Question

Найдите: cos α , если tg α = -4/3 , π/2 < α < π

Answer 1

π/2<α<π(2 четверть)⇒cosα<0
$1+tg^2\alpha=\frac{1}{cos^2\alpha} \\ \\ 1+\frac{16}{9}=\frac{1}{cos^2\alpha} \\ \\ \frac{25}{9}=\frac{1}{cos^2\alpha} \\ \\ 25cos^2\alpha=9 \\ \\ cos^2\alpha=\frac{9}{25} \\ \\ cos\alpha=-\frac{3}{5}=-0,6$