Biết \(\cos \left( {a - \frac{b}{2}} \right) = \frac{1}{2}\) và \(\sin \left( {a - \frac{b}{2}} \right) > 0;\sin \left( {\frac{a}

* Hướng dẫn giải

Ta có

\(\begin{array}{l}
\frac{{24\sqrt 3  - 7}}{{50}}.\left\{ \begin{array}{l}
\cos \left( {a - \frac{b}{2}} \right) = \frac{1}{2}\\
\sin \left( {a - \frac{b}{2}} \right) > 0
\end{array} \right. \Rightarrow \sin \left( {a - \frac{b}{2}} \right) = \sqrt {1 - {{\cos }^2}\left( {a - \frac{b}{2}} \right)}  = \frac{{\sqrt 3 }}{2}\\
\left\{ \begin{array}{l}
\sin \left( {\frac{a}{2} - b} \right) = \frac{3}{5}\\
\cos \left( {\frac{a}{2} - b} \right)
\end{array} \right. \Rightarrow \cos \left( {\frac{a}{2} - b} \right) = \sqrt {1 - {{\sin }^2}\left( {\frac{a}{2} - b} \right)}  = \frac{4}{5}\\
cos\frac{{a + b}}{2} = \cos \left( {a - \frac{b}{2}} \right)\cos \left( {\frac{a}{2} - b} \right) + \sin \left( {a - \frac{b}{2}} \right)\sin \left( {\frac{a}{2} - b} \right) = \frac{1}{2}.\frac{4}{5} + \frac{3}{5}.\frac{{\sqrt 3 }}{2} = \frac{{3\sqrt 3  + 4}}{{10}}.\\
\cos \left( {a + b} \right) = 2{\cos ^2}\frac{{a + b}}{2} - 1 = \frac{{24\sqrt 3  - 7}}{{50}}.
\end{array}\)