Question

(24*cos 15)/корень 6 +корень 2

Answer 1

$cos^2a= frac{1+cos2a}{2}quad Rightarrow quad cos^215^circ = frac{1+cos30^circ }{2}= frac{1+frac{sqrt3}{2}}{2}= frac{2+sqrt3}{4} \\cos15^circ = sqrt{frac{2+sqrt3}{4}}=frac{sqrt{2+sqrt3} }{2} = frac{sqrt{4+2sqrt3}}{2sqrt2}= frac{sqrt{(1+sqrt3)^2}}{2sqrt2}= frac{1+sqrt3}{2sqrt2} \\\ frac{24cos15^circ }{sqrt6+sqrt2}= frac{24, (1+sqrt{3})}{2sqrt2, (sqrt6+sqrt2)} = frac{12, (1+sqrt3)}{sqrt2cdot sqrt2, (sqrt3+1)} = frac{12}{2}=6$