Cho giá trị của sin x = 1/5, pi/2 < x < pi

Lời giải

a) Sai                                b) Đúng                           c) Sai                               d) Sai

\(\begin{array}{l}\frac{\pi }{2} < x < \pi  \Rightarrow \frac{{2\pi }}{2} < 2x < 2\pi  \Rightarrow \pi  < 2x < 2\pi  \Rightarrow \sin 2x < 0.\\\frac{\pi }{2} < x < \pi  \Rightarrow \cos x < 0\\\sin x = \frac{1}{5} \Rightarrow \cos x =  - \sqrt {1 - {{\sin }^2}x}  =  - \frac{{2\sqrt 6 }}{5}.\\\sin 2x = 2\sin x \cdot \cos x = 2 \cdot \frac{1}{5} \cdot \left( { - \frac{{2\sqrt 6 }}{5}} \right) =  - \frac{{4\sqrt 6 }}{5}.\\\frac{\pi }{2} < x < \pi  \Rightarrow \frac{{2\pi }}{2} < 2x < 2\pi  \Rightarrow \pi  < 2x < 2\pi  \Rightarrow \cos 2x > 0.\\\cos 2x = 1 - 2{\sin ^2}x = 1 - 2 \cdot \frac{1}{{25}} = \frac{{23}}{{25}}.\\\tan 2x = \frac{{\sin 2x}}{{\cos 2x}} = \frac{{ - \frac{{4\sqrt 6 }}{5}}}{{\frac{{23}}{{25}}}} =  - \frac{{20\sqrt 6 }}{3}.\cot 2x = \frac{{\cos 2x}}{{\sin 2x}} = \frac{{\frac{{23}}{{25}}}}{{ - \frac{{4\sqrt 6 }}{5}}} =  - \frac{{23\sqrt 6 }}{{120}}.\end{array}\)