Question

вычислить sin 2a если sin a = 3/5 и п/2 <a<пcos a = -4/5 и п<a<3п/2 срочно мне нужны ответы помогите!!!

Answer 1

$a)sinalpha=frac35; fracpi2 textless alpha textless pi;\ b)cosalpha=-frac45; pi textless alpha textless frac{3pi}2;\ sin2alpha-?;\ a)sin2alpha=2cdotsinalphacdotcosalpha;\ fracpi2 textless alpha textless pi== textgreater sinalpha textgreater 0cupcosalpha textless 0== textgreater \ sinalpha=frac35,cosalpha=-sqrt{1-sin^2alpha}=-sqrt{1-left(frac{3}{5}right)^2}=-sqrt{1-frac{9}{25}}=\ =-sqrt{frac{25-9}{25}}=-sqrt{frac{16}{25}}=-frac45;\ sin2alpha=2cdotfrac35cdotleft(-frac45right)=-frac{24}{25};$
$a)sin2alpha=2cdotsinalphacdotcosalpha;\ pi textless alpha textless frac{3pi}{2}== textgreater sinalpha textless 0cupcosalpha textless 0== textgreater \ cosalpha=-frac45,sinalpha=-sqrt{1-cos^2alpha}=-sqrt{1-left(-frac{4}{5}right)^2}=-sqrt{1-frac{16}{25}}=\ =-sqrt{frac{25-16}{25}}=-sqrt{frac{9}{25}}=-frac35;\ sin2alpha=2cdotleft(-frac35right)cdotleft(-frac45right)=frac{24}{25};$