Cho cos x = 4/5 , x ∈ ( − pi/ 2 ; 0 ) . Tính sin 2x .

Chọn B

Vì \[{\rm{ }}x \in \left( { - \frac{\pi }{2};0} \right)\] nên \(\sin x < 0\). Ta có:

\[{\sin ^2}x + {\cos ^2}x = 1 \Leftrightarrow {\sin ^2}x = 1 - {\cos ^2}x \Leftrightarrow {\sin ^2}x = 1 - {\left( {\frac{4}{5}} \right)^2} \Leftrightarrow {\sin ^2}x = \frac{9}{{25}} \Leftrightarrow \left[ \begin{array}{l}\sin x =  - \frac{3}{5}\left( n \right)\\\sin x = \frac{3}{5}\left( l \right)\end{array} \right.\]

Khi đó:  \(\sin 2x = 2\sin x\cos x = 2.\left( {\frac{{ - 3}}{5}} \right).\frac{4}{5} =  - \frac{{24}}{{25}}\)