Chứng minh \(\left( {\sin 2x + 2{{\cos }^2}x} \right)\left( {\tan 2x + \tan x} \right) = \dfrac{{2\sin 3x}}{{\cos x -
Chứng minh \(\left( {\sin 2x + 2{{\cos }^2}x} \right)\left( {\tan 2x + \tan x} \right) = \dfrac{{2\sin 3x}}{{\cos x - \sin x}}\)
Ta có \(\left( {\sin 2x + 2{{\cos }^2}x} \right)\left( {\tan 2x + \tan x} \right) = \left( {2\sin x\cos x + 2{{\cos }^2}x} \right)\left( {\dfrac{{\sin 2x}}{{\cos 2x}} + \dfrac{{\sin x}}{{\cos x}}} \right)\)
\( = 2\cos x\left( {\cos x + \sin x} \right).\dfrac{{\sin 2x\cos x + \cos 2x\sin x}}{{\cos 2x.\cos x}}\)
\( = 2\left( {\cos x + \sin x} \right).\dfrac{{\sin x\left( {2{{\cos }^2}x + \cos 2x} \right)}}{{\cos 2x}}\)
\( = 2\left( {\cos x + \sin x} \right).\dfrac{{\sin x\left( {1 + \cos 2x + \cos 2x} \right)}}{{{{\cos }^2}x - {{\sin }^2}x}}\)
\( = 2\left( {\cos x + \sin x} \right).\dfrac{{\sin x\left( {1 + 2\cos 2x} \right)}}{{\left( {\cos x - \sin x} \right)\left( {\cos x + \sin x} \right)}}\)
\( = 2.\dfrac{{\sin x\left[ {1 + 2\left( {1 - 2{{\sin }^2}x} \right)} \right]}}{{\cos x - \sin x}} = \dfrac{{2\left( {3\sin x - 4{{\sin }^3}x} \right)}}{{\cos x - \sin x}} = \dfrac{{2\sin 3x}}{{\cos x - \sin x}}\) (đpcm)
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