Упр.645 ГДЗ Никольский Потапов 9 класс (Алгебра)

1. Докажите справедливость равенства:
   а) $$\cos \frac{5\pi}{12}+\cos \frac{7\pi}{12}=0$$
   б) $$\sin \frac{3\pi}{5}-\sin \frac{2\pi}{5}=0$$
   в) $$\cos \frac{9\pi}{14}+\cos \frac{5\pi}{14}=0$$
   г) $$\sin \frac{3\pi}{10}-\sin \frac{7\pi}{10}=0$$

Воспользуемся формулами суммы и разности тригонометрических функций:

$$\cos \alpha+\cos \beta=2\cos \frac{\alpha+\beta}{2}\cos \frac{\alpha-\beta}{2}$$
$$\sin \alpha-\sin \beta=2\cos \frac{\alpha+\beta}{2}\sin \frac{\alpha-\beta}{2}$$

1. $$\cos \frac{5\pi}{12}+\cos \frac{7\pi}{12}=2\cos \frac{\frac{5\pi}{12}+\frac{7\pi}{12}}{2}\cos \frac{\frac{5\pi}{12}-\frac{7\pi}{12}}{2}$$
   $$=2\cos \frac{\pi}{2}\cos \left(-\frac{\pi}{12}\right)=2\cdot 0 \cdot \cos \left(-\frac{\pi}{12}\right)=0.$$

2. $$\sin \frac{3\pi}{5}-\sin \frac{2\pi}{5}=2\cos \frac{\frac{3\pi}{5}+\frac{2\pi}{5}}{2}\sin \frac{\frac{3\pi}{5}-\frac{2\pi}{5}}{2}$$
   $$=2\cos \frac{\pi}{2}\sin \frac{\pi}{10}=2\cdot 0 \cdot \sin \frac{\pi}{10}=0.$$

3. $$\cos \frac{9\pi}{14}+\cos \frac{5\pi}{14}=2\cos \frac{\frac{9\pi}{14}+\frac{5\pi}{14}}{2}\cos \frac{\frac{9\pi}{14}-\frac{5\pi}{14}}{2}$$
   $$=2\cos \frac{\pi}{2}\cos \frac{\pi}{7}=2\cdot 0 \cdot \cos \frac{\pi}{7}=0.$$

4. $$\sin \frac{3\pi}{10}-\sin \frac{7\pi}{10}=2\cos \frac{\frac{3\pi}{10}+\frac{7\pi}{10}}{2}\sin \frac{\frac{3\pi}{10}-\frac{7\pi}{10}}{2}$$
   $$=2\cos \frac{\pi}{2}\sin \left(-\frac{\pi}{5}\right)=2\cdot 0 \cdot \sin \left(-\frac{\pi}{5}\right)=0.$$

Ответ

$$\cos \frac{5\pi}{12}+\cos \frac{7\pi}{12}=0,\quad \sin \frac{3\pi}{5}-\sin \frac{2\pi}{5}=0,\quad \cos \frac{9\pi}{14}+\cos \frac{5\pi}{14}=0,\quad \sin \frac{3\pi}{10}-\sin \frac{7\pi}{10}=0.$$