Упр.643 ГДЗ Никольский Потапов 9 класс (Алгебра)

Задача

а) cos 5пи/12+cos пи/12
б) cos 7пи/12-cos пи/12
в) sin 5пи/12+sin пи/12
г) sin 7пи/12-sin пи/12

Подробный ответ

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)

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