Найти cos 2α, если tg α = 3/4 π меньше α меньше 3π/2

[latex]tg\alpha=\frac34\\1+tg^2\alpha=\frac1{\cos^2\alpha}\\\cos^2\alpha=\frac1{1+tg^2\alpha}\\\cos\alpha=\pm\sqrt{\frac1{1+tg^2\alpha}}=\pm\sqrt{\frac1{1+\frac9{16}}}=\pm\sqrt{\frac1{\frac{25}{16}}}=\pm\sqrt{\frac{16}{25}}=\pm\frac45\\\pi\ \textless \ \alpha\ \textless \ \frac{3\pi}2\Rightarrow \cos\alpha\ \textless \ 0\\\cos\alpha=-\frac45\\\cos2\alpha=2\cos^2\alpha-1=2\cdot\frac{16}{25}-1=\frac{32}{25}-1=\frac7{25}[/latex]