Question

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

Answer 1

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