Using Lagrange's theorem, prove that sin (x+h)- sin x = h

Using Lagrange's theorem, prove that $\sin\bigl(x+h\bigr)-\sin x = h\cos\xi$, where $x\lt \xi\lt x+h$.

Step 1: Choose the function and interval

Take $f(t) = \sin t$ on $\bigl[x,\,x+h\bigr]$, assuming $h>0$; for $h\lt 0$ the same argument applies with the interval $\bigl[x+h,\,x\bigr]$.

Step 2: The hypotheses hold

$\sin t$ is continuous and differentiable on the whole line, with $f'(t) = \cos t$.

Step 3: Apply the theorem

There exists $\xi$ strictly between $x$ and $x+h$ with

\[ f\bigl(x+h\bigr)-f(x) = f'(\xi)\Bigl[\bigl(x+h\bigr)-x\Bigr], \]

\[ \sin\bigl(x+h\bigr)-\sin x = h\cos\xi. \]

Step 4: Verification

\[x = 0,\ h = \frac{\pi}{2}:\ \sin\frac{\pi}{2}-\sin0 = 1 = \frac{\pi}{2}\cos\xi \;\Longrightarrow\; \cos\xi = \frac{2}{\pi} \approx 0.6366\]

\[\xi = \arccos\frac{2}{\pi} \approx 0.8807 \in \left(0,\frac{\pi}{2}\right) \;\;\checkmark\]

\[\text{Exact identity for comparison: } \sin(x+h)-\sin x = 2\cos\left(x+\frac h2\right)\sin\frac h2 \;\;\checkmark\]