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
$$ \sin 2x + \sin 4x = \sin(3x-x) +\sin(3x+x) = 2 \sin 3x \cos x \\ $$ and $$ \sin 6x = 2 \sin 3x \cos 3x $$ so $$ \sin 2x + \sin 4x +\sin 6x = 2\sin 3x (\cos x + \cos 3x) $$ but $$ \cos x + \cos 3x = \cos(2x-x) + \cos (2x+x) = 2\cos x \cos 2x $$
Let me try. $$\sin 2x + \sin 6x + \sin 4x = 2\sin 4x \cos 2x + 2\sin 2x\cos 2x = 2\cos 2x (\sin 4x + \sin 2x) = 4\cos 2x \sin 3x \cos x$$
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.
From Euler's identity, $sin(nx) = \frac{1}{2i}(e^{inx} - e^{-inx})$ and $cos(nx) = \frac{1}{2}(e^{inx} + e^{-inx})$ Therefore:
$$4cos(x)cos(2x)sin(3x) = \frac{1}{2i}(e^{ix}+e^{-ix})(e^{2ix}+e^{-2ix})(e^{3ix}-e^{-3ix})$$
$$=\frac{1}{2i}(e^{3ix} + e^{-ix} + e^{ix} + e^{-3ix})(e^{3ix} - e^{-3ix})$$
$$=\frac{1}{2i}(e^{6ix} - e^{-6ix} + e^{4ix} - e^{-4ix} + e^{2ix} - e^{-2ix})$$
$$=sin(2x)+sin(4x)+sin(6x)$$