1. Вычислите:
- a) \( \sin\frac{7\pi}{3} = \sin\left(2\pi + \frac{\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \)
- б) \( \cos\left(-\frac{5\pi}{4}\right) = \cos\frac{5\pi}{4} = \cos\left(\pi + \frac{\pi}{4}\right) = -\cos\frac{\pi}{4} = -\frac{\sqrt{2}}{2} \)
- в) \( \text{tg}\left(-\frac{13\pi}{6}\right) = -\text{tg}\frac{13\pi}{6} = -\text{tg}\left(2\pi + \frac{\pi}{6}\right) = -\text{tg}\frac{\pi}{6} = -\frac{\sqrt{3}}{3} \)
- г) \( \text{ctg}13.5\pi = \text{ctg}\left(13\pi + \frac{\pi}{2}\right) = \text{ctg}\frac{\pi}{2} = 0 \)
- д) \( \sin 58^\circ \cos 13^\circ - \cos 58^\circ \sin 13^\circ = \sin(58^\circ - 13^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2} \)
- е) \( \cos\frac{\pi}{12} \cos\frac{7\pi}{12} - \sin\frac{\pi}{12} \sin\frac{7\pi}{12} = \cos\left(\frac{\pi}{12} + \frac{7\pi}{12}\right) = \cos\frac{8\pi}{12} = \cos\frac{2\pi}{3} = -\frac{1}{2} \)
Ответ: а) \(\frac{\sqrt{3}}{2}\); б) \(-\frac{\sqrt{2}}{2}\); в) \(-\frac{\sqrt{3}}{3}\); г) 0; д) \(\frac{\sqrt{2}}{2}\); е) \(-\frac{1}{2}\).