Упр.28.14 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)
- Вычислите:
а) $$\frac{\cos 68^\circ-\cos 22^\circ}{\sin 22^\circ-\cos 22^\circ}$$;
б) $$\frac{\sin \frac{7\pi}{18}-\sin \frac{\pi}{9}}{\cos \frac{7\pi}{18}-\cos \frac{\pi}{9}}$$;
в) $$\frac{\sin 130^\circ+\cos 110^\circ}{\cos 130^\circ+\cos 110^\circ}$$;
г) $$\frac{\sin \frac{5\pi}{18}+\sin \frac{11\pi}{9}}{\cos \frac{5\pi}{18}+\cos \frac{11\pi}{9}}$$.
Используем формулы:
$$\cos \alpha-\cos \beta=-2\sin \frac{\alpha+\beta}{2}\sin \frac{\alpha-\beta}{2},$$
$$\sin \alpha-\sin \beta=2\cos \frac{\alpha+\beta}{2}\sin \frac{\alpha-\beta}{2},$$
$$\sin \alpha+\sin \beta=2\sin \frac{\alpha+\beta}{2}\cos \frac{\alpha-\beta}{2},$$
$$\cos \alpha+\cos \beta=2\cos \frac{\alpha+\beta}{2}\cos \frac{\alpha-\beta}{2}.$$
а)
$$\frac{\cos 68^\circ-\cos 22^\circ}{\sin 22^\circ-\cos 22^\circ} = \frac{-2\sin 45^\circ \sin 23^\circ}{2\sin 23^\circ \cos 45^\circ} = -\frac{\sin 45^\circ}{\cos 45^\circ} = -\tg 45^\circ = -1.$$
б)
$$\frac{\sin \frac{7\pi}{18}-\sin \frac{\pi}{9}}{\cos \frac{7\pi}{18}-\cos \frac{\pi}{9}} = \frac{2\cos \frac{5\pi}{36}\sin \frac{\pi}{4}}{-2\sin \frac{5\pi}{36}\sin \frac{\pi}{4}} = -\frac{\cos \frac{5\pi}{36}}{\sin \frac{5\pi}{36}} = -\ctg \frac{5\pi}{36}.$$
$$\ctg \frac{5\pi}{36}=\ctg \frac{\pi}{4}=1,$$
значит,
$$\frac{\sin \frac{7\pi}{18}-\sin \frac{\pi}{9}}{\cos \frac{7\pi}{18}-\cos \frac{\pi}{9}}=-1.$$
в)
$$\frac{\sin 130^\circ+\cos 110^\circ}{\cos 130^\circ+\cos 110^\circ} = \frac{\sin 130^\circ+\sin 20^\circ}{\cos 130^\circ+\cos 110^\circ} = \frac{2\sin 75^\circ \cos 55^\circ}{2\cos 120^\circ \cos 10^\circ} = \frac{\sin 120^\circ}{\cos 120^\circ} = \tg 120^\circ = -\sqrt{3}.$$
г)
$$\frac{\sin \frac{5\pi}{18}+\sin \frac{11\pi}{9}}{\cos \frac{5\pi}{18}+\cos \frac{11\pi}{9}} = \frac{2\sin \frac{27\pi}{36}\cos \frac{17\pi}{36}}{2\cos \frac{27\pi}{36}\cos \frac{17\pi}{36}} = \tg \frac{27\pi}{36} = \tg \frac{3\pi}{4} = -1.$$
Ответ: а) $$-1$$; б) $$-1$$; в) $$-\sqrt{3}$$; г) $$-1$$.
