Appendix: Why is \(4 \sin x \sin(60^\circ - x) \sin(60^\circ + x) = \sin 3x\\ \)?
Proof:
Using compound angle formula,
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = 4\sin x \left(\sin 60^\circ \cos x - \cos 60^\circ \sin x\right)\left(\sin 60^\circ \cos x + \cos 60^\circ \sin x\right)\)
Simplifying,
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = 4\sin x \left(\dfrac{\sqrt 3}2 \cos x - \dfrac12 \sin x\right)\left(\dfrac{\sqrt 3}2 \cos x + \dfrac12 \sin x\right) \)
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = \sin x \left(\sqrt 3 \cos x - \sin x\right)\left(\sqrt 3 \cos x + \sin x\right)\)
Now, by the identity \(a^2 - b^2 = (a - b)(a + b)\),
\(4\sin x\sin(60^\circ - x) \sin(60^\circ + x) = \sin x (3\cos^2 x - \sin^2 x)\)
Starting from the right hand side, expanding using triple angle formula,
(if you haven't learnt that yet, you can try expanding it with compound angle formula and double angle formula.)
\(\sin 3x = 3\sin x - 4\sin^3 x\)
Now, it suffices to show that \(\sin x (3\cos^2 x -\sin^2 x) = 3\sin x - 4\sin^3 x\)
To do so, we use the identity \(\cos^2 x = 1-\sin^2 x\).
\(\sin x (3\cos^2 x -\sin^2 x) = \sin x(3(1 - \sin^2 x) - \sin^2 x) = \sin x (3 - 3\sin^2 x - \sin^2 x) = \sin x (3 - 4\sin^2 x) = 3\sin x - 4\sin^3 x\)
Therefore, the original identity \(4 \sin x \sin(60^\circ - x) \sin(60^\circ + x) = \sin 3x\\ \) is true.
