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]