Is the sequence $(\sin^{n}t)_{n \in {\Bbb{N}}}$ totally bounded in the space $C[0,1]$ with the supremum metric?
By the Arzelà–Ascoli theorem, this sequence is totally bounded if and only if it is uniformly bounded and equicontinuous. It is uniformly bounded by $1$, so we can search for equicontinuity instead.
I believe that the sequence is equicontinuous, but I could not find a way to prove it.
Arzelà-Ascoli and equicontinuity are useless. Let $A=\{\sin^n\mid n\in\Bbb N\}.$ $(\sin^n)$ converges uniformly to $0$ on $[0,1]$ (because $0\le\sin t\le\sin 1<1$), so $\bar A=A\cup\{0\}$ is compact, which proves that $A$ is totally bounded.