Proving the identity: $\sin3x + \sin x = 2\sin2x\cos x$ I need some help proving this identity: 
$$\sin3x  + \sin x = 2\sin2x\cos x$$
I don't know where to start. I thought about expanding $\sin 3x$ into $\sin (2x + x)$ but I don't think that does me any good. Any hints would be appreciated.
Thanks!
 A: Consider $\sin(2x + x)$ and $\sin(2x-x)$. 
A: The easy way:
$$\sin3x  + \sin x = 2\sin2x\cos x$$
$$\frac {e^{i3x} - e^{-i3x}} {2i} + \frac {e^{ix} - e^{-ix}} {2i} = 2 \frac {e^{i2x} - e^{-i2x}} {2i} \frac {e^{ix} + e^{-ix}} {2}$$
$$\frac {e^{i3x} - e^{-i3x} + e^{ix} - e^{-ix}} {2i} = \frac {(e^{i2x} - e^{-i2x}) (e^{ix} + e^{-ix})} {2i}$$
$$\frac {e^{i3x} - e^{-i3x} + e^{ix} - e^{-ix}} {2i} = \frac {e^{i3x} + e^{ix} -e^{-ix} - e^{-i3x}} {2i}$$
And done, thank you Mr Euler.
A: HINT:
$$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$
More generally, $$\sin C\pm \sin D=2\sin\frac{C\pm D}2\cos\frac{C\mp D}2$$ 
A: Some ideas:
$$\sin(2x+x)=\sin2x\cos x+\sin x\cos2x=2\sin x\cos^2x+\sin x(\cos^2x-\sin^2x)=$$
$$=3\sin x\cos^2x-\sin^3x\ldots$$
Thus
$$\sin3x+\sin x=\sin x(3\cos^2x-\sin^2x+1)\stackrel{\text{trigonometric Pythagoras}}=4\sin x\cos^2x=$$
$$=2(2\sin x\cos x)\cos x=\ldots$$
A: $$\sin 3x+\sin x=2\sin 2x\cos x$$
$$LHS
=\sin(2x+x)+\sin x\\
=\sin2xcosx+\sin x\cos2x+\sin x\\
=2(\sin x \cos x)\cos x+\sin(cos^2x-\sin^2x)+\sin x\\
=2\sin x \cos^2x+\sin x \cos^2x-\sin^3x+\sin x\\
=\cos^2x(2\sin x+\sin x)-\sin^3x+\sin x\\ 
=(1-\sin^2x)3\sin x-\sin^3x+\sin x\\
=3\sin x-\sin^3x-\sin^3x+\sin x\\
=4\sin x-4\sin^3x\\
=4(\sin x-\sin x(1-\cos^2x)\\
=4(\sin x \cos^2x)\\
=4\sin x \cos^2x$$
RHS
$$=2(2\sin x \cos x)\cos x\\
=4\sin x \cos^2x$$
A: $sin\,3x = sin\,(2x + x) = sin\,2x \, cos\,x + sin\,x\,con\,2x = sin\,2x \, cos\,x + sin\, x \, (1-2sin^2\,x) = \,\,\,\,\,\,\,\,\,\,\,\,\,\;\,\,\,\,\,  sin\,2x \, cos\,x \, +sin\,x\, (2cos^2\,x -1) = sin\,2x \, cos\,x + 2sin\,x\,cos^2\,x - sin\,x = 2sin\,2x\,cos\,x - sin\,x $
