Đáp án:
$=\dfrac{24\sqrt{3}-7}{50}. $
Giải thích các bước giải:
Ta có :
$ \left\{ \begin{array}{l} \cos \left( a-\dfrac{b}{2} \right)=\dfrac{1}{2} \\ \sin \left( a-\dfrac{b}{2} \right) > 0 \end{array} \right. $ $ \Rightarrow \sin \left( a-\dfrac{b}{2} \right)=\sqrt{1-{{\cos }^{2}}\left( a-\dfrac{b}{2} \right)}=\dfrac{\sqrt{3}}{2} $ .
$ \left\{ \begin{array}{l} \sin \left( \dfrac{a}{2}-b \right)=\dfrac{3}{5} \\ \cos \left( \dfrac{a}{2}-b \right) \end{array} \right. $ $ \Rightarrow \cos \left( \dfrac{a}{2}-b \right)=\sqrt{1-{{\sin }^{2}}\left( \dfrac{a}{2}-b \right)}=\dfrac{4}{5} $ .
$ cos\dfrac{a+b}{2}=\cos \left( a-\dfrac{b}{2} \right)\cos \left( \dfrac{a}{2}-b \right)+\sin \left( a-\dfrac{b}{2} \right)\sin \left( \dfrac{a}{2}-b \right) $ $ =\dfrac{1}{2}.\dfrac{4}{5}+\dfrac{3}{5}.\dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}+4}{10}. $
$ \cos \left( a+b \right)=2{{\cos }^{2}}\dfrac{a+b}{2}-1=\dfrac{24\sqrt{3}-7}{50}. $
