Question

помогите решить хоть что-то​

Answer 1

$4)Cos^{2}120^{0}-Cos^{2}135^{0}=Cos^{2}(90^{0}+30^{0})-Cos^{2}(90^{0}+45^{0})=Sin^{2}30^{0}-Sin^{2}45^{0}=(\frac{1}{2})^{2}-(\frac{1}{\sqrt{2}})^{2}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}=-025\\\\Otvet:\boxed{-0,25}$

$Sin^{2}150^{0}-Sin^{2}120^{0}=Sin^{2}(180^{0}-30^{0})-Sin^{2}(90^{0}+30^{0})=Sin^{2}30^{0}-Cos^{2}30^{0}=(\frac{1}{2})^{2}-(\frac{\sqrt{3}}{2})^{2}=\frac{1}{4}-\frac{3}{4}=-\frac{1}{2}=-0,5\\\\Otvet:\boxed{-0,5}$

$5)Sin^{2}(180^{0}-\alpha)-1+2Cos^{2}\alpha = (Sin^{2}\alpha-1)+2Cos\alpha^{2}\alpha=-Cos^{2}\alpha+2Cos^{2}\alpha=Cos^{2}\alpha\\\\Otvet:\boxed{Cos^{2}\alpha }$

$2Sin^{2}\alpha+Cos^{2}(180^{0}-\alpha)-1=2Sin^{2}\alpha+(Cos^{2}\alpha-1)=2Sin^{2}\alpha-Sin^{2}\alpha=Sin^{2} \alpha\\\\Otvet:\boxed{Sin^{2}\alpha}$