Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{{3\pi }}{2} < a < 2\pi \Rightarrow \left\{ \begin{array}{l}
\sin a < 0\\
\cos a > 0
\end{array} \right.\\
\sin a < 0 \Rightarrow \sin a = - \sqrt {1 - {{\cos }^2}a} = - \sqrt {1 - {{\left( {\dfrac{4}{5}} \right)}^2}} = - \dfrac{3}{5}\\
\sin 2a = 2\sin a.\cos a = 2.\left( { - \dfrac{3}{5}} \right).\dfrac{4}{5} = - \dfrac{{24}}{{25}}\\
\sin \left( {a + \dfrac{\pi }{6}} \right) = \sin a.\cos \dfrac{\pi }{6} + \cos a.\sin \dfrac{\pi }{6} = - \dfrac{3}{5}.\dfrac{{\sqrt 3 }}{2} + \dfrac{4}{5}.\dfrac{1}{2} = \dfrac{{4 - 3\sqrt 3 }}{{10}}\\
\tan \dfrac{a}{2} = \dfrac{{\sin \dfrac{a}{2}}}{{\cos \dfrac{a}{2}}} = \dfrac{{{{\sin }^2}\dfrac{a}{2}}}{{\sin \dfrac{a}{2}.\cos \dfrac{a}{2}}} = \dfrac{{\dfrac{{1 - \cos a}}{2}}}{{\dfrac{1}{2}\sin a}} = \dfrac{{1 - \cos a}}{{\sin a}} = \dfrac{{1 - \dfrac{4}{5}}}{{ - \dfrac{3}{5}}} = - \dfrac{1}{3}\\
\tan \left( {\dfrac{\pi }{4} - 2a} \right) = \dfrac{{\sin \left( {\dfrac{\pi }{4} - 2a} \right)}}{{\cos \left( {\dfrac{\pi }{4} - 2a} \right)}} = \dfrac{{\sin \dfrac{\pi }{4}.\cos 2a - \cos \dfrac{\pi }{4}.\sin 2a}}{{\cos \dfrac{\pi }{4}.\cos 2a + \sin \dfrac{\pi }{4}.\sin 2a}} = \dfrac{{\cos 2a - \sin 2a}}{{\cos 2a + \sin 2a}} = \dfrac{{\left( {2{{\cos }^2}a - 1} \right) - \sin 2a}}{{\left( {2{{\cos }^2}a - 1} \right) + \sin 2a}} = - \dfrac{{31}}{{17}}
\end{array}\)
