[tex]\it b\in \Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow sinb=\sqrt{1-cos^2b} =\sqrt{1-\dfrac{1}{9}}=\sqrt{\dfrac{8}{9}}=\dfrac{2\sqrt2}{3}\\ \\ \\ a\in \Big(\dfrac{\pi}{2},\ \pi\Big) \Rightarrow cosa=-\sqrt{1-sin^2a} =-\sqrt{1-\dfrac{4}{9}}=-\sqrt{\dfrac{5}{9}}=-\dfrac{\sqrt5}{3}[/tex]
[tex]\it sin(a+b)=sinacosb+sinbcosa=\dfrac{2}{3}\cdot(-\dfrac{1}{3})+\dfrac{2\sqrt2}{3}\cdot(-\dfrac{\sqrt5}{3})=-\dfrac{2}{9}(1+\sqrt{10})[/tex]
