`b)` $B = \frac{\sin 2\alpha + \sin 4\alpha + \sin 6\alpha}{1 + \cos 2\alpha + \cos 4\alpha}$

 `+)` Công thức:

$\sin x + \sin y = 2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}$ 

Có:

$\sin 2\alpha + \sin 6\alpha = 2 \sin \frac{2\alpha + 6\alpha}{2} \cos \frac{2\alpha - 6\alpha}{2}$

$= 2 \sin 4\alpha \cos (-2\alpha) = 2 \sin 4\alpha \cos 2\alpha$ 

`->` Tử số:

$\sin 2\alpha + \sin 4\alpha + \sin 6\alpha$

$= 2 \sin 4\alpha \cos 2\alpha + \sin 4\alpha$

$= \sin 4\alpha (2 \cos 2\alpha + 1)$

`+)` Công thức: $1 + \cos 4\alpha = 2 \cos^2 2\alpha$

 `->` Mẫu số: 

$1 + \cos 2\alpha + \cos 4\alpha = \cos 2\alpha + 2 \cos^2 2\alpha$

$= \cos 2\alpha (1 + 2 \cos 2\alpha)$

`=>`$B = \frac{\sin 4\alpha (2 \cos 2\alpha + 1)}{\cos 2\alpha (1 + 2 \cos 2\alpha)}$

$= \frac{\sin 4\alpha}{\cos 2\alpha}$

$= \frac{2 \sin 2\alpha \cos 2\alpha}{\cos 2\alpha}$

$= 2 \sin 2\alpha$