Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$

This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve the problem. I've tried the following: $$\begin{align} 4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ})&=16\sin^2(12^{\circ})\cos^2(12^{\circ})+8\sin^2(12^{\circ})
\cos(12^{\circ})\\
\\
&=8\sin^2(12^{\circ})\cos(12^{\circ})\Big(2\cos(12^{\circ} ) + 1\Big)\end{align}$$
As you can see, I was trying to simplify the expression so that it'll contain only $\sin(12^{\circ})$ and $\cos(12^{\circ})$, since I thought by unifying the angles I would have a bigger chance of solving it. However, I couldn't find a way to make any further progress. Can someone show me the way?

We can avoid values of $\cos36^\circ,\sin18^\circ$ to derive at the identity.
Let $x=6^\circ$
using Prosthaphaeresis Formulas
$$4\sin^24x+4\sin4x\sin2x=4\sin4x(\sin4x+\sin2x)=8\sin4x\sin3xcos x$$
$$\implies\sin24^\circ\sin18^\circ\cos6^\circ=\sin24^\circ\cos(90-18)^\circ\sin84^\circ$$
Using Prove that: $\sin\beta\sin\left(\dfrac\pi3-\beta\right)\sin\left(\dfrac\pi3+\beta\right)=\frac {1}{4}\sin 3\beta$,
$$4\sin24^\circ\sin(60+24)^\circ\sin(60-24)^\circ=\sin(3\cdot24)^\circ $$
$$\implies4\sin24^\circ\sin(60+24)^\circ=\dfrac{\sin72^\circ}{\sin36^\circ}=\dfrac{\cos(90-72)^\circ}{2\sin18^\circ\cos18^\circ}=?$$
