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