If `sin(theta+alpha)=aa n dsin(theta+beta)=b ,` prove that `cos2(alpha-beta)-4a bcos(alpha-beta)=1-2a^2-2b^2`

Explanation

Given,

`sin(theta+alpha)=a->(i)`

and `sin(theta+beta)=b->(ii)`

thus,
`cos(theta+alpha)=sqrt(1-sin^2(theta+alpha))`

`=>sqrt(1-a^2)`

so `cos(theta+alpha)=sqrt(1-a^2)`
and
`cos(theta+beta)=sqrt(1-sin^2(theta+beta))`

...