Let $f(x)$ be a real transcendental function with algebraic coefficients. So $f(x)$ and $x$ are algebraically independent. Let $\alpha$ be a transcendental number, are the numbers $$\alpha+f(\alpha),\ \alpha f(\alpha),\ \alpha/f(\alpha)$$ transcendental?
It is clear these numbers are transcendental if $f(\alpha)$ is algebraic, but what if $f(\alpha)$ is transcendental? It seems in this case these numbers are also transcendental because $f(x)$ and $x$ are algebraically independent, but I'm not able to show this. I'm sorry maybe the description above is not satisfactoty. Here is a specific version: If $\alpha$ is transcendental, then are $$\alpha+\cos(\alpha), \ \alpha\cos(\alpha),\ \alpha/\cos(\alpha)\\ \alpha+e^{\alpha},\ \alpha e^{\alpha},\ \alpha/e^{\alpha}$$ transcendental?



Transcendence on $ \alpha+f(\alpha), \alpha f(\alpha) $ and $ \alpha/f(\alpha) $ where $ \alpha$ is transcendental
Here are two counterexamples to the specific question at the end.


1. Let $\alpha$ be the unique real solution to $xe^x=1$. This number is also called the [omega constant](https://en.wikipedia.org/wiki/Omega_constant#Transcendence). It is transcendental by the Lindemann-Weierstrass Theorem which (in particular) says that if $x$ is nonzero and algebraic then $e^x$ is not algebraic. So if $\alpha$ were algebraic then LW would imply that $e^\alpha=1/\alpha$ is not algebraic, a contradiction.
2. Let $\alpha$ be the unique real solution to $x=\cos(x)$. This number is also called the [Dottie number](https://en.wikipedia.org/wiki/Dottie_number). It is transcendental, again by the Lindemann-Weierstrass Theorem. Indeed, suppose $\alpha$ is algebraic. Then so is $i\alpha$, and so $e^{i\alpha}$ is not algebraic by LW. But $e^{i\alpha}=\cos(\alpha)+i\sin(\alpha)=\alpha+i\sqrt{1-\alpha^2}$.


