Prove $f_a\circ f_b = f_{ab}$

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.

• What does "(def)" mean? – DonAntonio Nov 8 at 14:47
• I meant that I'm noting the expression using t(x) – radoo Nov 8 at 14:52
• 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 $g_a(x)=x^a$. – metamorphy Nov 8 at 18:19