$\lim_{n\to \infty}\sin^n(x) = 0 \:\:\:\: \forall x\:\in\:[0,\pi/2) \\$
Since the pointwise limit is continuous $\forall x\:\in\:[0,\pi/2) $ therefore by definition of uniform convergence we have : $$\lim_{n\to \infty}\{\sup_{0\leq x<\pi/2} |f_n(x)-f(x)|\} = \lim_{n\to \infty}\{\sup_{0\leq x<\pi/2}|f_n(x)-0|\}\:\: \:= \lim_{n\to \infty}\{\sup_{0\leq x < \pi/2} |\sin^n(x)|\}\:\: \:=\:\lim_{n\to \infty}1\:\:=\:\:1\\\neq0$$
Thus the function should not be uniformly convergent.But it is uniformly convergent on the given interval. That means the supremum must be $0$ instead of $1.$ But i don't see why that should be the case. Can somebody please help and clarify the doubt.