Упр.1024 ГДЗ Колягин Ткачёва 10 класс (Алгебра)

1) cos 135°; 2) cos 120°; 3) cos 150°; 4) cos 240°.

Используем формулу сложения:

$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$$

1. $$\cos 135^\circ=\cos(90^\circ+45^\circ)$$

   $$\cos 135^\circ=\cos 90^\circ\cos 45^\circ-\sin 90^\circ\sin 45^\circ$$

   $$=0\cdot \frac{\sqrt2}{2}-1\cdot \frac{\sqrt2}{2}=-\frac{\sqrt2}{2}.$$

2. $$\cos 120^\circ=\cos(60^\circ+60^\circ)$$

   $$\cos 120^\circ=\cos 60^\circ\cos 60^\circ-\sin 60^\circ\sin 60^\circ$$

   $$=\frac12\cdot \frac12-\frac{\sqrt3}{2}\cdot \frac{\sqrt3}{2}=\frac14-\frac34=-\frac12.$$

3. $$\cos 150^\circ=\cos(90^\circ+60^\circ)$$

   $$\cos 150^\circ=\cos 90^\circ\cos 60^\circ-\sin 90^\circ\sin 60^\circ$$

   $$=0\cdot \frac12-1\cdot \frac{\sqrt3}{2}=-\frac{\sqrt3}{2}.$$

4. $$\cos 240^\circ=\cos(180^\circ+60^\circ)$$

   $$\cos 240^\circ=\cos 180^\circ\cos 60^\circ-\sin 180^\circ\sin 60^\circ$$

   $$=-1\cdot \frac12-0\cdot \frac{\sqrt3}{2}=-\frac12.$$

Ответ

$$\cos 135^\circ=-\frac{\sqrt2}{2},\quad \cos 120^\circ=-\frac12,\quad \cos 150^\circ=-\frac{\sqrt3}{2},\quad \cos 240^\circ=-\frac12.$$