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Ta có
Vế trái $= \sin 3x .\frac{3\sin x -\sin3x}{4}+\cos 3x.\frac{\cos 3x +3\cos x}{4}$
$=\frac{3}{4}\left ( \sin 3x\sin x+\cos 3x \cos x \right )+\frac{1}{4}\left (\cos^2 3x-\sin^2 3x \right )$
$=\frac{3}{4}\cos2x+\frac{1}{4}\cos6x$
$=\cos^3 2x$, đpcm.
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