Cho \(\cos a = - \frac{1}{2}\) và \(\frac{\pi }{2} < \alpha < \pi \).Tính \(\cos \left( {a + \frac{\pi }{3}} \right)\).

Chọn A

Vì \(\frac{\pi }{2} < a < \pi \) nên \(\sin a > 0\). Ta có:

\[{\sin ^2}a + {\cos ^2}a = 1 \Leftrightarrow {\sin ^2}a = 1 - {\cos ^2}a \Leftrightarrow {\sin ^2}a = 1 - {\left( { - \frac{1}{2}} \right)^2} \Leftrightarrow {\sin ^2}a = \frac{3}{4} \Leftrightarrow \left[ \begin{array}{l}\sin a = \frac{{\sqrt 3 }}{2}\left( n \right)\\\sin a =  - \frac{{\sqrt 3 }}{2}\left( l \right)\end{array} \right.\]

Khi đó:  \(\cos \left( {a + \frac{\pi }{3}} \right) = \cos a.\cos \frac{\pi }{3} - \sin a.\sin \frac{\pi }{3} = \left( { - \frac{1}{2}} \right).\frac{1}{2} - \frac{{\sqrt 3 }}{2}.\frac{{\sqrt 3 }}{2} =  - 1\)