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?

From DeMoivre's Theorem:
$$(\cos(12°) + i \sin(12°))^5 = \cos(60°) + i\sin(60²) = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
For brevity, let $c = \cos(12°)$ and $s = \sin(12°)$.
$$(c + is)^5 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
$$c^5 + 5isc^4 - 10s^2c^3 - 10is^3c^2 + 5s^4c + is^5 = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$
Splitting this equation into separate real and imaginary parts gives:
$$c^5 - 10s^2c^3 + 5s^4c = \frac{1}{2}$$
$$5sc^4 - 10s^3c^2 + s^5 = \frac{\sqrt{3}}{2}$$
But we can use the Pythagorean identity $s^2 + c^2 = 1$ to write each equation in terms of only one variable.
$$c^5 - 10(1-c^2)c^3 + 5(1-c^2)^2c = \frac{1}{2}$$
$$5s(1-s^2)^2 - 10s^3(1-s^2) + s^5 = \frac{\sqrt{3}}{2}$$
Or, after expanding and combining terms:
$$16c^5 - 20c^3 + 5c - \frac{1}{2} = 0$$
$$16s^5 - 20s^3 + 5s - \frac{\sqrt{3}}{2} = 0$$
Now, just plug the coefficients into the Quintic Formula, and...oh, wait, there isn't one.
Fortunately, we can use the Rational Root Theorem to determine that $c = \frac{1}{2}$ is a solution to the first question.  This can't be the actual value of $\cos(12°)$ because $\cos$ is decreasing on $[0, 180°]$ and so $\frac{\sqrt{3}}{2} = \cos(30°) < \cos(12°) < \cos(0) = 1$, but it does mean that we can factor out $c - \frac{1}{2}$ to reduce the equation to a quartic.
$$16c^4 + 8c^3 - 16c^2 - 8c + 1 = 0$$
Which AFAICT doesn't factor, but maybe we don't need to find an exact value for $c$, as the value that we ultimately want is:
$$4\sin^2(24°)+4\sin(24°)\sin(12°)$$
$$= 4(2cs)^2 + 4(2cs)s$$
$$=16c^2s^2 + 8cs^2$$
$$=16c^2(1-c^2) + 8c(1-c^2)$$
$$=16c^2 - 16c^4 + 8c - 8c^3$$
$$=- 16c^4 - 8c^3 + 16c^2 + 8c$$
$$=-(16c^4 + 8c^3 - 16c^2 - 8c)$$
$$=-(-1)$$
$$=1$$
Gee, it sure was lucky that the target expression happened to work out to be almost the same polynomial we found earlier.
