Proving that $\sin(54°)\sin(66°) = \sin(48°)\sin(96°)$

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.

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Trigonometry

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