Let $(u_n)_{n\in \Bbb N}$ be a sequence defined by
$ u_0=1$ and the recurrence $$\forall n \in \mathbb N , u_{n+1}=\sin(u_n)$$
I would like to demonstrate that $ 0<u_n<\frac{\pi}{2} $ for all $n\in \mathbb N$.
I started by $u_0$ which is positive and $0<\sin(u_0)<1< \pi/2 $
Is it enough to generalize for $u_{n+1}$ ?
$\sin(x)$ is a function that only returns values between $-1$ and $1$ for real $x$.
In addition, $0<\sin(1)<1$, and $\sin(x)>0$ for $0<x<\displaystyle \frac{\pi}{2}$.
Therefore, $0<u_n\leq1<\displaystyle \frac{\pi}{2}$.