Упр.1314 ГДЗ Алимов 10-11 класс (Алгебра)

2) cos(пи/6+a) + cos(пи/6-a) = (корень 3) cosa.

1. $$\sin\left(a+\frac{\pi}{3}\right)-\sin\left(a-\frac{\pi}{3}\right)$$
   $$=2\sin\frac{\left(a+\frac{\pi}{3}\right)-\left(a-\frac{\pi}{3}\right)}{2}\cdot \cos\frac{\left(a+\frac{\pi}{3}\right)+\left(a-\frac{\pi}{3}\right)}{2}$$
   $$=2\sin\frac{2\pi/3}{2}\cdot \cos\frac{2a}{2}=2\sin\frac{\pi}{3}\cos a$$
   $$=2\cdot \frac{\sqrt{3}}{2}\cos a=\sqrt{3}\cos a.$$

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

Ответ

1) $$\sqrt{3}\cos a$$
2) $$\sqrt{3}\cos a$$