11π3π. 6) cos + cos 12 4

6) Используем формулу суммы косинусов:$$cos(x) + cos(y) = 2cos(\frac{x+y}{2})cos(\frac{x-y}{2})$$

В нашем случае $$x = \frac{11\pi}{12}$$ и $$y = \frac{3\pi}{4}$$.

$$cos(\frac{11\pi}{12}) + cos(\frac{3\pi}{4}) = 2cos(\frac{\frac{11\pi}{12} + \frac{3\pi}{4}}{2})cos(\frac{\frac{11\pi}{12} - \frac{3\pi}{4}}{2}) = 2cos(\frac{\frac{20\pi}{12}}{2})cos(\frac{\frac{2\pi}{12}}{2}) = 2cos(\frac{5\pi}{6})cos(\frac{\pi}{12})$$.

Поскольку $$cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$$, то получаем:

$$2cos(\frac{5\pi}{6})cos(\frac{\pi}{12}) = 2 \cdot (-\frac{\sqrt{3}}{2}) \cdot cos(\frac{\pi}{12}) = -\sqrt{3}cos(\frac{\pi}{12})$$.

Ответ: $$- \sqrt{3}cos(\frac{\pi}{12})$$.