Question

помогите пожалуйста!​

Answer 1

$\displaystyle\bf\\1)\\\\Cos^{2} 3\alpha -Sin^{2} 3\alpha =Cos(3\alpha \cdot 2)=Cos6\alpha \\\\2)\\\\0 < \frac{\alpha }{2} < 90^\circ \ \ \Rightarrow \ \ Cos\frac{\alpha }{2} > 0\\\\\\Cos\frac{\alpha }{2} =\sqrt{1-Sin^2\frac{\alpha }{2} } =\sqrt{1-\Big(\frac{4}{5}\Big)^{2} } =\sqrt{1-\frac{16}{25} } =\sqrt{\frac{9}{25} } =\frac{3}{5}\\\\\\Sin\alpha =2Sin\frac{\alpha }{2} Cos\frac{\alpha }{2} =2\cdot \frac{4}{5} \cdot\frac{3}{5} =\frac{24}{25} =0,96\\\\3)$

$\displaystyle\bf\\\frac{Cos\alpha Sin\alpha }{Cos^{2} \alpha -Sin^{2} \alpha } =\frac{Cos\alpha Sin\alpha }{Cos2\alpha } =\frac{2\cdot Cos\alpha Sin\alpha }{2\cdot Cos2\alpha } =\\\\\\=\frac{Sin2\alpha }{2Cos2\alpha } =0,5 tg2\alpha$