[tex]sin \;60^o= \frac{ \sqrt{3} }{2} \\ \\ cos\;45^o= \frac{ \sqrt{2} }{2} \\ \\ 0,5= \frac{1}{2} \\ \\ Rezolvare: \\ \\ \frac{ \sqrt{2}}{3}\;sin\;60^o = \frac{ \sqrt{2}}{3}* \frac{ \sqrt{3} }{2}= \frac{ \sqrt{2}*\sqrt{3} }{2*3}= \boxed{\frac{\sqrt{6} }{6}} \\ \\ 0,5\;cos\;45^o = \frac{1}{2}* \frac{ \sqrt{2} }{2}= \frac{1* \sqrt{2} }{2*2}= \boxed{ \frac{\sqrt{2} }{4}}[/tex]
