(i) sin2(α+β2)−sin2(α−β2)=sin2αsinβ\sin^2\left(\alpha + \dfrac{\beta}{2}\right) - \sin^2\left(\alpha - \dfrac{\beta}{2}\right) = \sin 2\alpha \sin\betasin2(α+2β)−sin2(α−2β)=sin2αsinβ