Let $f_a:[1, \infty)\to[1,\infty)$, $$f_a(x)=\frac{(x+\sqrt{x^2-1})^a+(x-\sqrt{x^2-1})^a}{2}$$
Prove $f_a\circ f_b=f_{ab}$, $\forall a, b \in (0, \infty)$.
My attempt:
$$f_a(x) = \frac{\bigg(\frac{1}{x-\sqrt{x^2-1}}\bigg)^a+(x-\sqrt{x^2-1})^a}{2}=\frac{(x-\sqrt{x^2-1})^{-a}+(x-\sqrt{x^2-1})^a}{2}$$ By the fact that $(x-\sqrt{x^2-1})(def)= t(x)$ is bijective on $[1, \infty)$, we use another function, $g$, with the property that $g\circ t=f$, $$g(x)=\frac{x^{-a}+x^a}{2}$$, and composing $g$ with $e^x$ (another bijective function) we get $\cosh(ax)$
I don't know what to do from here, I'm not even sure if i'm on the right path, I was trying to do something like this answer to a previous question of mine.