Yes, the Cauchy Mean Value Theorem (CMVT) can be applied to $f(x) = \cos x$ and $g(x) = \sin x$ on the interval $[0, \pi]$, provided the conditions for the theorem are satisfied.

The Cauchy Mean Value Theorem states that if $f(x)$ and $g(x)$ are two functions that are:

1. Continuous on the closed interval $[a, b]$,
2. Differentiable on the open interval $(a, b)$,

then there exists some $c \in (a, b)$ such that: $\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}.$

Verifying the conditions:

1. Continuity: Both $\cos x$ and $\sin x$ are continuous on $[0, \pi]$.

2. Differentiability: Both $\cos x$ and $\sin x$ are differentiable on $(0, \pi)$.

Since both conditions are met, we can apply the Cauchy Mean Value Theorem.

Applying the theorem:

• $f(x) = \cos x$ and $g(x) = \sin x$.
• The derivatives are:
  • $f'(x) = -\sin x$,
  • $g'(x) = \cos x$.

We seek a $c \in (0, \pi)$ such that: $\frac{-\sin c}{\cos c} = \frac{\cos(\pi) - \cos(0)}{\sin(\pi) - \sin(0)}.$

Now, calculate the right-hand side: $\frac{\cos(\pi) - \cos(0)}{\sin(\pi) - \sin(0)} = \frac{-1 - 1}{0 - 0} = \frac{-2}{0}.$ However, the expression $\frac{-2}{0}$ is undefined.

Conclusion:

The Cauchy Mean Value Theorem cannot be applied in this case because the denominator on the right-hand side (i.e., $\sin(\pi) - \sin(0) = 0 - 0 = 0$) leads to division by zero. Therefore, the conditions for the theorem are not fully satisfied, and the CMVT does not hold for $f(x) = \cos x$ and $g(x) = \sin x$ on $[0, \pi]$.

Would you like further details or explanations on this?

Here are 5 related questions to explore:

1. Can we apply the Mean Value Theorem to $f(x) = \cos x$ on $[0, \pi]$?
2. What happens if we use different boundary conditions in the CMVT?
3. How does Rolle’s Theorem relate to the Cauchy Mean Value Theorem?
4. Can the CMVT be applied if $g(x)$ is not differentiable?
5. How do we interpret division by zero in the context of the CMVT?

Tip: Always check the behavior of the denominator in CMVT, as division by zero invalidates the result.