(24*cos 15)/корень 6 +корень 2

Решите задачу:

$cos^2a= \frac{1+cos2a}{2}\quad \Rightarrow \quad cos^215^\circ = \frac{1+cos30^\circ }{2}= \frac{1+\frac{\sqrt3}{2}}{2}= \frac{2+\sqrt3}{4} \\\\cos15^\circ = \sqrt{\frac{2+\sqrt3}{4}}=\frac{\sqrt{2+\sqrt3} }{2} = \frac{\sqrt{4+2\sqrt3}}{2\sqrt2}= \frac{\sqrt{(1+\sqrt3)^2}}{2\sqrt2}= \frac{1+\sqrt3}{2\sqrt2} \\\\\\ \frac{24cos15^\circ }{\sqrt6+\sqrt2}= \frac{24\, (1+\sqrt{3})}{2\sqrt2\, (\sqrt6+\sqrt2)} = \frac{12\, (1+\sqrt3)}{\sqrt2\cdot \sqrt2\, (\sqrt3+1)} = \frac{12}{2}=6$