Đáp án:
\(\begin{array}{l}
\cos a = \dfrac{{2\sqrt 2 }}{3}\\
\sin \left( {a - \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 - 4}}{6}\\
\cos \left( {a - \dfrac{\pi }{6}} \right) = \dfrac{{2\sqrt 6 + 1}}{6}
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
0 < a < \dfrac{\pi }{2} \Rightarrow \cos a > 0\\
{\sin ^2}a + {\cos ^2}a = 1\\
\Leftrightarrow {\left( {\dfrac{1}{3}} \right)^2} + {\cos ^2}a = 1\\
\Leftrightarrow \dfrac{1}{9} + {\cos ^2}a = 1\\
\Leftrightarrow {\cos ^2}a = \dfrac{8}{9}\\
\cos a > 0 \Rightarrow \cos a = \dfrac{{2\sqrt 2 }}{3}\\
\sin \left( {a - \dfrac{\pi }{4}} \right) = \sin a.\cos \dfrac{\pi }{4} - \cos a.\sin \dfrac{\pi }{4}\\
= \dfrac{1}{3}.\dfrac{{\sqrt 2 }}{2} - \dfrac{{2\sqrt 2 }}{3}.\dfrac{{\sqrt 2 }}{2} = \dfrac{{\sqrt 2 - 4}}{6}\\
\cos \left( {a - \dfrac{\pi }{6}} \right) = \cos a.\cos \dfrac{\pi }{6} + \sin a.\sin \dfrac{\pi }{6}\\
= \dfrac{{2\sqrt 2 }}{3}.\dfrac{{\sqrt 3 }}{2} + \dfrac{1}{3}.\dfrac{1}{2} = \dfrac{{2\sqrt 6 + 1}}{6}
\end{array}\)
