Prove $\sin{2x}+\sin{4x}+\sin{6x}=4\cos{x}\cos{2x}\sin{3x}$
I have reached the point where the LHS equation has turned into $2\cos{x}\cos2x\sin{x}(2\sin2x+1)$
But I have no idea how to turn $\sin{x}(2\sin2x+1)$ into $2\sin3x$
A quicker method if it exists would be greatly appreciated
Thanks in advance
Best Answer
To solve this problem, we can use the identities:
$$ \sin A + \sin B = 2\sin \frac{A+B}{2} \cos \frac{A - B}{2}, $$
$$ \cos A + \cos B = 2\cos \frac{A+B}{2} \cos \frac{A - B}{2}, $$
and
$$ \sin 2\phi = 2\sin \phi \cos \phi. $$
Going back to the question,
$ \text{LHS} = \sin 2x + \sin 4x + \sin 6x \\ = 2\sin 3x \cos x + \sin 6x \\ = 2\sin 3x \cos x + 2\sin 3x \cos 3x \\ = 2\sin 3x (\cos x + \cos 3x) \\ = 2\sin 3x \times 2\cos 2x \cos x \\ = 4\cos x \cos 2x \sin 3x \\ = \text{RHS}. $
Hence, proved.