Question

`sqrt(2)cosA=cosB+cos^3B`, `sqrt(2)sinA=sinB-sin^3B` , find `sin(A-B)`

Answer 1

`sqrt2cos A = cosB + cos^3B`
`=>sqrt2cos A = cosB(1 + cos^2B)->(1)`
`sqrt2sinA = sinB - sin^3B`
`=>sqrt2sinA = sinB(1 - sin^2B)`
`=>sqrt2sinA = sinBcos^2B`
`=>cos^2B = sqrt2sinA /sinB`
Putting value of `cos^2B` in (1),
`=>sqrt2cos A = cosB(1 + (sqrt2sinA) /sinB) `
`=>sqrt2cosAsinB = sinBcosB+sqrt2sinAcosB`
`=>sqrt2(sinAcosB-cosAsinB) = -sinBcosB`
`=>sqrt2(sin(A-B)) = (-sin2B)/2`
`=>sin(A-B) = (-sin2B)/(2sqrt2).`