Question

Решите задание по снимку

Answer 1

Объяснение:

$sin\alpha =\frac{2\sqrt{2} }{3} \ \ \ \ \alpha \in (90^0;180^0)\\cos^260^0*cos\alpha -cos^230^0*cos(2\alpha)=(\frac{1}{2})^2*cos\alpha-(\frac{\sqrt{3} }{2})^2*c0s(2\alpha )=\\=\frac{cos\alpha }{4} -\frac{3*(cos^2\alpha -sin^2\alpha )}{4} =\frac{cos\alpha -3*cos^2\alpha +3*sin^2\alpha }{4}.\\ sin\alpha =\frac{2\sqrt{2} }{3}=\frac{\sqrt{8} }{3}\ \ \ \ sin^2\alpha =(\frac{\sqrt{8} }{3})^2=\frac{8}{9}\ \ \ \ cos^2\alpha =1-sin^2\alpha =1-\frac{8}{9} =\frac{1}{9}$

$cos\alpha =б\sqrt{\frac{1}{9} }=б\frac{1}{3}\ \ \alpha \in(90^0;180^0) \ \ \ \ \Rightarrow\ \ \ \ cos\alpha =-\frac{1}{3}.\ \ \ \ \Rightarrow$

$\frac{cos\alpha -3*cos^2\alpha +3*sin^2\alpha }{4}=\frac{-\frac{1}{3}-\frac{3*1}{9}+\frac{3*8}{9} }{4}=\frac{-\frac{1}{3}-\frac{1}{3} +\frac{8}{3} }{4} =\frac{\frac{6}{3} }{4}=\frac{2}{4}=\frac{1}{2} =0,5.$