Question

Найдите sina, sin2a, если tga = -3, pi/2 < a < pi

Answer 1

$tg^2a+1= \frac{1}{cos^2a} \\ \frac{1}{cos^2a} = (-3)^2+1 ; \frac{1}{cos^2a} = 10; cos^2a= \frac{1}{10}$
т.к. a∈(0.5π; π), то:
$cosa=- \frac{1}{ \sqrt{10}} = - \frac{ \sqrt{10}}{10}=-0.1 \sqrt{10} \\ sina=tga*cosa=-3*- \frac{ \sqrt{10} }{10} =0.3 \sqrt{10} \\ sin2a=2*sina*cosa=-2*0.1 \sqrt{10} *0.3 \sqrt{10} =-0.06 \sqrt{10}$