If $${\cos ^{ - 1}}x - {\cos ^{ - 1}}{y \over 2} = \alpha ,$$ then $$4{x^2} - 4xy\cos \alpha + {y^2}$$ is equal to :

The trigonometric equation $${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$ has a solution for :

$${\cot ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) - {\tan ^{ - 1}}\left( {\sqrt {\cos \alpha } } \right) = x,$$ then sin x is equal to :