Răspuns:
Explicație pas cu pas:
[tex]sinx=\frac{12}{13},~din~sin^{2}x+cos^{2}x=1,~=>~(\frac{12}{13})^{2}+cos^{2}x=1, ~=>~cos^{2}x=1-(\frac{12}{13})^{2},~=>~cos^{2}x=\frac{169-144}{169},~=>~cos^{2}x=\frac{25}{169},~=>~cosx=-\frac{5}{13},~deoarece~x~apartine~ cadr.II.\\ctgx=\frac{cosx}{sinx}=-\frac{5}{13}: \frac{12}{13}=-\frac{5}{12}.\\sin2x=2*sinx*cosx=2*\frac{12}{13}*(-\frac{5}{13})=-\frac{120}{169}\\cos2x=cos^{2}x-sin^{2}x=(-\frac{5}{13})^{2}-(\frac{12}{13})^{2}=\frac{25}{169}-\frac{144}{169}=-\frac{119}{169}.[/tex]