Ответ:

Объяснение:

[tex]\displaystyle\\cos\alpha =-0,6\ \ \ \ \pi < \alpha < \frac{3\pi }{2} \ \ \ \ \ sin(60^0-\alpha )=?\\\\sin60^0*cos\alpha -cos60^0*sin\alpha =sin(60^0-\alpha ).\ \ \ \ \Rightarrow\\\\ sin^2\alpha +cos^2\alpha =1 \\\\ sin^2\alpha =1-cos^2\alpha =1-(-0,6)^2=1-36=0,64.\\\\ sin\alpha =б\sqrt{0,64}=б0,8 .\\\\ \pi < \alpha < \frac{3\pi }{2} \ \ \ \ \Rightarrow\\\\ sin\alpha = -0,8.\\\\[/tex]

[tex]sin(60^0-\alpha )=sin60^0*(-0,6)-cos60^0*(-0,8)=\frac{\sqrt{3} }{2}*(-0,6)-\frac{1}{2}*(-0,8)=\\\\ =-0,3\sqrt{3} +0,4\approx-0,12.[/tex]