Упр.524 ГДЗ Алимов 10-11 класс (Алгебра)
1) cos 75° = cos (90° — а);
2) sin 150° = sin (90° + а);
3) sin 150° = sin (180° — а);
4) cos 310° = cos (270° + а);
5) sin 5пи/4 = sin (пи + а);
6) tg пи/5 = tg(пи/2 -a);
7) cos 7пи/4 = cos (3пи/2 + а);
8) ctg 11пи/6 = ctg(2пи-a).
$$\cos 75^\circ=\cos(90^\circ-\alpha)$$
Так как косинусы равны, то
$$75^\circ=90^\circ-\alpha$$
$$\alpha=90^\circ-75^\circ=15^\circ$$
$$\sin 150^\circ=\sin(90^\circ+\alpha)$$
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
$$310^\circ=270^\circ+\alpha$$
$$\alpha=310^\circ-270^\circ=40^\circ$$
$$\sin \frac{5\pi}{4}=\sin(\pi+\alpha)$$
Тогда
$$\frac{5\pi}{4}=\pi+\alpha$$
$$\alpha=\frac{5\pi}{4}-\pi=\frac{5\pi}{4}-\frac{4\pi}{4}=\frac{\pi}{4}$$
$$\tg \frac{\pi}{5}=\tg\left(\frac{\pi}{2}-\alpha\right)$$
Следовательно,
$$\frac{\pi}{5}=\frac{\pi}{2}-\alpha$$
$$\alpha=\frac{\pi}{2}-\frac{\pi}{5}=\frac{5\pi}{10}-\frac{2\pi}{10}=\frac{3\pi}{10}$$
$$\cos \frac{7\pi}{4}=\cos\left(\frac{3\pi}{2}+\alpha\right)$$
Отсюда
$$\frac{7\pi}{4}=\frac{3\pi}{2}+\alpha$$
$$\alpha=\frac{7\pi}{4}-\frac{3\pi}{2}=\frac{7\pi}{4}-\frac{6\pi}{4}=\frac{\pi}{4}$$
$$\ctg \frac{11\pi}{6}=\ctg(2\pi-\alpha)$$
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) $$\frac{\pi}{6}$$.
