Упр.21.3 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) sin(a)cos(4a)+cos(a)sin(4a);
2) cos(17°)cos(43°)-sin(17°)sin(43°);
3) cos(3?/8)cos(?/8)-sin(3?/8)sin(?/8);
4) sin(a)sin(a+?)+cos(a)cos(a+?);
5) sin(53°)cos(7°)-cos(53°)sin(-7°);
6) sin(a+?)cos(a-?)-sin(a-?)cos(a+?);
7) (sin(a)cos(?)+cos(a)sin(?))^2+(cos(a)cos(?)-sin(a)sin(?))^2;
8) (sin(20°)cos(5°)-cos(20°)sin(5°))/(cos(10°)cos(5°)-sin(10°)sin(5°));
9) cos(a+?)+2sin(a)sin(?).

1. $$\sin a \cos 4a+\cos a \sin 4a=\sin(a+4a)=\sin 5a.$$

2. $$\cos 17^\circ \cos 43^\circ-\sin 17^\circ \sin 43^\circ=\cos(17^\circ+43^\circ)=\cos 60^\circ=\frac12.$$

3. $$\cos \frac{3\pi}{8}\cos \frac{\pi}{8}-\sin \frac{3\pi}{8}\sin \frac{\pi}{8}=\cos\left(\frac{3\pi}{8}+\frac{\pi}{8}\right)=\cos \frac{\pi}{2}=0.$$

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

5. $$\sin 53^\circ \cos 7^\circ-\cos 53^\circ \sin(-7^\circ)=\sin 53^\circ \cos 7^\circ+\cos 53^\circ \sin 7^\circ$$

   $$=\sin(53^\circ+7^\circ)=\sin 60^\circ=\frac{\sqrt3}{2}.$$

6. $$\sin(a+\beta)\cos(a-\beta)-\sin(a-\beta)\cos(a+\beta)$$

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

7. $$\bigl(\sin a \cos \beta+\cos a \sin \beta\bigr)^2+\bigl(\cos a \cos \beta-\sin a \sin \beta\bigr)^2$$

   $$=\sin^2(a+\beta)+\cos^2(a+\beta)=1.$$

8. $$\frac{\sin 20^\circ \cos 5^\circ-\cos 20^\circ \sin 5^\circ}{\cos 10^\circ \cos 5^\circ-\sin 10^\circ \sin 5^\circ}$$

   $$=\frac{\sin(20^\circ-5^\circ)}{\cos(10^\circ+5^\circ)}=\frac{\sin 15^\circ}{\cos 15^\circ}=\tg 15^\circ.$$

9. $$\cos(a+\beta)+2\sin a \sin \beta$$

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

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

Ответ

1. $$\sin 5a$$
2. $$\frac12$$
3. $$0$$
4. $$\cos \beta$$
5. $$\frac{\sqrt3}{2}$$
6. $$\sin 2\beta$$
7. $$1$$
8. $$\tg 15^\circ$$
9. $$\cos(a-\beta)$$