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

A

- Hướng dẫn giải

Ta thấy \(\left\{ \begin{array}{l}
\cos \alpha  = \sqrt {1 - {{\sin }^2}\alpha }  = \sqrt {1 - \frac{1}{3}} \\
0 < \cos \alpha  < 1\left( {do\,\,0 < \alpha  < \frac{\pi }{2}} \right)
\end{array} \right.\)

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