3}(1 + \cos(2\pi + \frac{5\pi}{6})) - 2\sqrt{3} = 2\sqrt{3}(1 - \frac{\sqrt{3}}{2}) - 2\sqrt{3} = 2\sqrt{3} - 3 - 2\sqrt{3} = -3$.
6) $\sqrt{50} - \sqrt{200}\sin^2 \frac{5\pi}{8} = 5\sqrt{2} - 10\sqrt{2} \cdot \frac{1 - \cos \frac{5\pi}{4}}{2} = 5\sqrt{2} - 5\sqrt{2}(1 - (-\frac{\sqrt{2}}{2})) = 5\sqrt{2} - 5\sqrt{2} - 5\sqrt{2} \cdot \frac{\sqrt{2}}{2} = -5$.
7) $\frac{22\sin 57^\circ \cos 57^\circ}{\sin 114^\circ} = \frac{11(2\sin 57^\circ \cos 57^\circ)}{\sin 114^\circ} = \frac{11\sin 114^\circ}{\sin 114^\circ} = 11$.
8) $\frac{5\sin 142^\circ}{\cos 71^\circ \cos 19^\circ} = \frac{5(2\sin 71^\circ \cos 71^\circ)}{\cos 71^\circ \sin 71^\circ} = 10$.
### Задания 34-37
34) $\frac{2(\cos^2 \alpha - \sin^2 \alpha)}{\cos 2\alpha} = \frac{2 \cos 2\alpha}{\cos 2\alpha} = 2$.
35) $\frac{\sin \alpha \cos \alpha \cos 2\alpha}{\cos(\frac{3\pi}{2} - 8\alpha)} = \frac{\frac{1}{2}\sin 2\alpha \cos 2\alpha}{-\sin 8\alpha} = \frac{\frac{1}{4}\sin 4\alpha}{-\sin 8\alpha} = \frac{\frac{1}{4}\sin 4\alpha}{-2\sin 4\alpha \cos 4\alpha} = -\frac{1}{8\cos 4\alpha}$.
36) $\frac{\sin(\pi + 4\alpha)}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos \alpha \cos 2\alpha} = \frac{-\sin 4\alpha}{\frac{1}{2} \sin \alpha \cos \alpha \cos 2\alpha} = \frac{-\sin 4\alpha}{\frac{1}{4} \sin 2\alpha \cos 2\alpha} = \frac{-\sin 4\alpha}{\frac{1}{8} \sin 4\alpha} = -8$.
37) $\frac{\sin 6\alpha}{\cos^2 3\alpha} \cdot \text{ctg } 3\alpha = \frac{2 \sin 3\alpha \cos 3\alpha}{\cos^2 3\alpha} \cdot \frac{\cos 3\alpha}{\sin 3\alpha} = \frac{2 \sin 3\alpha \cos^2 3\alpha}{\cos^2 3\alpha \sin 3\alpha} = 2$.
