Cho sin α = 2 √ 2/ 3 với pi/2 < α < pi . Tính sin ( α − pi/ 4 ) .

Tính \(cos\alpha \) .Ta có \[{\sin ^2}\alpha  + co{s^2}\alpha  = 1 \Rightarrow co{s^2}\alpha  = 1 - {\sin ^2}\alpha  = 1 - \frac{8}{9} = \frac{1}{9}\]

\( \Rightarrow cos\alpha  =  \pm \frac{1}{3}\) . Vì \(\frac{\pi }{2} < \alpha  < \pi  \Rightarrow cos\alpha  =  - \frac{1}{3}\)

Ta có \(\sin \left( {\alpha- \frac{\pi }{4}} \right) = \sin \alpha \cos \frac{\pi }{4} - \cos \alpha \sin \frac{\pi }{4}\)=\( = \frac{{2\sqrt 2 }}{3}.\frac{{\sqrt 2 }}{2} - \frac{{ - 1}}{3}.\frac{{\sqrt 2 }}{2} = \,\frac{{4 + \sqrt 2 }}{6}\)