Найдите cos2x , если tgx=- 3/4

[latex]tg x=-frac{3}{4};\ cos2x-?\ cos2x=cos^2x-sin^2x=frac{cos^2x-sin^2x}{1}=frac{cos^2x-sin^2x}{cos^2x+sin^2x}=\ =frac{frac{cos^2x}{cos^2x}-frac{sin^2x}{cos^2x}}{frac{cos^2x}{cos^2x}+frac{sin^2x}{cos^2x}}=frac{1-tg^2x}{1+tg^2x}=\ =frac{1-(-frac{3}{4})^2}{1+(-frac{3}{4})^2}=frac{1-frac{9}{16}}{1+frac{9}{16}}=frac{frac{16-9}{16}}{frac{16+9}{16}}=frac{frac{7}{16}}{frac{25}{16}}=frac{7}{25}=frac{28}{100}=0.28[/latex]