A challenging trig identity

Question

Prove that, for $g$ a nonnegative integer,

\begin{eqnarray}3\left(\frac{64}{75}\sin^2\frac{\pi}{5}\sin^2\frac{2\pi}{5}\right)^{1-g}\left(\sin^{2(1-g)}\frac{\pi}{5}+\sin^{2(1-g)}\frac{2\pi}{5}\right)=\sum_{p=1}^2\frac{4\sin^2\frac{p\pi}{5}}{5}\left(37+62\cos\frac{2p\pi}{5}+56\cos\frac{4p\pi}{5}\right)^g.\end{eqnarray}

I came up with this identity in my research and have a proof which I think is an overkill. I would appreciate any high-school level proof.

Answer 1

Converting to radicals, \begin{align} \sin^2\tfrac\pi5&= \tfrac58-\tfrac{\sqrt5}8 ,\\ \sin^2\tfrac{2\pi}5&= \tfrac58+\tfrac{\sqrt5}8 ,\\ \cos\tfrac{2\pi}5&= -\tfrac 14+\tfrac{\sqrt5}4 ,\\ \cos\tfrac{4\pi}5&= -\tfrac 14-\tfrac{\sqrt5}4 ,\\ \cos\tfrac{8\pi}5&= -\tfrac 14+\tfrac{\sqrt5}4 \end{align}

\begin{align} 3(\tfrac 1{30})^{1-g}& \Big((5-\sqrt5)^{1-g}+(5+\sqrt5)^{1-g}\Big) = \\ \tfrac 1{10}& \Big( (5-\sqrt5)(\tfrac{15}2+\tfrac{3\sqrt5}2)^g+ (5+\sqrt5)(\tfrac{15}2-\tfrac{3\sqrt5}2)^g \Big) ,\\ 3(30)^{g-1}& \Big((5-\sqrt5)^{1-g}+(5+\sqrt5)^{1-g}\Big) = \\ \tfrac 1{10} (\tfrac 32)^g & \Big( (5-\sqrt5)(5+\sqrt5)^g+ (5+\sqrt5)(5-\sqrt5)^g \Big) ,\\ \tfrac1{10}(3^g 10^g)& \Big((5-\sqrt5)^{1-g}+(5+\sqrt5)^{1-g}\Big) = \\ \tfrac 1{10} (\tfrac 32)^g & \Big( (5-\sqrt5)(5+\sqrt5)^g+ (5+\sqrt5)(5-\sqrt5)^g \Big) ,\\ \end{align}

\begin{align} 10^g \Big((5-\sqrt5)^{1-g}&+(5+\sqrt5)^{1-g}\Big) = \\ 2^{-g} & \Big( (5-\sqrt5)(5+\sqrt5)^g+ (5+\sqrt5)(5-\sqrt5)^g \Big) ,\\ \end{align}

\begin{align} 10^g \Big((5-\sqrt5)\Big(\frac1{5-\sqrt5}\Big)^g &+(5+\sqrt5)\Big(\frac1{5+\sqrt5}\Big)^g\Big) = \\ 2^{-g} & \Big( (5-\sqrt5)(5+\sqrt5)^g+ (5+\sqrt5)(5-\sqrt5)^g \Big) ,\\ \end{align}

\begin{align} 10^g \Big((5-\sqrt5)\Big(\frac{5+\sqrt5}{20}\Big)^g &+(5+\sqrt5)\Big(\frac{5-\sqrt5}{20}\Big)^g\Big) = \\ 2^{-g} & \Big( (5-\sqrt5)(5+\sqrt5)^g+ (5+\sqrt5)(5-\sqrt5)^g \Big) , \end{align}

as expected.