Упр.24.8 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1. Преобразуйте в произведение: 1) $$1-2\sin(a)$$; 2) $$\sqrt{3}-2\cos(a)$$; 3) $$\sqrt{2}+2\cos(a)$$.

1) $$1-2\sin a=2\left(\frac12-\sin a\right)=2\left(\sin\frac{\pi}{6}-\sin a\right)$$

$$=2\cdot 2\sin\left(\frac{\pi}{12}-\frac a2\right)\cos\left(\frac{\pi}{12}+\frac a2\right)=4\sin\left(\frac{\pi}{12}-\frac a2\right)\cos\left(\frac{\pi}{12}+\frac a2\right).$$

2) $$\sqrt3-2\cos a=-2\left(\cos a-\frac{\sqrt3}{2}\right)=-2\left(\cos a-\cos\frac{\pi}{6}\right)$$

$$=-2\cdot(-2)\sin\left(\frac a2-\frac{\pi}{12}\right)\sin\left(\frac a2+\frac{\pi}{12}\right)=4\sin\left(\frac a2-\frac{\pi}{12}\right)\sin\left(\frac a2+\frac{\pi}{12}\right).$$

3) $$\sqrt2+2\cos a=2\left(\frac{\sqrt2}{2}+\cos a\right)=2\left(\cos\frac{\pi}{4}+\cos a\right)$$

$$=2\cdot 2\cos\left(\frac{\pi}{8}+\frac a2\right)\cos\left(\frac{\pi}{8}-\frac a2\right)=4\cos\left(\frac{\pi}{8}+\frac a2\right)\cos\left(\frac{\pi}{8}-\frac a2\right).$$

Ответ:
$$1-2\sin a=4\sin\left(\frac{\pi}{12}-\frac a2\right)\cos\left(\frac{\pi}{12}+\frac a2\right),$$
$$\sqrt3-2\cos a=4\sin\left(\frac a2-\frac{\pi}{12}\right)\sin\left(\frac a2+\frac{\pi}{12}\right),$$
$$\sqrt2+2\cos a=4\cos\left(\frac{\pi}{8}+\frac a2\right)\cos\left(\frac{\pi}{8}-\frac a2\right).$$