Hello!!
Let $a_n=\sin (\sin (\sin (\ldots \sin 1 ))), n\in \mathbb{N}$.
We can write the sequence in the recursive form $a_n=\sin (a_{n-1})$ with $a_0=1$, right?
I want to show that the sequence is monotone and bounded.
It holds that $|\sin (x)|\leq |x|$. So, we have that $|a_n|=|\sin (a_{n-1})|\leq |a_{n-1}|=|\sin (a_{n-2})|\leq |a_{n-2}|\leq ...\leq |a_1|=|\sin (a_0)|\leq |a_0|=1\Rightarrow |a_n|\leq 1$, so the sequence is bounded. Is this correct?
How can we show that the sequence is monotone?