<s>[INST] 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?
 [/INST] I prove: $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) - 1=0$ without using special values at $18^\circ$ or $36^\circ$
Start:
$$\begin{align} \text{LHS} &=4\sin^2(24^{\circ})+\frac{2\sin(24^{\circ})\sin(24^{\circ})}{\cos(12^\circ)} - 1\\
\\
&=4\sin^2(24^\circ)\cdot\frac{\cos(12^\circ)+\cos(60^\circ)}{\cos(12^\circ)} -1\\
\\
&=\frac{8\sin^2(24^\circ)\cos(24^\circ)\cos(36^\circ)}{\cos(12^\circ)}-1\\
\\
&=\frac{4\sin(24^\circ)\sin(48^\circ)\cos(36^\circ)}{\cos(12^\circ)}-1\\
\\
&=8\sin(12^\circ)\sin(48^\circ)\cos(36^\circ)-1\\
\\
&=4\left(\cos(36^\circ)-\frac{1}2 \right)\cos(36^\circ)-1\\
\\
&=4\cos^2(36^\circ)-2\cos(36^\circ)-1\\
\\
&=\frac{\left(4\cos^2(36^\circ)-2\cos(36^\circ)-1\right)\left(\cos(36^\circ)+1\right)}{\cos(36^\circ)+1}\\
\\
&=\frac{4\cos^3(36^\circ)-3\cos(36^\circ)+2\cos^2(36^\circ)-1}{\cos(36^\circ)+1}\\
\\
&=\frac{\cos(108^\circ)+\cos(72^\circ)}{\cos(36^\circ)+1}\\
\\
&=\frac{-\cos(72^\circ)+\cos(72^\circ)}{\cos(36^\circ)+1}\\
\\
&=0
\end{align}$$
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