Question

√6 cos a если sin a=1/√3 -π/2 ≤ a ≤π/2

Answer 1

$Cos\alpha =\sqrt{1-Sin^{2}\alpha}=\sqrt{1-(\frac{1}{\sqrt{3}})^{2}}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}\\\\\sqrt{6}Cos\alpha=\sqrt{6}*\sqrt{\frac{2}{3}}=\sqrt{6*\frac{2}{3}}=\sqrt{4}=2\\\\Otvet:\boxed{2}$