Đáp ánTa có:\(\left. \matrix{
a = 2\cos {{\alpha + \beta } \over 2}\cos {{\alpha - \beta } \over 2} \hfill \cr
b = 2\sin {{\alpha + \beta } \over 2}\sin {{\alpha - \beta } \over 2} \hfill \cr} \right\} \)\(\Rightarrow ab = 2\sin (\alpha + \beta )co{s^2}{{\alpha - \beta } \over 2}\) Mặt khác: \({a^2} + {b^2} = 4{\cos ^2}{{\alpha - \beta } \over 2}\)Do đó: \(\sin (\alpha + \beta ) = {{2ab} \over {{a^2} + {b^2}}}\)
