Cho biết \(\sin \alpha = \frac{3}{5},\frac{\pi }{2} < \alpha < \pi \). Khi đó:

a) \(\cos \alpha < 0\)

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

c) \(\tan \alpha = \frac{3}{4}\)

d) \(\tan \left( {\alpha + \frac{\pi }{3}} \right) = \frac{{48 - \sqrt 3 }}{{11}}\)

Cho \(\sin x = \frac{1}{5},\frac{\pi }{2} < x < \pi \). Tính \(\cot 2x\).

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

Chọn D

Vì \[A,\,\,B,\,\,C\] là các góc của tam giác \[ABC\] nên \[A + B + C = {180^o} \Rightarrow C = {180^o} - \left( {A + B} \right).\]

Do đó \[C\] và \[\left( {A + B} \right)\] là 2 góc bù nhau.

\[ \Rightarrow \sin C = \sin \left( {A + B} \right);\,\,\cos C =  - \cos \left( {a + b} \right);\,\,\tan C =  - \tan \left( {A + B} \right);\,\,\cot C = \cot \left( {A + B} \right).\]