6.4. Вычислите значение выражения.

Решение:

Вычисляем значения тригонометрических функций:

1. \( \cos\frac{2\pi}{3} = -\frac{1}{2} \)
2. \( \cos\frac{11\pi}{6} = \cos(2\pi - \frac{\pi}{6}) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
3. \( \cos 135° = \cos(180° - 45°) = -\cos 45° = -\frac{\sqrt{2}}{2} \)
4. \( \sin \pi = 0 \)
5. \( \cos(-\frac{\pi}{6}) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
6. \( \cos 5\pi = \cos(4\pi + \pi) = \cos\pi = -1 \)
7. \( \sin(-150°) = -\sin 150° = -\sin(180° - 30°) = -\sin 30° = -\frac{1}{2} \)
8. \( \sin(-600°) = -\sin 600° = -\sin(360° + 240°) = -\sin 240° = -\sin(180° + 60°) = -(-\sin 60°) = \sin 60° = \frac{\sqrt{3}}{2} \)
9. \( \sin\frac{5\pi}{6} = \sin(\pi - \frac{\pi}{6}) = \sin\frac{\pi}{6} = \frac{1}{2} \)
10. \( \sin\frac{5\pi}{3} = \sin(2\pi - \frac{\pi}{3}) = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2} \)
11. \( \sin 225° = \sin(180° + 45°) = -\sin 45° = -\frac{\sqrt{2}}{2} \)
12. \( \cos(-510°) = \cos 510° = \cos(360° + 150°) = \cos 150° = \cos(180° - 30°) = -\cos 30° = -\frac{\sqrt{3}}{2} \)
13. \( \text{ctg}\frac{7\pi}{6} = \text{ctg}(\pi + \frac{\pi}{6}) = \text{ctg}\frac{\pi}{6} = \sqrt{3} \)
14. \( \text{ctg} 270° = 0 \)
15. \( \text{ctg}\frac{7\pi}{4} = \text{ctg}(2\pi - \frac{\pi}{4}) = -\text{ctg}\frac{\pi}{4} = -1 \)
16. \( \text{tg} 150° = \text{tg}(180° - 30°) = -\text{tg} 30° = -\frac{\sqrt{3}}{3} \)
17. \( \text{ctg}\frac{7\pi}{3} = \text{ctg}(2\pi + \frac{\pi}{3}) = \text{ctg}\frac{\pi}{3} = \frac{\sqrt{3}}{3} \)
18. \( \text{tg}(-\frac{4\pi}{3}) = -\text{tg}\frac{4\pi}{3} = -\text{tg}(\pi + \frac{\pi}{3}) = -\text{tg}\frac{\pi}{3} = -\sqrt{3} \)
19. \( \text{tg}(-\frac{7\pi}{6}) = -\text{tg}\frac{7\pi}{6} = -\text{tg}(\pi + \frac{\pi}{6}) = -\text{tg}\frac{\pi}{6} = -\frac{\sqrt{3}}{3} \)
20. \( \text{tg}(-\frac{8\pi}{3}) = -\text{tg}\frac{8\pi}{3} = -\text{tg}(2\pi + \frac{2\pi}{3}) = -\text{tg}\frac{2\pi}{3} = -(-\sqrt{3}) = \sqrt{3} \)
21. \( \sin 60° - \text{tg} 120° = \frac{\sqrt{3}}{2} - (-\sqrt{3}) = \frac{\sqrt{3}}{2} + \sqrt{3} = \frac{3\sqrt{3}}{2} \)
22. \( \text{tg} 45° \cdot \sin 60° \cdot \text{ctg} 30° = 1 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} \)
23. \[ \frac{2+2\sin 60°\cos 150°}{\sin 150°} = \frac{2+2(\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2})}{\frac{1}{2}} = \frac{2 - \frac{3}{2}}{\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]
24. \[ \sin \pi + \cos \pi + \text{tg} \pi = 0 + (-1) + 0 = -1 \]
25. \[ \frac{\sin \frac{\pi}{2} - \cos \pi + \text{tg} \frac{\pi}{4}}{2\sin \frac{\pi}{6} - \sin \frac{3\pi}{2}} = \frac{1 - (-1) + 1}{2(\frac{1}{2}) - (-1)} = \frac{1+1+1}{1+1} = \frac{3}{2} \]
26. \[ \sin 270° + \text{ctg} 750° = -1 + \text{ctg}(2 \cdot 360° + 30°) = -1 + \text{ctg} 30° = -1 + \sqrt{3} \]
27. \[ \text{tg} 630° + \sin 420° = \text{tg}(360°+270°) + \sin(360°+60°) = \text{tg} 270° + \sin 60° = \text{не опр.} + \frac{\sqrt{3}}{2} \text{ (задание некорректно, т.к. tg 270° не существует)} \]
28. \[ \text{ctg} 120° + \cos(-600°) = \text{ctg}(180°-60°) + \cos(720°-120°) = -\text{ctg} 60° + \cos 120° = -\frac{\sqrt{3}}{3} - \frac{1}{2} \]
29. \[ (2 - 2\sin 300°)(2 - 2\cos 330°) = (2 - 2(-\frac{\sqrt{3}}{2}))(2 - 2(\frac{\sqrt{3}}{2})) = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]
30. \[ \cos(-\frac{2\pi}{3}) + \sin\frac{5\pi}{6} = \cos\frac{2\pi}{3} + \sin\frac{5\pi}{6} = -\frac{1}{2} + \frac{1}{2} = 0 \]