Operation $\circ$ on $\mathbb{R}: x\circ y=x\sqrt{1+y^2}+y\sqrt{1+x^2}$

Question

I have the following operation $\circ$ on $\mathbb{R}: x\circ y=x\sqrt{1+y^2}+y\sqrt{1+x^2}, \forall x, y\in \mathbb{R}$. My question: are the groups $(\mathbb{R},\circ)$ and $(\mathbb{R},+)$ isomorphic? $"+"$ is the standard addition operation. Thank you!

EDIT: As I said in the comments section below, $(\mathbb{R},\circ)$ is a group.

Answer 1

Yes they are:

$$ (\sinh u)\circ (\sinh v) = \sinh(u)\cosh(v)+\sinh(v)\cosh(u) = \sinh(u+v) $$ and $\sinh$ is a bijective function on $\mathbb{R}$. That also gives:

$$\underbrace{x\circ x\circ\ldots\circ x}_\text{n times}=\sinh\left(n\cdot\text{arcsinh}(x)\right)=\frac{\left(x+\sqrt{1+x^2}\right)^n-\left(-x+\sqrt{1+x^2}\right)^{n}}{2},$$ so: $$ \underbrace{x\circ x\circ\ldots\circ x}_\text{(2n+1) times}=i^{2n-1}\, T_{2n+1}(ix),\qquad \underbrace{x\circ x\circ\ldots\circ x}_\text{2n times}=i^{2n-1}\sqrt{1+x^2}\,U_{2n-1}(ix)$$ where $T_n,U_n$ are Chebyshev polynomials.