21. Найдите значение выражения 19/(cos^2 37° + 1 + cos^2 53°).

21. $\frac{19}{\cos^2 37^\circ + 1 + \cos^2 53^\circ} = \frac{19}{\cos^2 37^\circ + \sin^2 37^\circ + 1} = \frac{19}{1+1} = 9.5$ 22. $\frac{59}{\cos^2 14^\circ + 3 + \cos^2 76^\circ} = \frac{59}{\cos^2 14^\circ + \sin^2 14^\circ + 3} = \frac{59}{1+3} = 14.75$ 23. $\frac{35 \cos 11^\circ}{\sin 79^\circ} + 7 = \frac{35 \cos 11^\circ}{\cos(90^\circ - 79^\circ)} + 7 = \frac{35 \cos 11^\circ}{\cos 11^\circ} + 7 = 35 + 7 = 42$ 24. $46 \text{tg} 7^\circ \cdot \text{tg} 83^\circ = 46 \text{tg} 7^\circ \cdot \text{ctg} 7^\circ = 46 \cdot 1 = 46$ 25. $\cos \alpha = -\frac{\sqrt{10}}{10}$, $\alpha \in (\frac{\pi}{2}; \pi)$ (2-я четверть, $\text{tg} \alpha < 0$) $\sin \alpha = \sqrt{1 - (-\frac{\sqrt{10}}{10})^2} = \sqrt{1 - \frac{10}{100}} = \sqrt{0.9} = \frac{3}{\sqrt{10}}$ $\text{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{3/\sqrt{10}}{-\sqrt{10}/10} = \frac{3}{\sqrt{10}} \cdot (-\frac{10}{\sqrt{10}}) = -\frac{30}{10} = -3$ 26. $\frac{32 \cos 26^\circ}{\sin 64^\circ} = \frac{32 \cos 26^\circ}{\cos(90^\circ - 64^\circ)} = \frac{32 \cos 26^\circ}{\cos 26^\circ} = 32$ 27. $\sqrt{50} \cos^2 \frac{9\pi}{8} - \sqrt{50} \sin^2 \frac{9\pi}{8} = \sqrt{50} (\cos^2 \frac{9\pi}{8} - \sin^2 \frac{9\pi}{8}) = \sqrt{50} \cos(\frac{18\pi}{8}) = \sqrt{50} \cos(\frac{9\pi}{4}) = \sqrt{50} \cos(\frac{\pi}{4}) = 5\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 5$ 28. $\sin \alpha = \frac{2\sqrt{6}}{5}$, $\alpha \in (\frac{\pi}{2}; \pi)$ (2-я четверть, $\cos \alpha < 0$) $\cos \alpha = -\sqrt{1 - (\frac{2\sqrt{6}}{5})^2} = -\sqrt{1 - \frac{4 \cdot 6}{25}} = -\sqrt{1 - \frac{24}{25}} = -\sqrt{\frac{1}{25}} = -0.2$ 29. $\cos 2\alpha = 1 - 2 \sin^2 \alpha = 1 - 2(-0.7)^2 = 1 - 2(0.49) = 1 - 0.98 = 0.02$. Значит, $2 \cos 2\alpha = 2 \cdot 0.02 = 0.04$ 30. $\cos \alpha = 0.6$, $\alpha \in (\pi; 2\pi)$ (4-я четверть, $\sin \alpha < 0$) $\sin \alpha = -\sqrt{1 - 0.6^2} = -\sqrt{1 - 0.36} = -\sqrt{0.64} = -0.8$ $\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot (-0.8) \cdot 0.6 = -0.96$ 31. $4\sqrt{2} \cos^2 \frac{15\pi}{8} - 2\sqrt{2} = 4\sqrt{2} \cos^2 \frac{15\pi}{8} - 2\sqrt{2} (\cos^2 \frac{15\pi}{8} + \sin^2 \frac{15\pi}{8}) = 2\sqrt{2} (\cos^2 \frac{15\pi}{8} - \sin^2 \frac{15\pi}{8}) = 2\sqrt{2} \cos(2 \cdot \frac{15\pi}{8}) = 2\sqrt{2} \cos \frac{15\pi}{4} = 2\sqrt{2} \cos(4\pi - \frac{\pi}{4}) = 2\sqrt{2} \cos(-\frac{\pi}{4}) = 2\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 2$