Step 1: Understanding the Concept:
Rolle's Theorem is a special case of the Mean Value Theorem. It states that if a function \(f\) is continuous on a closed interval \([a,b]\), differentiable on the open interval \((a,b)\), and \(f(a)=f(b)\), then there exists at least one number \(c\) in \((a,b)\) such that \(f'(c)=0\).
Step 2: Key Formula or Approach:
1. Verify that the function \(f(x)=\sin x\) satisfies the three conditions of Rolle's Theorem on the interval \([0, \pi]\).
2. Find the derivative of the function, \(f'(x)\).
3. Set the derivative equal to zero, \(f'(c)=0\), and solve for \(c\).
4. Check that the value of \(c\) lies within the open interval \((0, \pi)\).
Step 3: Detailed Explanation:
The function is \(f(x)=\sin x\) and the interval is \([0, \pi]\).
1. Verify the conditions:
- Continuity: The function \(f(x)=\sin x\) is continuous for all real numbers, so it is continuous on \([0, \pi]\).
- Differentiability: The function \(f(x)=\sin x\) is differentiable for all real numbers, so it is differentiable on \((0, \pi)\).
- Equal endpoints: We check the value of the function at the endpoints \(a=0\) and \(b=\pi\).
- \(f(0) = \sin(0) = 0\)
- \(f(\pi) = \sin(\pi) = 0\)
Since \(f(0)=f(\pi)=0\), this condition is satisfied.
All three conditions of Rolle's Theorem are met. Therefore, there must exist a \(c \in (0, \pi)\) such that \(f'(c)=0\).
2. Find the derivative:
\[ f'(x) = \frac{d}{dx}(\sin x) = \cos x \]
3. Solve for c:
We set \(f'(c)=0\):
\[ \cos(c) = 0 \]
4. Check the interval:
We need to find the value(s) of \(c\) in the open interval \((0, \pi)\) where \(\cos(c)=0\).
The cosine function is zero at odd multiples of \(\frac{\pi}{2}\). The only such value within the interval \((0, \pi)\) is:
\[ c = \frac{\pi}{2} \]
Since \(0 < \frac{\pi}{2} < \pi\), this is the value guaranteed by Rolle's Theorem.
Step 4: Final Answer:
The value of c is \(\frac{\pi}{2}\), which corresponds to option (C).