Đáp án:
\[A = \tan 2x\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{\sin x + \sin 2x + \sin 3x}}{{\cos x + \cos 2x + \cos 3x}}\\
= \dfrac{{\left( {\sin 3x + \sin x} \right) + \sin 2x}}{{\left( {\cos 3x + \cos x} \right) + \cos 2x}}\\
= \dfrac{{2.\sin \dfrac{{3x + x}}{2}.\cos \dfrac{{3x - x}}{2} + \sin 2x}}{{2.cos\dfrac{{3x + x}}{2}.\cos \dfrac{{3x - x}}{2} + \cos 2x}}\\
= \dfrac{{2\sin 2x.\cos x + \sin 2x}}{{2.\cos 2x.\cos x + \cos 2x}}\\
= \dfrac{{\sin 2x\left( {2\cos x + 1} \right)}}{{\cos 2x\left( {2\cos x + 1} \right)}}\\
= \dfrac{{\sin 2x}}{{\cos 2x}} = \tan 2x
\end{array}\)
