Упр.643 ГДЗ Никольский Потапов 9 класс (Алгебра)

1. Вычислите: а) $$\cos\frac{5\pi}{12}+\cos\frac{\pi}{12}$$; б) $$\cos\frac{7\pi}{12}-\cos\frac{\pi}{12}$$; в) $$\sin\frac{5\pi}{12}+\sin\frac{\pi}{12}$$; г) $$\sin\frac{7\pi}{12}-\sin\frac{\pi}{12}$$.

Воспользуемся формулами суммы и разности синусов и косинусов:

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

а)

$$\cos \frac{5\pi}{12}+\cos \frac{\pi}{12} =2\cos \frac{\frac{5\pi}{12}+\frac{\pi}{12}}{2}\cos \frac{\frac{5\pi}{12}-\frac{\pi}{12}}{2}$$
$$=2\cos \frac{6\pi}{12}\cos \frac{4\pi}{12} =2\cos \frac{\pi}{4}\cos \frac{\pi}{6} =2\cdot \frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2} =\frac{\sqrt{6}}{2}$$

б)

$$\cos \frac{7\pi}{12}-\cos \frac{\pi}{12} =-2\sin \frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}\sin \frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}$$
$$=-2\sin \frac{8\pi}{12}\sin \frac{6\pi}{12} =-2\sin \frac{2\pi}{3}\sin \frac{\pi}{2} =-2\cdot \frac{\sqrt{3}}{2}\cdot 1 =-\sqrt{3}$$

в)

$$\sin \frac{5\pi}{12}+\sin \frac{\pi}{12} =2\sin \frac{\frac{5\pi}{12}+\frac{\pi}{12}}{2}\cos \frac{\frac{5\pi}{12}-\frac{\pi}{12}}{2}$$
$$=2\sin \frac{6\pi}{12}\cos \frac{4\pi}{12} =2\sin \frac{\pi}{2}\cos \frac{\pi}{6} =2\cdot 1\cdot \frac{\sqrt{3}}{2} =\sqrt{3}$$

г)

$$\sin \frac{7\pi}{12}-\sin \frac{\pi}{12} =2\sin \frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}\cos \frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}$$
$$=2\sin \frac{6\pi}{12}\cos \frac{8\pi}{12} =2\sin \frac{\pi}{2}\cos \frac{2\pi}{3} =2\cdot 1\cdot \left(-\frac{1}{2}\right) =-1$$

Ответ

а) $$\frac{\sqrt{6}}{2}$$; б) $$-\sqrt{3}$$; в) $$\sqrt{3}$$; г) $$-1$$.