Biết sin 2alpha = -4/5, pi/2 < α < 3pi /2. Khi đó:

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

a) Đúng. Vì \(\frac{\pi }{2}\) < α <\(\frac{{3\pi }}{2}\) nên cosα < 0.

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

c) Đúng. Ta có hệ phương trình:\(\left\{ {\begin{array}{*{20}{c}}{{{\sin }^2}\alpha + {{\cos }^2}\alpha = 1}\\{2\sin \alpha \cos \alpha = \frac{{ - 4}}{5}}\end{array}} \right.\) ⇔ \(\left\{ {\begin{array}{*{20}{c}}{\frac{4}{{25{{\cos }^2}\alpha }} + {{\cos }^2}\alpha = 1}\\{\sin \alpha = \frac{{ - 2}}{{5\cos \alpha }}}\end{array}} \right.\)

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

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

d) Sai. cosα = \(\frac{{ - 1}}{{\sqrt 5 }}\), sinα = \(\frac{2}{{\sqrt 5 }}\).