Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
x \in \left( {\pi ;\,\,\frac{{3\pi }}{2}} \right) \Rightarrow \left\{ \begin{array}{l}
\sin x < 0\\
\cos x < 0
\end{array} \right.\\
\cos x < 0 \Rightarrow \cos x = - \sqrt {1 - {{\sin }^2}x} = - \frac{5}{{13}}\\
\cos 2x = 2{\cos ^2}x - 1 = \frac{{ - 119}}{{169}}\\
\pi < x < \frac{{3\pi }}{2} \Rightarrow \frac{\pi }{2} < \frac{x}{2} < \frac{{3\pi }}{4} \Rightarrow \sin \frac{x}{2} > 0\\
\cos x = 1 - 2{\sin ^2}\frac{x}{2} \Rightarrow \sin \frac{x}{2} = \frac{3}{{\sqrt {13} }}\\
\sin 2x = 2\sin x.\cos x = \frac{{120}}{{169}}\\
\cot \left( {2x - \frac{\pi }{4}} \right) = \frac{{\cos \left( {2x - \frac{\pi }{4}} \right)}}{{\sin \left( {2x - \frac{\pi }{4}} \right)}}\\
= \frac{{\cos 2x.\cos \frac{\pi }{4} + \sin 2x.\sin \frac{\pi }{4}}}{{\sin 2x.cos\frac{\pi }{4} - \cos 2x.\sin \frac{\pi }{4}}}\\
= \frac{1}{{239}}
\end{array}\)
