[m]cos2 α =cos^2 α -sin^2 α [/m]
[m]sin2 α =2sin α cos α [/m]
Тогда
[m]1-cos2 α+sin2 α =sin^2 α +cos^2 α -(cos^2 α -sin^2 α) +2sin α cos α =[/m]
[m]=2sin^2 α +2sin α cos α =2sin α (cos α +sin α )[/m]
По формулам приведения:
[m]sin(2π- α )=-sin α [/m]
[m]\frac{1+cos2 α -sin2 α }{1-sin(2π- α )}=\frac{2sin α (cos α +sin α )}{cos α -(-sin α )}=\frac{2sin α (cos α +sin α )}{cos α +sin α}=2sinα[/m]
При
[m]sin α=-\frac{\sqrt{3}}{2}[/m]
получаем ответ
[m]-\sqrt{3}[/m]