Question

știind ca tg x =3 sa se calculeze sin 2x RAPID VĂ ROG

Answer 1

Salut,

$sin(2x)=\dfrac{sin(2x)}{1}=\dfrac{sin(2x)}{sin^2x+cos^2x}=\dfrac{2\cdot sinx\cdot cosx}{sin^2x+cos^2x}=\\\\=\dfrac{\dfrac{2\cdot sinx\cdot cosx}{cos^2x}}{\dfrac{sin^2x+cos^2x}{cos^2x}}=\dfrac{2tgx}{tg^2x+1}=\dfrac{2\cdot 3}{3^2+1}=\dfrac{6}{10}=\dfrac{3}{5}.$

Green eyes.