Prove $\sin2x+\sin4x+\sin6x=4\cos x\cos2x\sin3x$ 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
 A: 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.
A: Let me try. $$\sin 2x + \sin 6x + \sin 4x = 2\sin 4x \cos 2x + 2\sin 2x\cos 2x = 2\cos 2x (\sin 4x + \sin 2x) = 4\cos 2x \sin 3x \cos x$$
A: From Euler's identity, $\sin(nx) = \frac{1}{2i}(e^{inx} - e^{-inx})$ and $\cos(nx) = \frac{1}{2}(e^{inx} + e^{-inx})$ Therefore:
$$4\cos(x)\cos(2x)\sin(3x) = \frac{1}{2i}(e^{ix}+e^{-ix})(e^{2ix}+e^{-2ix})(e^{3ix}-e^{-3ix})$$
$$=\frac{1}{2i}(e^{3ix} + e^{-ix} + e^{ix} + e^{-3ix})(e^{3ix} - e^{-3ix})$$
$$=\frac{1}{2i}(e^{6ix} - e^{-6ix} + e^{4ix} - e^{-4ix} + e^{2ix} - e^{-2ix})$$
$$=\sin(2x)+\sin(4x)+\sin(6x)$$
A: $$
\sin 2x + \sin 4x = \sin(3x-x) +\sin(3x+x) = 2 \sin 3x \cos x \\
$$
and
$$
\sin 6x = 2 \sin 3x \cos 3x
$$
so
$$
\sin 2x + \sin 4x +\sin 6x = 2\sin 3x (\cos x + \cos 3x)
$$
but
$$
\cos x + \cos 3x = \cos(2x-x) + \cos (2x+x) = 2\cos x \cos 2x
$$
