Giả sử \(E = \tan x \cdot \tan \left( {\frac{\pi }{3} - x} \right) \cdot \tan \left( {\frac{\pi }{3} + x} \right)\). Biết E có dạng tan(k∙x). Khi đó k bằng bao nhiêu?

Đáp án: 3

\(E = \tan x \cdot \tan \left( {\frac{\pi }{3} - x} \right) \cdot \tan \left( {\frac{\pi }{3} + x} \right) = \tan x\frac{{\sin \left( {\frac{\pi }{3} - x} \right) \cdot \sin \left( {\frac{\pi }{3} + x} \right)}}{{\cos \left( {\frac{\pi }{3} - x} \right) \cdot \cos \left( {\frac{\pi }{3} + x} \right)}}\)

\(E = \tan x\frac{{\frac{{ - 1}}{2}\left( {\cos \frac{{2\pi }}{3} - \cos 2x} \right)}}{{\frac{1}{2}\left( {\cos \frac{{2\pi }}{3} + \cos 2x} \right)}} = \tan x\frac{{ - \left( {\frac{{ - 1}}{2} - \cos 2x} \right)}}{{\frac{{ - 1}}{2} + \cos 2x}}\)

\(E = \tan x\frac{{2\cos 2x + 1}}{{2\cos 2x - 1}} = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}}}{{\cos x}} \cdot \frac{{2\left( {1 - 2{{\sin }^2}x} \right) + 1}}{{2\left( {2{{\cos }^2}x - 1} \right) - 1}} = \frac{{{\mathop{\rm s}\nolimits} {\rm{inx}}\left( { - 4{{\sin }^2}x + 3} \right)}}{{\cos x\left( {4{{\cos }^2}x - 3} \right)}}\)

\(E = \frac{{3\sin x - 4\sin {}^3x}}{{4{{\cos }^3}x - 3\cos x}} = \frac{{\sin 3x}}{{\cos 3x}} = \tan 3x\).