Упр.27.4 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) 1-2sin(?); 3) v2+2cos(?);
2) v3-2cos(?); 4) v3-tg(?).

1. $$1-2\sin\alpha=2\left(\sin\frac{\pi}{6}-\sin\alpha\right)$$
   $$=4\sin\left(\frac{\pi}{12}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{12}+\frac{\alpha}{2}\right).$$

2. $$\sqrt{3}-2\cos\alpha=2\left(\cos\frac{\pi}{6}-\cos\alpha\right)$$
   $$=4\sin\left(\frac{\alpha}{2}-\frac{\pi}{12}\right)\sin\left(\frac{\alpha}{2}+\frac{\pi}{12}\right).$$

3. $$\sqrt{2}+2\cos\alpha=2\left(\cos\frac{\pi}{4}+\cos\alpha\right)$$
   $$=4\cos\left(\frac{\pi}{8}+\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{8}-\frac{\alpha}{2}\right).$$

4. $$\sqrt{3}-\tg\alpha=\tg\frac{\pi}{3}-\tg\alpha$$
   $$=\frac{\sin\left(\frac{\pi}{3}-\alpha\right)}{\cos\frac{\pi}{3}\cos\alpha}$$
   $$=\frac{2\sin\left(\frac{\pi}{3}-\alpha\right)}{\cos\alpha}.$$

Ответ

$$1-2\sin\alpha=4\sin\left(\frac{\pi}{12}-\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{12}+\frac{\alpha}{2}\right);$$
$$\sqrt{3}-2\cos\alpha=4\sin\left(\frac{\alpha}{2}-\frac{\pi}{12}\right)\sin\left(\frac{\alpha}{2}+\frac{\pi}{12}\right);$$
$$\sqrt{2}+2\cos\alpha=4\cos\left(\frac{\pi}{8}+\frac{\alpha}{2}\right)\cos\left(\frac{\pi}{8}-\frac{\alpha}{2}\right);$$
$$\sqrt{3}-\tg\alpha=\frac{2\sin\left(\frac{\pi}{3}-\alpha\right)}{\cos\alpha}.$$