Chứng minh 4cosx/3.cos[(π+x)/3].cos[(π-x)/3]=cosx.

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1. $4\cos\dfrac{x}{3}.\cos\dfrac{\pi+x}{3}\cos\dfrac{\pi-x}{3}$

   $=4\cos\dfrac{x}{3}.\dfrac{1}{2}\Big( \cos\dfrac{2\pi}{3}+\cos\dfrac{2x}{3}\Big)$

   $=2\cos\dfrac{x}{3}\cos\dfrac{2x}{3}+2\cos\dfrac{x}{3}.\cos\dfrac{2\pi}{3}$

   $=\cos x+\cos\dfrac{-x}{3}-\cos\dfrac{x}{3}$

   $=\cos x$

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2. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   2\cos x.\cos y = \cos \left( {x + y} \right) + \cos \left( {x – y} \right)\\
   4\cos \dfrac{x}{3}.\cos \dfrac{{\pi  + x}}{3}.\cos \dfrac{{\pi  – x}}{3}\\
    = 2\cos \dfrac{x}{3}.\left( {2\cos \dfrac{{\pi  + x}}{3}.\cos \dfrac{{\pi  – x}}{3}} \right)\\
    = 2\cos \dfrac{x}{3}.\left[ {\cos \left( {\dfrac{{\pi  + x}}{3} + \dfrac{{\pi  – x}}{3}} \right) + \cos \left( {\dfrac{{\pi  + x}}{3} – \dfrac{{\pi  – x}}{3}} \right)} \right]\\
    = 2\cos \dfrac{x}{3}.\left( {\cos \dfrac{{2\pi }}{3} + \cos \dfrac{{2x}}{3}} \right)\\
    = 2\cos \dfrac{x}{3}.\left( { – \dfrac{1}{2} + \cos \dfrac{{2x}}{3}} \right)\\
    =  – \cos \dfrac{x}{3} + 2\cos \dfrac{x}{3}.\cos \dfrac{{2x}}{3}\\
    =  – \cos \dfrac{x}{3} + \cos \left( {\dfrac{x}{3} + \dfrac{{2x}}{3}} \right) + \cos \left( {\dfrac{x}{3} – \dfrac{{2x}}{3}} \right)\\
    =  – \cos \dfrac{x}{3} + \cos x + \cos \left( { – \dfrac{x}{3}} \right)\\
    =  – \cos \dfrac{x}{3} + \cos x + \cos \dfrac{x}{3}\\
    = \cos x
   \end{array}\)

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