$\cos\;\dfrac{\pi}{12}$ $\dfrac{\sqrt{2}+\sqrt{6}}{4}$ $\dfrac{\sqrt{2}-\sqrt{6}}{4}$ $\dfrac{\sqrt{6}-\sqrt{2}}{4}$

Задание

Вычисли.

\(\cos\;\dfrac{\pi}{12}\)

Выбери верный вариант.

\(\dfrac{\sqrt{2}+\sqrt{6}}{4}\)
\(\dfrac{\sqrt{2}-\sqrt{6}}{4}\)
\(\dfrac{\sqrt{6}-\sqrt{2}}{4}\)