Cho hai góc nhọn \(a\) và \(b\) với \(\tan a = \frac{1}{7}\) và \(\tan b = \frac{3}{4}\). Tính \(a + b\).

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

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

Chọn D

Ta có \(\sin \frac{A}{2}{\cos ^3}\frac{B}{2} - \sin \frac{B}{2}{\cos ^3}\frac{A}{2} = 0 \Leftrightarrow \frac{{\sin \frac{A}{2}}}{{{{\cos }^2}\frac{A}{2}}} = \frac{{\sin \frac{B}{2}}}{{{{\cos }^3}\frac{B}{2}}}\).

\( \Leftrightarrow \tan \frac{A}{2}\left( {1 + {{\tan }^2}\frac{A}{2}} \right) = \tan \frac{B}{2}\left( {1 + {{\tan }^2}\frac{B}{2}} \right) \Leftrightarrow \tan \frac{A}{2} = \tan \frac{B}{2} \Leftrightarrow \frac{A}{2} = \frac{B}{2} \Leftrightarrow A = B\).