Cho các góc α , β thỏa mãn pi/ 2 < α , β < π , sin α = 1/3 , cos β = − 2/3 . Tính sin ( α + β ) .

Do \(\frac{\pi }{2} < \alpha ,\beta  < \pi  \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\cos \alpha  < 0}\\{\sin \beta  > 0}\end{array}} \right.\).

Ta có \(\cos \alpha  =  - \sqrt {1 - {{\sin }^2}\alpha }  =  - \sqrt {1 - \frac{1}{9}}  =  - \frac{{2\sqrt 2 }}{3};\sin \beta  = \sqrt {1 - {{\cos }^2}\beta }  = \sqrt {1 - \frac{4}{9}}  = \frac{{\sqrt 5 }}{3}\).

Suy ra \(\sin (\alpha  + \beta ) = \sin \alpha  \cdot \cos \beta  + \cos \alpha  \cdot \sin \beta  = \frac{1}{3} \cdot \left( { - \frac{2}{3}} \right) + \left( { - \frac{{2\sqrt 2 }}{3}} \right) \cdot \frac{{\sqrt 5 }}{3} =  - \frac{{2 + 2\sqrt {10} }}{9}\).

Vậy \(\sin (\alpha  + \beta ) =  - \frac{{2 + 2\sqrt {10} }}{9}\).