№2. b) Найдите значение выражения 1) cos 225° 2) sin 7π/6 3) tg 8π/3 4) cos² (8π/3) - sin² (8π/3)

Решение:

1. \( \cos 225° \)
   \[ \cos 225° = \cos(180° + 45°) = -\cos 45° = -\frac{\sqrt{2}}{2} \]
2. \( \sin \frac{7\pi}{6} \)
   \[ \sin \frac{7\pi}{6} = \sin\left(\pi + \frac{\pi}{6}\right) = -\sin \frac{\pi}{6} = -\frac{1}{2} \]
3. \( \text{tg} \frac{8\pi}{3} \)
   \[ \text{tg} \frac{8\pi}{3} = \text{tg}\left(2\pi + \frac{2\pi}{3}\right) = \text{tg} \frac{2\pi}{3} = -\sqrt{3} \]
4. \( \cos^2 \left(\frac{8\pi}{3}\right) - \sin^2 \left(\frac{8\pi}{3}\right) \)
   Используем формулу \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \). Здесь \( \alpha = \frac{8\pi}{3} \).
   \[ \cos^2 \left(\frac{8\pi}{3}\right) - \sin^2 \left(\frac{8\pi}{3}\right) = \cos\left(2 \cdot \frac{8\pi}{3}\right) = \cos \frac{16\pi}{3} \]
   \[ \cos \frac{16\pi}{3} = \cos\left(5\pi + \frac{\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos \frac{\pi}{3} = -\frac{1}{2} \]

Ответ: 1) \( -\frac{\sqrt{2}}{2} \); 2) \( -\frac{1}{2} \); 3) \( -\sqrt{3} \); 4) \( -\frac{1}{2} \).