Biết sin 2 α = − 4/5 , pi/ 2 < α < 3 pi/ 2 . Khi đó: a) cos α < 0

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

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

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