### Evaluation of $R(e^{-2\pi/5})$

In the last post we established the transformation formula $$\left[\left[\frac{\sqrt{5} + 1}{2}\right]^{5} + R^{5}(e^{-2\alpha})\right]\left[\left[\frac{\sqrt{5} + 1}{2}\right]^{5} + R^{5}(e^{-2\beta})\right] = 5\sqrt{5}\left[\frac{\sqrt{5} + 1}{2}\right]^{5}\tag{1}$$ under the condition $\alpha\beta = \pi^{2}/5$.If we put $\alpha = \pi$ then $\beta = \pi/5$ and since we already know the value of $R(e^{-2\pi})$ we can use equation $(1)$ to evaluate $R(e^{-2\pi/5})$. But in order to do that we need to calculate $R^{5}(e^{-2\pi})$ first.

We have from an earlier post $$R(e^{-2\pi}) = \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2}$$