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

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

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 \[\sin \left( {\alpha  + \beta } \right) = \sin \alpha .\cos \beta  + \cos \alpha .\sin \beta  = \frac{1}{3}.\left( { - \frac{2}{3}} \right) + \left( { - \frac{{2\sqrt 2 }}{3}} \right).\frac{{\sqrt 5 }}{3} =  - \;\frac{{2 + 2\sqrt {10} }}{9}\].

Vậy \[\sin \left( {\alpha  + \beta } \right) =  - \;\frac{{2 + 2\sqrt {10} }}{9}\].