Ta có:
\(\sin x + \sin 3x + \sin 5x \\= \sin x + \sin 5x + \sin 3x\\= 2\sin {{x + 5x} \over 2}.\cos {{x - 5x} \over 2} + \sin 3x \\= 2\sin 3x + \cos 2x + \sin 3x\\= \sin 3x (2\cos 2x + 1) \, \, \, \, (1)\)
\( \cos x + \cos3x + \cos5x \\= \cos x + \cos5x +\cos3x \\= 2\cos3x . \cos2x + \cos3x \\= \cos3x (2\cos2x + 1) \, \, \, (2)\)
Từ (1) và (2) ta có:
\(A = {{\sin 3x} \over {\cos 3x}} = \tan 3x\)
Vậy \(A= \tan 3x.\)
