estimation of a unit circle - how to show a relationship

Question

It has been eons since I've done any trigonometry, but I just can't prove how this following relationship holds for $n = 4, 8, 16, 32, \dots$

The relation is: $$ 2 \biggl( \! \frac{A_{2n}}{n} \! \biggr)^2 = \, 1 - \Biggl( \sqrt{1 - \frac{2A_n^{\phantom{X}}}{n}} \, \Biggr)^{\!2} $$

Any help would be appreciated.

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Progress:

In this case, our $A_n$ can be given by the rule: $\frac{n \sin(\frac{2\pi}{n})}{2}$

We can expand $\text{LHS} = 2 \sin^2(\frac{\pi}{n})$

and $\text{RHS} = 1 - \sqrt{1-\sin^2(\frac{2\pi}{n})}$

Answer 1

I am not sure that I properly understand; forgive me if I am stupid.

If $$\text{RHS} = 1 - \sqrt{1-\sin^2(\frac{2\pi}{n})}=1-\cos(\frac{2\pi}{n})$$ but $\cos(2x)=1-2\sin^2(x)$ which makes $$\text{RHS} =2 \sin^2(\frac{\pi}{n})=\text{LHS}$$