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For any $u<v$ in the interval, the mean value theorem, applicable by differentiability implies continuity, givesAll pairs of values are equal. Thus $\boxed{f\text{ is a constant function}}$, the constancy theorem.

The derivative of $h(x)=f(x)-\sin x$ vanishes. By the constancy theorem on the interval $I$,This need not hold on a disconnected domain. On $I=(-\infty ,0)\cup (0,\infty )$, define $f(x)=\sin x$ for $x<0$ and $f(x)=\sin x+1$ for $x>0$. The function is differentiable throughout $I$ with $f^{\prime }(x)=\cos x$, but no single $C$ works on both components.

The constancy theorem states that a differentiable function on an interval whose derivative vanishes everywhere is constant. At any included endpoints, continuity there is understood.

Indeed, take arbitrary $x<y$ in the interval. The mean value theorem applies on $[x,y]$ and gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. Every pair of values is equal, proving the assertion. The interval hypothesis ensures that the whole segment $[x,y]$ lies in the domain.