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:
$$\sin(3x) = 2\sin(x)(1 - \sin^2(x)) + \sin(x) - 2\sin^3(x) = 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) = 3\sin(x) - 4\sin^3(x)$$