Differentiability at a piecewise junction point

First note $f(0) = \sin(0) = 0$. For $h < 0$, $f(h) = \sin(h)$, so the left-hand difference quotient is $\frac{\sin(h) - 0}{h} = \frac{\sin h}{h}$. Taking the limit as $h \to 0^-$ gives $\lim_{h\to 0^-} \frac{\sin h}{h} = 1$, using the well-known limit $\lim_{h\to 0}\frac{\sin h}{h}=1$.

For $h > 0$, $f(h) = h$, so the right-hand difference quotient is $\frac{h - 0}{h} = 1$, giving $\lim_{h\to 0^+} 1 = 1$. Since the left-hand and right-hand derivative limits both equal $1$, the two-sided limit $\lim_{h\to 0} \frac{f(h)-f(0)}{h}$ exists and equals $1$. Therefore $f$ is differentiable at $x=0$, with $f'(0) = 1$ — the pieces join smoothly with matching slopes, so there is no corner there.