Đáp án:

\[\dfrac{{7 + 24\sqrt 3 }}{{50}}\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
{\sin ^2}x + {\cos ^2}x = 1\\
\sin 2x = 2.\sin x.\cos x\\
\cos 2x = 2{\cos ^2}x - 1\\
\cos \left( {x - y} \right) = \cos x.\cos y + \sin x.\sin y\\
*)\,\,\,{\sin ^2}\left( {a - \dfrac{b}{2}} \right) + {\cos ^2}\left( {a - \dfrac{b}{2}} \right) = 1\\
 \Leftrightarrow {\sin ^2}\left( {a - \dfrac{b}{2}} \right) + {\left( {\dfrac{1}{2}} \right)^2} = 1\\
 \Leftrightarrow {\sin ^2}\left( {a - \dfrac{b}{2}} \right) = \dfrac{3}{4}\\
\sin \left( {a - \dfrac{b}{2}} \right) > 0 \Rightarrow \sin \left( {a - \dfrac{b}{2}} \right) = \dfrac{{\sqrt 3 }}{2}\\
 \Rightarrow \left\{ \begin{array}{l}
\sin \left( {2a - b} \right) = 2.\sin \left( {a - \dfrac{b}{2}} \right).\cos \left( {a - \dfrac{b}{2}} \right) = \dfrac{{\sqrt 3 }}{2}\\
\cos \left( {2a - b} \right) = 2{\cos ^2}\left( {a - \dfrac{b}{2}} \right) - 1 = \dfrac{1}{2}
\end{array} \right.\\
*)\,\,\,{\sin ^2}\left( {\dfrac{a}{2} - b} \right) + {\cos ^2}\left( {\dfrac{a}{2} - b} \right) = 1\\
 \Leftrightarrow {\left( {\dfrac{3}{5}} \right)^2} + {\cos ^2}\left( {\dfrac{a}{2} - b} \right) = 1\\
 \Leftrightarrow {\cos ^2}\left( {\dfrac{a}{2} - b} \right) = \dfrac{{16}}{{25}}\\
\cos \left( {\dfrac{a}{2} - b} \right) > 0 \Rightarrow \cos \left( {\dfrac{a}{2} - b} \right) = \dfrac{4}{5}\\
 \Rightarrow \left\{ \begin{array}{l}
\sin \left( {a - 2b} \right) = 2.\sin \left( {\dfrac{a}{2} - b} \right).\cos \left( {\dfrac{a}{2} - b} \right) = \dfrac{{24}}{{25}}\\
\cos \left( {a - 2b} \right) = 2.{\cos ^2}\left( {\dfrac{a}{2} - b} \right) - 1 = \dfrac{7}{{25}}
\end{array} \right.\\
*)\\
\cos \left( {a + b} \right) = \cos \left[ {\left( {2a - b} \right) - \left( {a - 2b} \right)} \right]\\
 = \cos \left( {2a - b} \right).\cos \left( {a - 2b} \right) + \sin \left( {2a - b} \right).\sin \left( {a - 2b} \right)\\
 = \dfrac{1}{2}.\dfrac{7}{{25}} + \dfrac{{\sqrt 3 }}{2}.\dfrac{{24}}{{25}}\\
 = \dfrac{{7 + 24\sqrt 3 }}{{50}}
\end{array}\)