Упр.42.43 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) sin(a)sin(?)(ctg(a)+ctg(?))=sin(a+?);
2) (2sin(a)cos(?)-sin(a-?))/(cos(a-?)-2sin(a)sin(?))=tg(a+?);
3) (v2cos(a)-2cos(?/4+a))/(2sin(?/4+a)-v2sin(a))=tg(a).

1. $$\sin a \cdot \sin \beta \left(\ctg a+\ctg \beta\right)=\sin a \cdot \sin \beta \left(\frac{\cos a}{\sin a}+\frac{\cos \beta}{\sin \beta}\right)$$

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

2. $$\frac{2\sin a \cos \beta-\sin(a-\beta)}{\cos(a-\beta)-2\sin a \sin \beta}$$

   $$=\frac{2\sin a \cos \beta-\left(\sin a \cos \beta-\cos a \sin \beta\right)}{\left(\cos a \cos \beta+\sin a \sin \beta\right)-2\sin a \sin \beta}$$

   $$=\frac{\sin a \cos \beta+\cos a \sin \beta}{\cos a \cos \beta-\sin a \sin \beta}=\frac{\sin(a+\beta)}{\cos(a+\beta)}=\tg(a+\beta).$$

3. $$\frac{\sqrt{2}\cos a-2\cos\left(\frac{\pi}{4}+a\right)}{2\sin\left(\frac{\pi}{4}+a\right)-\sqrt{2}\sin a}$$

   $$=\frac{\sqrt{2}\cos a-2\left(\cos\frac{\pi}{4}\cos a-\sin\frac{\pi}{4}\sin a\right)}{2\left(\sin\frac{\pi}{4}\cos a+\cos\frac{\pi}{4}\sin a\right)-\sqrt{2}\sin a}$$

   $$=\frac{\sqrt{2}\cos a-\left(\sqrt{2}\cos a-\sqrt{2}\sin a\right)}{\left(\sqrt{2}\cos a+\sqrt{2}\sin a\right)-\sqrt{2}\sin a}$$

   $$=\frac{\sqrt{2}\sin a}{\sqrt{2}\cos a}=\tg a.$$

Ответ

$$1)\ \sin a \cdot \sin \beta \left(\ctg a+\ctg \beta\right)=\sin(a+\beta);$$
$$2)\ \frac{2\sin a \cos \beta-\sin(a-\beta)}{\cos(a-\beta)-2\sin a \sin \beta}=\tg(a+\beta);$$
$$3)\ \frac{\sqrt{2}\cos a-2\cos\left(\frac{\pi}{4}+a\right)}{2\sin\left(\frac{\pi}{4}+a\right)-\sqrt{2}\sin a}=\tg a.$$