How to simplify these trigonometry terms? (involving sin, cos, tan, cot)

Question

Please, simplify these trigonometry term: $$ \frac{\sin 2x + \sin 4x + \sin 6x}{\cos 2x + \cos 4x + \cos 6x}=? $$

Answer 1

HINT:

Use $\displaystyle\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$ on $\displaystyle\sin2x+\sin6x$

and $\displaystyle\cos C+\cos D=2\cos\frac{C+D}2\cos\frac{C-D}2$ on $\displaystyle\cos2x+\cos6x$

Answer 2

As the angles are in Arithmetic Progression,

using this solution,

$\displaystyle\sin2x+\sin4x+\sin6x=\frac{2\sin x(\sin2x+\sin4x+\sin6x)}{2\sin x}=\frac{\cos x-\cos7x}{2\sin x}$

Now as $\displaystyle\cos C-\cos D=2\sin\frac{C+D}2\sin\frac{D-C}2,$ $\displaystyle\cos x-\cos7x=2\sin4x\sin3x$

Similarly, $\displaystyle\cos2x+\cos4x+\cos6x=\frac{\sin7x-\sin x}{2\sin x}$

Now as $\displaystyle\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2, \sin7x-\sin x=2\sin3x\cos4x$