Question

Помогите пожалуйста решить, подробно

Answer 1

$\displaystyle\bf\\Cos^{2} \Big(5\pi +\alpha \Big)+Ctg\alpha \cdot Sin2\alpha =Cos^{2}\Big[4\pi +(\pi +\alpha )\Big] +\frac{Cos\alpha }{Sin\alpha } \cdot 2Sin\alpha Cos\alpha =\\\\\\=Cos^{2}\Big(\pi +\alpha \Big)+2Cos^{2} \alpha =Cos^{2} \alpha +2Cos^{2} \alpha =3Cos^{2} \alpha \\\\\\\alpha =\frac{3\pi }{4} \\\\\\3Cos^{2} \alpha =3Cos^{2} \frac{3\pi }{4} =3Cos^{2} \Big(\pi -\frac{\pi }{4} \Big)=3Cos^{2}\frac{\pi }{4} =3\cdot\Big(\frac{1}{\sqrt{2} } \Big)^{2} =3\cdot\frac{1}{2}=1,5$