TITLE: Uniform Convergence of $(\sin x)^n$
QUESTION [6 upvotes]: Is the sequence of functions $(\sin(x))^n$ uniformly convergent in $[0,\pi]$? Can you give me a hint or solution, I have already already prove that it is UC in $[0,1]$ but I don't know how to proceed in $[1,\pi]$.

REPLY [4 votes]: It is not uniformly convergent: At $\pi/2$ we have $\sin(\pi/2)=1$ and thus 
$$
\sin(\pi/2)^n\to 1
$$
What happens everywhere else where $\sin(x)<1$?