Check the conditions of Cauchy's theorem for f(x) = sin x

Check the conditions of Cauchy's theorem for $f(x) = \sin x$ and $\varphi(x) = \cos x$ on $\left[0,\dfrac{\pi}{2}\right]$, apply it and find $\xi$.

Step 1: The theorem

If $f$ and $\varphi$ are continuous on $[a,b]$, differentiable on $(a,b)$, and $\varphi'\neq0$ there, then for some $\xi\in(a,b)$

\[ \frac{f(b)-f(a)}{\varphi(b)-\varphi(a)} = \frac{f'(\xi)}{\varphi'(\xi)}. \]

Step 2: Check the hypotheses

Both functions are smooth everywhere, so continuity and differentiability are automatic. The extra condition is on $\varphi'$: $\varphi'(x) = -\sin x \lt 0 \ \text{ for } x\in\left(0,\frac{\pi}{2}\right)$.

so $\varphi'$ never vanishes inside the interval. Also $\varphi\left(\tfrac{\pi}{2}\right)-\varphi(0) = 0-1 = -1\neq0$, so the left-hand side is defined.

Step 3: The left-hand side

\[ \frac{\sin\frac{\pi}{2}-\sin0}{\cos\frac{\pi}{2}-\cos0} = \frac{1-0}{0-1} = -1. \]

Step 4: The right-hand side, and $\xi$

\[ \frac{f'(\xi)}{\varphi'(\xi)} = \frac{\cos\xi}{-\sin\xi} = -\cot\xi = -1 \;\Longrightarrow\; \cot\xi = 1 \;\Longrightarrow\; \xi = \frac{\pi}{4}. \]

Step 5: Verification

\[\frac{f'(\pi/4)}{\varphi'(\pi/4)} = \frac{\sqrt2/2}{-\sqrt2/2} = -1 \;\;\checkmark\]

\[\text{Lagrange applied separately gives different points: } \cos\xi_{1} = \frac{2}{\pi},\ \xi_{1} \approx 0.8807; \ \sin\xi_{2} = \frac{2}{\pi},\ \xi_{2} \approx 0.6901\]

\[\text{Neither equals } \frac{\pi}{4} \approx 0.7854, \text{ which is why Cauchy's theorem is a genuinely stronger statement} \;\;\checkmark\]