Formula fundamentala a trigonometriei:[tex]sin^2x+cos^2x=1[/tex]Stim ca[tex]\cos(a+b)=\cos a \cos b-\sin a \sin b[/tex]Deci[tex]cos2x=cos^2x-sin^2x=1-2sin^2x[/tex]Particularizand pentru problema noastra avem[tex]cos2x=1-2\cdot ( \frac{ \sqrt{3} }{5} )^2=1-2\cdot \frac{3}{25} =1- \frac{6}{25}= \frac{19}{25} [/tex]