Lemma: $\sin 3x = 3\sin(x) - 4\sin^3(x)$
Proof: Use the angle addition formula on the argument of the expression:
$\sin 3x = \sin (2x + x)$
Then:
$\sin(2x + x) = \sin(2x)\cos(x) + \sin(x)\cos(2x)$
Use the double-angle identites (for your reference, $\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(2x) = 1 - 2\sin^2(x)$:
$\sin(3x) = 2\sin(x)\cos(x)\cos(x) + \sin(x)(1 - 2\sin^2(x))$
Simplify:
$\sin(3x) = 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^3(x)$
Use Pythagoras's Identity on the last $\cos^2$ factor to write the entire RHS in terms of sine:
$\sin(3x) = 2\sin(x)(1 - \sin^2(x)) + \sin(x) - 2\sin^3(x) = 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 3\sin(x) - 4\sin^3(x)$