Question

Обчислити sin2a i tg2a якщо sina=0.6 і п/2<a<п​

Answer 1

$sin(2\alpha) =-0,96\\\\tg(2\alpha) = -\frac{24}{7}$

Пошаговое объяснение:

$sin(\alpha) = 0,6\ \ \ \alpha \in (\frac{\pi}{2}; \pi ) \ \Rightarrow \ cos(\alpha) < 0\\\\cos(\alpha) = -\sqrt{1-sin^2(\alpha)} = -\sqrt{1-0,36} = - \sqrt{0,64} = -0,8\\\\sin(2\alpha) = 2sin(\alpha)cos(\alpha) = 2\cdot0,6\cdot(-0,8)=-0,96\\\\cos(2\alpha) = cos^2(\alpha) - sin^2(\alpha)=0,64-0,36 = 0,28\\\\tg(2\alpha) = \frac{sin(\alpha)}{cos(2\alpha)} = \frac{-0,96}{0,28} = -\frac{96}{28} = -\frac{24}{7}$