Finding the value guaranteed by the Mean Value Theorem

First compute the average rate of change of $f$ over $[0,\pi]$: since $f(0) = \sin(0) = 0$ and $f(\pi) = \sin(\pi) = 0$, the slope of the secant line is $\frac{0-0}{\pi - 0} = 0$. So the MVT guarantees some $c$ in $(0,\pi)$ with $f'(c) = 0$.

Since $f'(x) = \cos(x)$, we need $\cos(c) = 0$ for $c \in (0,\pi)$. The only solution in this open interval is $c = \frac{\pi}{2}$, which corresponds to the point where the tangent line to $\sin(x)$ is horizontal, parallel to the (horizontal) secant line.

The tangent line at the guaranteed point is parallel to the secant line. Since the secant line connecting the endpoints was horizontal with slope zero, the tangent line at c equals pi over two must also be horizontal. This matches the geometric meaning of the Mean Value Theorem: the tangent at c has the same slope as the secant through the endpoints. Therefore the answer is that the two lines are parallel.