Biết \(0 < a,b < \frac{\pi }{2},a + b = \frac{\pi }{4}\) và \(\tan a\tan b = 3 - 2\sqrt 2 \). Khi đó:

a) \(\tan a + \tan b = - 2 + 2\sqrt 2 {\rm{. }}\)

b) \(\tan a = - 1 + \sqrt 2 \)

c) \(\tan b = - 1 - \sqrt 2 \)

d) \(\tan a - \tan b = - 2 - 2\sqrt 2 {\rm{. }}\)

Ta có: \(\sin x\sin 2x\sin 3x = (\sin 3x\sin x)\sin 2x = \frac{1}{2}(\cos 2x - \cos 4x)\sin 2x\)

\(\begin{array}{l} = \frac{1}{2}(\sin 2x\cos 2x - \sin 2x\cos 4x) = \frac{1}{2}\left[ {\frac{1}{2}\sin 4x - \frac{1}{2}(\sin 6x - \sin 2x)} \right]\\ = \frac{1}{4}(\sin 4x - \sin 6x + \sin 2x)\end{array}\)

\(\begin{array}{*{20}{l}}A&{ = \frac{{\sin x + \sin 2x + \sin 3x + \sin 4x}}{{\cos x + \cos 2x + \cos 3x + \cos 4x}}}\\{}&{ = \frac{{(\sin 3x + \sin x) + (\sin 4x + \sin 2x)}}{{(\cos 3x + \cos x) + (\cos 4x + \cos 2x)}} = \frac{{2\sin 2x\cos x + 2\sin 3x\cos x}}{{2\cos 2x\cos x + 2\cos 3x\cos x}}}\\{}&{ = \frac{{2\cos x(\sin 2x + \sin 3x)}}{{2\cos x(\cos 2x + \cos 3x)}} = \frac{{2\sin \frac{{5x}}{2}\cos \frac{x}{2}}}{{2\cos \frac{{5x}}{2}\cos \frac{x}{2}}} = \tan \frac{{5x}}{2}}\end{array}\)