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 $[1,\pi]$.
Hint: find the pointwise limit of this sequence of functions on $[0,\pi]$. If a sequence is uniformly convergent, then its limit is a continuous function. Is it continuous in this case?
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 $\sin(x)<1$?