Cho \(\sin \alpha = {1 \over {\sqrt 3 }},\,\,\left( {0 < \alpha < {\pi \over 2}} \right)\,\). Tính \(\cos \left( {\alpha + {\pi \over 3}} \right) = ?\)
Giải chi tiết:

Ta có:
\({\sin ^2}\alpha + {\cos ^2}\alpha = 1 \Leftrightarrow {\left( {{1 \over {\sqrt 3 }}} \right)^2} + {\cos ^2}\alpha = 1 \Leftrightarrow {\cos ^2}\alpha = {2 \over 3} \Leftrightarrow \left[ \matrix{ \cos \alpha = \sqrt {{2 \over 3}} \hfill \cr \cos \alpha = - \sqrt {{2 \over 3}} \hfill \cr} \right.\)
Vì \(0 < \alpha < {\pi \over 2} \Rightarrow \cos \alpha > 0 \Rightarrow \cos \alpha = \sqrt {{2 \over 3}} \)
\(\cos \left( {\alpha + {\pi \over 3}} \right) = \cos \alpha \cos {\pi \over 3} - \sin \alpha \sin {\pi \over 3} = \sqrt {{2 \over 3}} .{1 \over 2} - {1 \over {\sqrt 3 }}.{{\sqrt 3 } \over 2} = {{\sqrt 6 - 3} \over 6}\)
Chọn: C.