Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 2{\cos ^2}x - 1 \Rightarrow {\cos ^2}x = \dfrac{{\cos 2x + 1}}{2}\\
B = 1 - \left( {{{\cos }^2}a + {{\cos }^2}b + {{\cos }^2}c} \right) + 2\cos a.\cos b.\cos c\\
= 1 - \left[ {\left( {\dfrac{{\cos 2a + 1}}{2}} \right) + \left( {\dfrac{{\cos 2b + 1}}{2}} \right) + {{\cos }^2}c} \right] + 2\cos a.\cos b.\cos c\\
= - \dfrac{1}{2}\left( {\cos 2a + \cos 2b} \right) + {\cos ^2}c + 2\cos a.\cos b.\cos c\\
= - \dfrac{1}{2}.2.cos\dfrac{{2a + 2b}}{2}.\cos \dfrac{{2a - 2b}}{2} + {\cos ^2}c + 2\cos a.\cos b.\cos c\\
= - \cos \left( {a + b} \right).\cos \left( {a - b} \right) + {\cos ^2}c + 2\cos a.\cos b.\cos c\\
= \cos \left( {180^\circ - a - b} \right).\cos \left( {a - b} \right) + {\cos ^2}c + 2\cos a.\cos b.\cos c\\
= \cos c.\cos \left( {a - b} \right) + {\cos ^2}c + 2\cos a.\cos b.\cos c\\
= \cos c.\left[ {\cos \left( {a - b} \right) + \cos c + 2\cos a.\cos b} \right]
\end{array}\)
