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 that, ` sin(theta+alpha)=a`
and `" "sin(theta+beta)=b" "...(i)`
`therefore" "cos(theta+alpha)=sqrt(1-a^(2)) and cos(theta+beta)=sqrt(1-b^(2))" "...(ii)`
`therefore" "cos(alpha-beta)=cos{theta+alpha-(theta+beta)}`
`" "=cos(theta+beta)cos(theta+alpha)+sin(theta+alpha)sin(theta+beta)`
`" "=sqrt(1-a^(2))sqrt(1-b^(2))+a*b=ab+sqrt((1-a)^(2)(1-b)^(2))`
` " "=ab+sqrt(1-a^(2)-b^(2)+a^(2)b^(2))`
and `" "cos(alpha-beta)=ab+sqrt(1-a^(2)-b^(2)+a^(2)b^(2))`
`" "=cos2(alpha-beta)-4abcos(alpha-beta)`
`" "=2cos^(2)(alpha-beta)-1-4abcos(alpha-beta)`
`" "= 2cos(alpha-beta)(cosalpha-beta-2ab)-1`
`" "=2(ab+sqrt(1-a^(2)-b^(2) +a^(2)b^(2)))(ab+sqrt(1-a^(2)-b^(2)+a^(2)b^(2))-2ab)-1`
`" "=2[(sqrt(1-a^(2)-b^(2)+a^(2)b^(2) +ab))(sqrt(1-a^(2)-b^(2)+a^(2)b ^(2))-ab)]-1`
`" "=2[1-a^(2)-b^(2)+a^(2)b^(2)-a^(2)b^(2)]-1`
`" "=2-2a^(2)-2b^(2)-1`
`" "=1-2a^(2)-2b^(2)" "` Hence proved.