Verified answer

Объяснение:

[tex]sin\alpha =-\frac{3}{5} \ \ \ \ \ \frac{3\pi }{2} < \alpha < 2\pi \ \ \ \ \ -\frac{25sin2\alpha }{4} =?\\sin^2\alpha +cos^2\alpha =1\\cos^2\alpha =1-sin^2\alpha =1-(-\frac{3}{5})^2=1-\frac{9}{25} =\frac{16}{25} .\\ cos\alpha =б\sqrt{\frac{16}{25} } =б\frac{4}{5} .\ \ \ \ \ \frac{3\pi }{2} < \alpha < 2\pi \ \ \ \ \ \Rightarrow\\cos\alpha =\frac{4}{5}.\\ -\frac{25sin2\alpha }{4} =-\frac{25*2*sin\alpha *cos\alpha }{4} =-\frac{25*2*(-\frac{3}{5})*\frac{4}{5} }{4}= \frac{5*5*2*3*4}{4*5*5} =2*3=6.[/tex]

Ответ: 6.