Biết \(\sin 2\alpha  =  - \frac{4}{5},\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{2}\).

a) \(\cos \alpha  < 0\).

b) \(2\sin \alpha \cos \alpha  =  - \frac{4}{5}\).

c) \(\cos \alpha  = \frac{{ - 2}}{{\sqrt 5 }},\sin \alpha  = \frac{1}{{\sqrt 5 }}\).

d) \(\cos \alpha  = \frac{{ - 1}}{{\sqrt 5 }},\sin \alpha  =  - \frac{2}{{\sqrt 5 }}\).

Vì \(\frac{\pi }{2} < \alpha  < \frac{{3\pi }}{2}\) nên \(\cos \alpha  < 0\). Ta có hệ: \(\left\{ {\begin{array}{*{20}{l}}{{{\sin }^2}\alpha  + {{\cos }^2}\alpha  = 1}\\{2\sin \alpha \cos \alpha  =  - \frac{4}{5}}\end{array}} \right.\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{4}{{25 {{\cos }^2}\alpha }} +  {{\cos }^2}\alpha  = 1}\\{ \sin \alpha  =  - \frac{2}{{5 \cos  \alpha }}}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{25{{\cos }^4}\alpha  - 25{{\cos }^2}\alpha  + 4 = 0}\\{\sin \alpha  =  - \frac{2}{{5\cos \alpha }}}\end{array}} \right.} \right.\)

\( \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\cos { ^2}\alpha  = \frac{4}{5}\\\cos { ^2}\alpha  = \frac{1}{5}\end{array} \right.\\\sin  \alpha  =  - \frac{2}{{5 \cos  \alpha }}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}\cos  \alpha  = \frac{{ - 2}}{{\sqrt 5 }}\\\cos  \alpha  = \frac{{ - 1}}{{\sqrt 5 }}\end{array} \right.\\\sin  \alpha  =  - \frac{2}{{5 \cos  \alpha }}\end{array} \right. \Rightarrow \left[ {\begin{array}{*{20}{l}}{\cos \alpha  = \frac{{ - 2}}{{\sqrt 5 }},\sin \alpha  = \frac{1}{{\sqrt 5 }}}\\{\cos \alpha  = \frac{{ - 1}}{{\sqrt 5 }},\sin \alpha  = \frac{2}{{\sqrt 5 }}}\end{array}} \right.\)

Đáp án:       a) Đúng,      b) Đúng,      c) Đúng,      d) Sai.