Đáp án: $\,\cos a = \dfrac{{\sqrt 5 }}{3};\sin \left( {a - \dfrac{\pi }{6}} \right) = \dfrac{{2\sqrt 3 - \sqrt 5 }}{6}$
Giải thích các bước giải:
$\begin{array}{l}
0 < a < \dfrac{\pi }{2} \Leftrightarrow 0 < \cos a < 1\\
Do:{\cos ^2}a + {\sin ^2}a = 1\\
\Leftrightarrow \cos a = \sqrt {1 - {{\sin }^2}a} = \sqrt {1 - \dfrac{4}{9}} = \dfrac{{\sqrt 5 }}{3}\\
\sin \left( {a - \dfrac{\pi }{6}} \right) = \sin a.\cos \dfrac{\pi }{6} - \cos a.\sin \dfrac{\pi }{6}\\
= \dfrac{2}{3}.\dfrac{{\sqrt 3 }}{2} - \dfrac{{\sqrt 5 }}{3}.\dfrac{1}{2}\\
= \dfrac{{2\sqrt 3 - \sqrt 5 }}{6}\\
Vậy\,\cos a = \dfrac{{\sqrt 5 }}{3};\sin \left( {a - \dfrac{\pi }{6}} \right) = \dfrac{{2\sqrt 3 - \sqrt 5 }}{6}
\end{array}$
