Proving that $\sin(54°)\sin(66°) = \sin(48°)\sin(96°)$ I'm trying to prove that $\sin(54°)\sin(66°) = \sin(48°)\sin(96°)$ but I don't really have a way to approach it. Most of what I tried was replacing $\sin(2x)$ with $2\sin(x)\cos(x)$ or changing sines with cosines but none of that has really simplified it.
Would appreciate any solution. 
 A: The LHS is
$$
\sin(54)\sin(66)=\cos(36)\cos(24). 
$$
The RHS is
$$
\begin{align}
\sin(48)\sin(96)&=2\sin(24)\cos(24)\sin(96)=\cos(24)\big(\cos(72)-\cos(120)\big)=\\
&=\cos(24)\left(\cos(72)+\frac12\right).
\end{align}
$$
Thus, proving that LHS=RHS is equivalent to proving that
$$
\color{red}{\cos(36)-\cos(72)=\frac12}.
$$
The cosine values in the last identity are related to the golden section via the regular pentagon as
$$
\begin{align}
\cos(36)&=\frac{\sqrt{5}+1}{4},\\
\cos(72)&=\frac{\sqrt{5}-1}{4},
\end{align}
$$
which makes the red identity true.
P.S. See also this question.
A: Using degrees as a measure unit,
$$ \sin(60+6)\sin(60-6) = \frac{1}{2}\left(\cos(12)-\cos(10\cdot 12)\right)$$
and
$$ \sin(4\cdot 12)\sin(8\cdot 12) = \frac{1}{2}\left(\cos(4\cdot 12)-\cos(12\cdot 12)\right)$$
hence the problem boils down to proving that
$$\cos(12)-\cos(48)+\cos(144)=-\frac{1}{2}$$
or, reverting to radians, proving that $x=\cos\left(\frac{2\pi}{30}\right)$ is a root of the polynomial
$$ 1+2\,T_1(x)-2\,T_4(x)+2\,T_{12}(x)\\=\color{red}{1 + 2 x - 128 x^2 + 1664 x^4 - 7168 x^6 + 13824 x^8 - 12288 x^{10} + 
 4096 x^{12}}. $$
Since the minimal polynomial of $\cos\left(\frac{2\pi}{30}\right)$ is $\color{blue}{1 - 8 x - 16 x^2 + 8 x^3 + 16 x^4}$, it is enough to check that the blue polynomial is a divisor of the red polynomial.
A: As $\sin54^\circ=\cos36^\circ$  and $$\sin48^\circ=2\cos24^\circ\sin24^\circ$$
the proposition reduces to $$\cos36^\circ=2\sin24^\circ\sin96^\circ=\cos(96-24)^\circ-\cos(96+24)^\circ=\cos72^\circ-\left(-\dfrac12\right)$$
Now use How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?.
