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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.

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  • $\begingroup$ What does "(def)" mean? $\endgroup$ – DonAntonio Nov 8 at 14:47
  • $\begingroup$ I meant that I'm noting the expression using t(x) $\endgroup$ – radoo Nov 8 at 14:52
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    $\begingroup$ The inverse of $r(x)=x\color{red}{+}\sqrt{x^2-1}$ is $r^{-1}(x)=(x+1/x)/2$. That is, $f_a=r^{-1}\circ g_a\circ r$, where $g_a(x)=x^a$. $\endgroup$ – metamorphy Nov 8 at 18:19

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