Tính giá trị cos alpha - pi/6

Vì ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$ nên ${\left( {\frac{1}{3}} \right)^2} + {\cos ^2}\alpha = 1⇔cos2α<spanstyle="mso-spacerun:yes;"> </span>=89 \Leftrightarrow {\cos ^2}\alpha  = \frac{8}{9} \Leftrightarrow \cos \alpha = \pm \frac{{2\sqrt 2 }}{3}$.

Mà $\frac{\pi }{2} < \alpha < \pi $ nên $\cos \alpha < 0$. Do đó $\cos \alpha = - \frac{{2\sqrt 2 }}{3}$.

Ta có $\cos \left( {\alpha - \frac{\pi }{6}} \right) = \cos \alpha \cos \frac{\pi }{6} + \sin \alpha \sin \frac{\pi }{6}$

                               $ = \left( { - \frac{{2\sqrt 2 }}{3}} \right) \cdot \frac{{\sqrt 3 }}{2} + \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} - \frac{{\sqrt 6 }}{3}$.