Binet's Formula with Cosines

Elsewhere we found cosines and sines of several angles related to the regular pentagon.

$\mbox{cos}(36^{\circ}) = \frac{(1 + \sqrt{5})}{4}.$

For the following I shall need one more:

$\mbox{cos}(108^{\circ}) = -\mbox{cos}(72^{\circ})=-\mbox{sin}(18^{\circ})=\frac{(1 - \sqrt{5})}{4}.$

Observe that $36^{\circ} = \frac{\pi}{5}$ and $108^{\circ} = \frac{3\pi}{5}$. On the other hand, $\frac{(1 + \sqrt{5})}{4}=\frac{\phi}{2}$, half of the Golden Ratio, while $\frac{(1 - \sqrt{5})}{4}=\frac{1}{2}[-\frac{1}{\phi}]$.

What I plan to do is to combine those values to rewrite the Binet's formula for the Fibonacci numbers:

$F_{n}=\frac{1}{\sqrt{5}}({\phi}^{n}-{\tau}^{n}),$

where $\tau$ is exactly $-\frac{1}{\phi}$.

The combination of the above obtained, according to [Grimaldi, p. 67], first by W. Hope-Jones in 1921, leads to

$F_{n}=\frac{1}{\sqrt{5}}(2^{n})\left[\mbox{cos}^{n}(\frac{\pi}{5})-\mbox{cos}^{n}(\frac{3\pi}{5})\right].$

References

1. R. Grimaldi, Fibonacci and Catalan Numbers: an Introduction, Wiley, 2012

Trigonometry

Copyright © 1996-2018 Alexander Bogomolny

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