Cho \(\sin \alpha  = \frac{1}{{\sqrt 3 }}\) với \(...

A

- Hướng dẫn giải

\(\begin{array}{l}
\cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \frac{{\sqrt 6 }}{3}\left( {0 < \alpha  < \frac{\pi }{2}} \right)\\
 \Rightarrow \cos \left( {\alpha  + \frac{\pi }{3}} \right) = \cos \alpha .\cos \frac{\pi }{3} - \sin \alpha .\sin \frac{\pi }{3}\\
 = \frac{{\sqrt 6 }}{3}.\frac{1}{2} - \frac{1}{{\sqrt 3 }}.\frac{{\sqrt 3 }}{2} = \frac{1}{{\sqrt 6 }} - \frac{1}{2}
\end{array}\)