Solution for $\displaystyle f'\left(x\right)=\cos\left(f\left(x\right)\right)$ Let $f$ be defined for $x \in \mathbb{R}$ by
$$
f\left(x\right)=2\text{arctan}\left(e^x\right)-\frac{\pi}{2}
$$
I've shown that $f$ is odd and satisfies for $x \in \mathbb{R}$
$$f'\left(x\right)=\cos\left(f\left(x\right)\right)$$
To prove it, i've used that
$$
\cos\left(f\left(x\right)\right)=2\sin\left(\text{arctan}\left(e^x\right)\right)\cos\left(\text{arctan}\left(e^x\right)\right)
$$
And then use that
$$
\cos\left(\text{arctan}\left(e^x\right)\right)=\frac{1}{\sqrt{1+e^{2x}}}
$$
I've two questions : 
$\bullet$ Is that the unique solution for $f(0)=0$ ?
I've tried to prove it by supposing to different solutions and trying to prove there are infact equals with trigonometric formula but it does not seem to work.
$\bullet$ Is there another way ( even wiser or faster ) to prove it ?
Thanks for those who take time to answer.

Do you ask if $f\left(x\right)=2\text{arctan}\left(e^x\right)-\frac{\pi}{2}$ is the unique solution of
$$
\begin{cases}
f'(x)=\cos(f(x))\\
f(0)=0
\end{cases}
$$
?
Since $\cos$ is Lipschitz continuous, you can use Picard-Lindelöff theorem to see that the solution is unique.
For your second question: If you have to show that your function is a solution of the ODE above, then the only way is to check that the equations hold for your function. There is no other way.
