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?
 A: Rather than manipulating to get things in terms of $12^\circ$, we'll try to get things in terms of $18^\circ$, because we (hopefully) know that $\sin(18^\circ)=\frac{\sqrt{5}-1}{4}$. Now, let's do it. First factor out $4\sin(24^\circ)$ and use sum-to-product on the remaining factor to get
$$8\sin(24^\circ)(\sin(18^\circ)\cos(6^\circ)).$$
Then use product-to-sum on the first and third terms to finish. I leave the details to you, as there is value in knowing how to do these manipulations on your own.
EDIT: In general, if there are trigonometric expressions involving multiples of $6^\circ$, there is a good chance the solution will involve $\sin(18^\circ)=\frac{\sqrt 5 -1}{4}$ and/or $\cos(36^\circ)=\frac{\sqrt 5+1}{4}$. If you'd like another example for more practice, try to show that
$$\sin(6^\circ)\sin(12^\circ)\sin(24^\circ)\sin(42^\circ)+\sin(12^\circ)\sin(24^\circ)\sin(42^\circ)=\frac{1}{16}.$$
(Source: ARML 2019)
A: 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:
$$\begin{align} \text{LHS} &=4\sin^2(24^{\circ})+\frac{2\sin(24^{\circ})\sin(24^{\circ})}{\cos(12^\circ)} - 1\\
\\
&=4\sin^2(24^\circ)\cdot\frac{\cos(12^\circ)+\cos(60^\circ)}{\cos(12^\circ)} -1\\
\\
&=\frac{8\sin^2(24^\circ)\cos(24^\circ)\cos(36^\circ)}{\cos(12^\circ)}-1\\
\\
&=\frac{4\sin(24^\circ)\sin(48^\circ)\cos(36^\circ)}{\cos(12^\circ)}-1\\
\\
&=8\sin(12^\circ)\sin(48^\circ)\cos(36^\circ)-1\\
\\
&=4\left(\cos(36^\circ)-\frac{1}2 \right)\cos(36^\circ)-1\\
\\
&=4\cos^2(36^\circ)-2\cos(36^\circ)-1\\
\\
&=\frac{\left(4\cos^2(36^\circ)-2\cos(36^\circ)-1\right)\left(\cos(36^\circ)+1\right)}{\cos(36^\circ)+1}\\
\\
&=\frac{4\cos^3(36^\circ)-3\cos(36^\circ)+2\cos^2(36^\circ)-1}{\cos(36^\circ)+1}\\
\\
&=\frac{\cos(108^\circ)+\cos(72^\circ)}{\cos(36^\circ)+1}\\
\\
&=\frac{-\cos(72^\circ)+\cos(72^\circ)}{\cos(36^\circ)+1}\\
\\
&=0
\end{align}$$
A: There are more elementary solutions, but using
cyclotomic polynomials,
Computer Algebra Systems for factorization, and Euler's formula for
the sine, then there is one method.
Define the function
$$ f(x) := 4\sin^2(2x) + 4\sin(2x)\sin(x)-1. $$
Now let $\,t := e^{ix}\,$ and use the Euler identity $\,\sin(x) = (t-1/t)/(2i).$
Express the function $f(x)$ in terms of $\,t\,$ and factorize to get
$$ f(x) = -(t^2-1/t^2)^2 - (t^2-1/t^2)(t-1/t) - 1 = 
-\frac{1+t-t^3-t^4-t^5+t^7+t^8}{t^4}. $$
Notice that the numerator is the $30$th cyclotomic polynomial and that
$12^\circ = 2\pi/30.\,$ Thus $\,f(x)=0\,$ for all of the primitive
$30$th roots of unity. That implies that not only is $\,f(12^\circ) = 0,\,$
but also $\,f(n\, 12^\circ) = 0\,$ for $\,n=1,7,11,13,17,19,23,29.$
A: 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: $\sin\beta\sin\left(\dfrac\pi3-\beta\right)\sin\left(\dfrac\pi3+\beta\right)=\frac {1}{4}\sin 3\beta$,
$$4\sin24^\circ\sin(60+24)^\circ\sin(60-24)^\circ=\sin(3\cdot24)^\circ $$
$$\implies4\sin24^\circ\sin(60+24)^\circ=\dfrac{\sin72^\circ}{\sin36^\circ}=\dfrac{\cos(90-72)^\circ}{2\sin18^\circ\cos18^\circ}=?$$
A: 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.
