# 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

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.

Let me try. $$\sin 2x + \sin 6x + \sin 4x = 2\sin 4x \cos 2x + 2\sin 2x\cos 2x = 2\cos 2x (\sin 4x + \sin 2x) = 4\cos 2x \sin 3x \cos x$$

$$\sin 2x + \sin 4x = \sin(3x-x) +\sin(3x+x) = 2 \sin 3x \cos x \\$$ and $$\sin 6x = 2 \sin 3x \cos 3x$$ so $$\sin 2x + \sin 4x +\sin 6x = 2\sin 3x (\cos x + \cos 3x)$$ but $$\cos x + \cos 3x = \cos(2x-x) + \cos (2x+x) = 2\cos x \cos 2x$$

From Euler's identity, $$\sin(nx) = \frac{1}{2i}(e^{inx} - e^{-inx})$$ and $$\cos(nx) = \frac{1}{2}(e^{inx} + e^{-inx})$$ Therefore:

$$4\cos(x)\cos(2x)\sin(3x) = \frac{1}{2i}(e^{ix}+e^{-ix})(e^{2ix}+e^{-2ix})(e^{3ix}-e^{-3ix})$$

$$=\frac{1}{2i}(e^{3ix} + e^{-ix} + e^{ix} + e^{-3ix})(e^{3ix} - e^{-3ix})$$

$$=\frac{1}{2i}(e^{6ix} - e^{-6ix} + e^{4ix} - e^{-4ix} + e^{2ix} - e^{-2ix})$$

$$=\sin(2x)+\sin(4x)+\sin(6x)$$