[m]2sin^2\frac{\alpha }{2}=1-cos\alpha[/m], то
[m]\sqrt{\frac{1}{2}-\frac{cos\alpha }{2}}=|sin\frac{\alpha }{2}|[/m]
2.
так как
(1/2)-(1/2)sin α =(1/2)*(cos^2( α /2)-2sin( α /2)*cos( α /2)+sin^2( α /2)=
=(1/2)(cos( α /2)-sin( α /2))^2, то
sqrt((1/2)-(1/2)sin α)=sqrt(1/2)*|cos( α /2)-sin( α /2)|
3)
=|cos2 α |
4)
=(1/2)*|sin3 α |