Упр.652 ГДЗ Никольский Потапов 9 класс (Алгебра)

а) 1 +2 sin a = 2(1/2 + sina) = 2(sinпи/6 + sina) =….
б) 1 — 2 cos a;
в) v3 — 2 sin a.

Воспользуемся формулами суммы и разности синусов и косинусов:

$$ \sin x+\sin y=2\sin \frac{x+y}{2}\cos \frac{x-y}{2}, $$
$$ \cos x-\cos y=-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}. $$

1. $$ 1+2\sin \alpha=2\left(\frac12+\sin \alpha\right)=2\left(\sin \frac{\pi}{6}+\sin \alpha\right) $$
   $$ =2\cdot 2\sin \frac{\frac{\pi}{6}+\alpha}{2}\cos \frac{\frac{\pi}{6}-\alpha}{2} =4\sin \frac{\pi+6\alpha}{12}\cos \frac{\pi-6\alpha}{12}. $$

2. $$ 1-2\cos \alpha=2\left(\frac12-\cos \alpha\right)=2\left(\cos \frac{\pi}{3}-\cos \alpha\right) $$
   $$ =2\cdot(-2)\sin \frac{\frac{\pi}{3}+\alpha}{2}\sin \frac{\frac{\pi}{3}-\alpha}{2} =-4\sin \frac{\pi+3\alpha}{6}\sin \frac{\pi-3\alpha}{6}. $$

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

Ответ

$$ \text{а) }4\sin \frac{\pi+6\alpha}{12}\cos \frac{\pi-6\alpha}{12};\quad \text{б) }-4\sin \frac{\pi+3\alpha}{6}\sin \frac{\pi-3\alpha}{6};\quad \text{в) }4\sin \frac{\pi-3\alpha}{6}\cos \frac{\pi+3\alpha}{6}. $$