Упр.21.4 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

Упростите выражение:

1. $$\cos(6a)\cos(2a)-\sin(6a)\sin(2a)$$;
2. $$\sin(12^\circ)\cos(18^\circ)+\sin(18^\circ)\cos(12^\circ)$$;
3. $$\sin(-15^\circ)\cos(75^\circ)+\cos(15^\circ)\sin(75^\circ)$$;
4. $$\cos(a+?)\cos(a-?)+\sin(a+?)\sin(a-?)$$;
5. $$\frac{\cos(64^\circ)\cos(4^\circ)+\sin(64^\circ)\sin(4^\circ)}{\sin(19^\circ)\cos(41^\circ)+\sin(41^\circ)\cos(19^\circ)}$$;
6. $$\cos(a-?)-2\sin(a)\sin(?)$$.

1. $$\cos 6a \cdot \cos 2a-\sin 6a \cdot \sin 2a=\cos(6a+2a)=\cos 8a.$$

2. $$\sin 12^\circ \cdot \cos 18^\circ+\sin 18^\circ \cdot \cos 12^\circ=\sin(12^\circ+18^\circ)=\sin 30^\circ=\frac12.$$

3. $$\sin(-15^\circ)\cdot \cos 75^\circ+\cos 15^\circ\cdot \sin 75^\circ$$

   $$=-\sin 15^\circ \cdot \cos 75^\circ+\cos 15^\circ\cdot \sin 75^\circ$$

   $$=\sin(75^\circ-15^\circ)=\sin 60^\circ=\frac{\sqrt3}{2}.$$

4. $$\cos(a+\beta)\cdot \cos(a-\beta)+\sin(a+\beta)\cdot \sin(a-\beta)$$

   $$=\cos\bigl((a+\beta)-(a-\beta)\bigr)=\cos 2\beta.$$

5. $$\frac{\cos 64^\circ \cdot \cos 4^\circ+\sin 64^\circ \cdot \sin 4^\circ}{\sin 19^\circ \cdot \cos 41^\circ+\sin 41^\circ \cdot \cos 19^\circ}$$

   $$=\frac{\cos(64^\circ-4^\circ)}{\sin(19^\circ+41^\circ)}=\frac{\cos 60^\circ}{\sin 60^\circ}=\ctg 60^\circ=\frac{1}{\sqrt3}.$$

6. $$\cos(a-\beta)-2\sin a \cdot \sin \beta$$

   $$=\bigl(\cos a \cdot \cos \beta+\sin a \cdot \sin \beta\bigr)-2\sin a \cdot \sin \beta$$

   $$=\cos a \cdot \cos \beta-\sin a \cdot \sin \beta=\cos(a+\beta).$$