Упр.524 ГДЗ Алимов 10-11 класс (Алгебра)
Найти значение острого угла $$a$$, если:
- $$\cos 75^\circ=\cos(90^\circ-a);$$
2) $$\sin 150^\circ=\sin(90^\circ+a);$$
3) $$\sin 150^\circ=\sin(180^\circ-a);$$
4) $$\cos 310^\circ=\cos(270^\circ+a);$$
5) $$\sin\frac{5\pi}{4}=\sin(\pi+a);$$
6) $$\operatorname{tg}\frac{\pi}{5}=\operatorname{tg}\left(\frac{\pi}{2}-a\right);$$
7) $$\cos\frac{7\pi}{4}=\cos\left(\frac{3\pi}{2}+a\right);$$
8) $$\operatorname{ctg}\frac{11\pi}{6}=\operatorname{ctg}(2\pi-a).$$
$$\cos 75^\circ=\cos(90^\circ-\alpha)$$
Так как косинусы равны, то при острых углах:
$$75^\circ=90^\circ-\alpha$$
$$\alpha=90^\circ-75^\circ=15^\circ$$
$$\sin 150^\circ=\sin(90^\circ+\alpha)$$
Тогда:
$$150^\circ=90^\circ+\alpha$$
$$\alpha=150^\circ-90^\circ=60^\circ$$
$$\sin 150^\circ=\sin(180^\circ-\alpha)$$
Следовательно:
$$150^\circ=180^\circ-\alpha$$
$$\alpha=180^\circ-150^\circ=30^\circ$$
$$\cos 310^\circ=\cos(270^\circ+\alpha)$$
Отсюда:
$$310^\circ=270^\circ+\alpha$$
$$\alpha=310^\circ-270^\circ=40^\circ$$
$$\sin \frac{5\pi}{4}=\sin(\pi+\alpha)$$
Получаем:
$$\frac{5\pi}{4}=\pi+\alpha$$
$$\alpha=\frac{5\pi}{4}-\pi=\frac{\pi}{4}$$
$$\tg \frac{\pi}{5}=\tg\left(\frac{\pi}{2}-\alpha\right)$$
Тогда:
$$\frac{\pi}{5}=\frac{\pi}{2}-\alpha$$
$$\alpha=\frac{\pi}{2}-\frac{\pi}{5}=\frac{5\pi}{10}-\frac{2\pi}{10}=\frac{3\pi}{10}$$
$$\cos \frac{7\pi}{4}=\cos\left(\frac{3\pi}{2}+\alpha\right)$$
Следовательно:
$$\frac{7\pi}{4}=\frac{3\pi}{2}+\alpha$$
$$\alpha=\frac{7\pi}{4}-\frac{3\pi}{2}=\frac{7\pi}{4}-\frac{6\pi}{4}=\frac{\pi}{4}$$
$$\ctg \frac{11\pi}{6}=\ctg(2\pi-\alpha)$$
Тогда:
$$\frac{11\pi}{6}=2\pi-\alpha$$
$$\alpha=2\pi-\frac{11\pi}{6}=\frac{12\pi}{6}-\frac{11\pi}{6}=\frac{\pi}{6}$$
Ответ: 1) $$15^\circ$$; 2) $$60^\circ$$; 3) $$30^\circ$$; 4) $$40^\circ$$; 5) $$\frac{\pi}{4}$$; 6) $$\frac{3\pi}{10}$$; 7) $$\frac{\pi}{4}$$; 8) $$\frac{\pi}{6}$$.
