Step 1: Set the range.
We compare $\sin x$ and $\sin^2 x$ on $0 \le x \le \pi$. Both start at $0$, rise to a peak at $x=\pi/2$, then fall back to $0$.
Step 2: Find where they meet.
They touch wherever $\sin x$ is $0$ or $1$. That happens at $x=0$, $x=\pi$ (both zero) and at $x=\pi/2$ (both equal $1$).
Step 3: Compare them in between.
For $0 < x < \pi$ the value $\sin x$ lies between $0$ and $1$. Squaring a number in that band makes it smaller, so $\sin^2 x < \sin x$ everywhere except at the three meeting points.
Step 4: Picture the curves.
So the $\sin x$ curve sits on top and the $\sin^2 x$ curve dips below it, the two hugging at the start, the middle peak and the end.
Step 5: Answer.
The graph showing this is option C. \[ \boxed{\text{Graph C}} \]
