Question

Срочно! Помогите пожалуйста!! 40 баллов!!!!!

Answer 1

$\displaystyle\bf\\\alpha \in\Big(\pi \ ; \ \frac{3\pi }{2} \Big) \ \ \ \Rightarrow \ \ \ Cos\alpha < 0 \ \ , \ Sin\alpha < 0\\\\\\1)\\\\1+tg^{2} =\frac{1}{Cos^{2} \alpha } \\\\\\Cos^{2} \alpha =\frac{1}{1+tg^{2} \alpha } =\frac{1}{1+(\frac{3}{4} )^{2} } =\frac{1}{1+\frac{9}{16} } =\frac{1}{\frac{25}{16} } =\frac{16}{25} \\\\\\Cos\alpha =-\sqrt{\frac{16}{25} } =-\frac{4}{5} =-0,8\\\\\\2)\\\\Sin2\alpha =2Sin\alpha Cos\alpha$

$\displaystyle\bf\\Sin\alpha =-\sqrt{1-Cos^{2}\alpha } =-\sqrt{1-(-0,8)^{2} } =-\sqrt{1-0,64} =\\\\\\=-\sqrt{0,36}=-0,6\\\\\\Sin2\alpha =2\cdot (-0,8)\cdot(-0,6)=0,96$