№2. a) Найдите значение выражения 1) cos 150° 2) sin 13π/4 3) tg 7π/3 4) cos² (2π/3) - sin² (2π/3)

Решение:

1. \( \cos 150° \)
   \[ \cos 150° = \cos(180° - 30°) = -\cos 30° = -\frac{\sqrt{3}}{2} \]
2. \( \sin \frac{13\pi}{4} \)
   \[ \sin \frac{13\pi}{4} = \sin\left(3\pi + \frac{\pi}{4}\right) = \sin\left(\pi + \frac{\pi}{4}\right) = -\sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2} \]
3. \( \text{tg} \frac{7\pi}{3} \)
   \[ \text{tg} \frac{7\pi}{3} = \text{tg}\left(2\pi + \frac{\pi}{3}\right) = \text{tg} \frac{\pi}{3} = \sqrt{3} \]
4. \( \cos^2 \left(\frac{2\pi}{3}\right) - \sin^2 \left(\frac{2\pi}{3}\right) \)
   Используем формулу \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \). Здесь \( \alpha = \frac{2\pi}{3} \).
   \[ \cos^2 \left(\frac{2\pi}{3}\right) - \sin^2 \left(\frac{2\pi}{3}\right) = \cos\left(2 \cdot \frac{2\pi}{3}\right) = \cos \frac{4\pi}{3} \]
   \[ \cos \frac{4\pi}{3} = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos \frac{\pi}{3} = -\frac{1}{2} \]

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