[tex]Aplicam formula: \\ sin^2x + cos^2x=1 \\ =\ \textgreater \ sin\,x = \sqrt{1-cos^2x} \\ =\ \textgreater \ cos\,x = \sqrt{1-sin^2x} \\ \\ a) \\ cos\,x= \frac{4}{5} \;\;\;=\ \textgreater \ \;\;\;sin\,x=\sqrt{1-( \frac{4}{5} )^2} =\sqrt{1- \frac{16}{25} }= \sqrt{\frac{9}{25} }= \boxed{\frac{3}{5}} \\ \\ b) \\ sin\,x= \frac{1}{3} \;\;\;=\ \textgreater \ \;\;\;cos\,x=\sqrt{1-( \frac{1}{3} )^2} =\sqrt{1- \frac{1}{9} }= \sqrt{\frac{8}{9} }= \boxed{\frac{2 \sqrt{2} }{3}} [/tex]
