How does one approximate the value of something like this?
Apparently Euler found the value of $\large \frac{2^i+2^{-i}}{2}\large $ [which equals $\cos {\left (\ln 2 \right )}$] to be close to $\frac{10}{13}$, which is correct upto the $5$th decimal place.
I came across this in P.J. Nahin's book "An Imaginary Tale: The Story of $\sqrt {-1}$"
The rational function $$\frac{18x}{x^2+4x+1}-2$$ is an extremely good approximation on $[\tfrac12,2]$ and substituting $x=2$ (or $x=\tfrac12$) gives $\tfrac{10}{13}$. I stumbled upon this by noting that $$\frac1{1-\cos(\log(x))}= \frac2{(x-1)^2}+\frac2{x-1}+\frac13+O((x-1)^4)$$ and therefore that $$ \frac2{(x-1)^2}+\frac2{x-1}+\frac13$$ is a good approximation of $$\frac1{1-\cos(\log(x))}.$$