cos⁶x-sin⁶x=1/4cos2x(4-sin²2x) cm 2 vế bằng nhau

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1. $VT=\cos^6x-\sin^6x$

   $=(\sin^2x-\cos^2x)(\sin^4x+\sin^2x\cos^2x+\cos^4x)$

   $=-\cos 2x[(\sin^2x+\cos^2x)^2-\sin^2xcos^2x]$

   $=-\cos 2x(1-\sin^2x\cos^2x)$

   $=-\cos 2x(1-\dfrac{1}{4}\sin^22x)$

   $=-\dfrac{1}{4}\cos 2x(4-\sin^22x)$

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2. Giải thích các bước giải:

   $VT=\cos^6x-\sin^6x\\
   =(\cos^3x)^2-(\sin^3x)^2\\
   =(\cos^3x+\sin^3x)(\cos^3x-\sin^3x)\\
   =(\cos x+\sin x)(\sin^2x-\sin x\cos x+\cos^2x)(\cos x-\sin x)(\cos^2x +\sin x\cos x+\sin^2 x)\\
   =(\cos x+\sin x)(1-\sin x\cos x)(\cos x-\sin x)(1 +\sin x\cos x)\\
   =\cos2x(1-\sin^2x\cos^2x)\\
   =\cos2x\left ( 1-\dfrac{1}{4}\sin^22x \right )\\
   =\dfrac{1}{4}\cos2x(4-\sin^2x)=VP\Rightarrow ĐPCM$ 

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