10√2 cos (−п/12) − 5√3 .

Решите задачу:

$cos\frac{\pi}{12}=cos\, 15^\circ =\sqrt{ \frac{1+cos30^\circ }{2} }=\sqrt{ \frac{1+\frac{\sqrt3}{2}}{2}} = \sqrt{ \frac{2+\sqrt3}{4}} = \frac{ \sqrt{2+\sqrt3} }{2} \\\\\\10\sqrt2\cdot cos(-\frac{\pi}{12})-5\sqrt3=10\sqrt2\cdot cos \frac{\pi}{12} -5\sqrt3=\\\\=10\sqrt2\cdot \frac{\sqrt{2+\sqrt3}}{2} -5\sqrt3=5\sqrt2\cdot \sqrt{2+\sqrt3}-5\sqrt3=\\\\=5\cdot \sqrt{2\cdot (2+\sqrt3)}-5\sqrt3=5\cdot \sqrt{4+2\sqrt3}-5\sqrt3=\\\\=5\cdot \sqrt{(1+\sqrt3)^2}-5\sqrt3=5\cdot (1+\sqrt3)-5\sqrt3=5+5\sqrt3-5\sqrt3=5$

$P.S.\; \; cos\frac{\pi}{12}= \frac{\sqrt{2+\sqrt3}}{2} = \frac{\sqrt2\cdot \sqrt{2+\sqrt3}}{2\sqrt2}=\frac{\sqrt{2\cdot (2+\sqrt3)}}{2\sqrt2} = \\\\=\frac{\sqrt{4+2\sqrt3}}{2\sqrt2}=\frac{\sqrt{(\sqrt3+1)^2}}{2\sqrt2} = \frac{\sqrt3+1}{2\sqrt2}$