Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{\pi }{2} < a < \pi \Rightarrow \left\{ \begin{array}{l}
\sin a > 0\\
\cos a < 0
\end{array} \right.\\
\cos a < 0 \Rightarrow \cos a = - \sqrt {1 - {{\sin }^2}a} = - \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} = - \dfrac{4}{5}\\
\sin 2a = 2\sin a.\cos a = 2.\dfrac{3}{5}.\left( { - \dfrac{4}{5}} \right) = - \dfrac{{24}}{{25}}\\
\sin \left( {\dfrac{\pi }{3} + a} \right) = \sin \dfrac{\pi }{3}.\cos a + \cos \dfrac{\pi }{3}.\sin a = \dfrac{{\sqrt 3 }}{2}.\left( { - \dfrac{4}{5}} \right) + \dfrac{1}{2}.\dfrac{3}{5} = \dfrac{{3 - 4\sqrt 3 }}{{10}}\\
\tan \left( {a - \dfrac{\pi }{4}} \right) = \dfrac{{\sin \left( {a - \dfrac{\pi }{4}} \right)}}{{\cos \left( {a - \dfrac{\pi }{4}} \right)}} = \dfrac{{\sin a.\cos \dfrac{\pi }{4} - \cos a.\sin \dfrac{\pi }{4}}}{{\cos a.\cos \dfrac{\pi }{4} + \sin a.\sin \dfrac{\pi }{4}}} = \dfrac{{\sin a - \cos a}}{{\sin a + \cos a}} = - 7\\
\dfrac{\pi }{2} < a < \pi \Rightarrow \dfrac{\pi }{4} < \dfrac{a}{2} < \dfrac{\pi }{2} \Rightarrow \cos \dfrac{a}{2} > 0\\
\cos a = 2{\cos ^2}\dfrac{a}{2} - 1 \Rightarrow \cos \dfrac{a}{2} = \dfrac{{\sqrt {10} }}{{10}}
\end{array}\)
