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

1) cos(6a)cos(2a)-sin(6a)sin(2a);
2) sin(12°)cos(18°)+sin(18°)cos(12°);
3) sin(-15°)cos(75°)+cos(15°)sin(75°);
4) cos(a+?)cos(a-?)+sin(a+?)sin(a-?);
5) (cos(64°)cos(4°)+sin(64°)sin(4°))/(sin(19°)cos(41°)+sin(41°)cos(19°));
6) cos(a-?)-2sin(a)sin(?).

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

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

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

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

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

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

Ответ

1) $$\cos 8a$$; 2) $$\frac12$$; 3) $$\frac{\sqrt3}{2}$$; 4) $$\cos 2\beta$$; 5) $$\frac{1}{\sqrt3}$$; 6) $$\cos(a+\beta)$$.